A Less Obvious Martingale Fallacy

[MMMMMM]

Here are some thoughts which have occurred to me over time regarding this betting system.

Many knowledgeable gamblers are familiar with the Martingale system of doubling up after each loss until a win, then restarting the series. On a game like roulette, betting red or black, this system results in a win of 1 unit each time a series is completed (example: bet 1 unit and lose, bet 2 units and lose, bet 4 units and win: which gives a net profit of 1 unit for the entire series).

Many understand that this system does not work in a casino where table limits exist, because eventually you will encounter a string of losses in which your next required bet in the series will exceeed the table limit.

Some good posters in other 2+2 forums have in times past asserted that it is only the table limits and bankroll limitations which prevent the Martingale from working. They believe that the system would in fact work if there were no table limits and if the gambler possessed an unlimited bankroll.

What they do not understand, is that even WITHOUT table limits, and WITH infinite bankrolls, this system still would not work in the long run.

It is easy to think that the system would work without such restrictions, because you get to keep playing until you win, at which point you have another unit of profit.

What is less obvious is that it is STILL all one long game.

If you intend to keep repeating the system, you cannot logically "take accounting" at only those times when you fancy to take accounting.

Counting your bankroll only at the times immediately after completion of a winning series is fallacious. If you were truly to play this game eternally, your chips would slowly go down, on average, due to the house edge. Yes, you would have some huge peaks and valleys, and you would eventually recoup bad losses after longer and longer series, but it would be the *average* pluses or minuses to your bankroll that would matter, not your arbitrarily chosen accounting points.

Let's say you could play this game until the universe itself winds down and comes to a complete grinding halt. Well, you would be expected to be down A LOT of money at that point.

You might have completed God-knows-how-many series by then, but this still would be true. This, of course, is because the house edge has been inexorably eating away at your bankroll despite your contrived betting pattern and your artificially chosen accounting points.

Another way of looking at it is to imagine a great matrix of all possible paths of outcomes. If the Red-or-Black wheel game were truly 50-50 odds (with no green zeros), then even with a doubling up Martingale system, the extremes of the matrix would balance the small accumulated wins, and you would break even in the long run. However, with the green zeros putting the advantage in the house favor, the matrix of possible paths is not weighted fairly. Hence some of your "get even" doubling up streaks will take longer than they would otherwise take with a 50-50 wheel, and some of your first bets of a new sequence will start with a loss whereas with a 50-50 wheel they would instead have started with a win.

Another way of looking at it is to imagine an immensely huge casino, so large you could never count all the gamblers in it, with each gambler seated at his or her own private roulette table and betting this system against the house.

At any given time after the first spin, the average bankroll of all the gamblers would be expected to be showing a loss--even though they are all playing the Martingale system.

Presuming a great computer system could instantaneously know the totals at each table and immediately sum the results, it would be ludicrous to take the accounting in any way other than simultaneously for all tables.

If you were the Pit Boss at the Cosmic Casino, and the Cosmic Casino Manager strolled by and asked you how the casino was doing today, it would be ludicrous for you to say, "Well, we have to wait until each player finishes their own individual series before I can tell you". If you tried it, the Cosmic Casino Manager would probably throw a fit and tell you that is ridiculous. Why, he might ask you, should we tally only when a player is winning? Would it make any sense to tally only when the players are losing? And, by the way, how did you get this job in the first place? Then, in a more kindly tone, he might say, "Look, it is more reasonable and far simpler to simply tally all the tables at once. Look, just press the "T" button here, for "Tally";-)).

Now suppose that Fast Eddy Felson were watching all of this from a catbird seat, as a special guest of the casino. Fast Eddie happens to believe in the Martingale as a winning system, as long as no limitations are present. He sees all the gamblers and gets a special offer from the Cosmic Casino Manager: Fast Eddie, you too can play this system against us. The only thing is, all tables are currently taken. But as our special esteemed guest, I will let you pick any table and I will have the patron occupying it removed. The only catch is you must start the series where he left off and take on his wins or losses up to this point.

Now, from his high vantage point, Fast Eddie can view acres and acres of tables, stretching out as far as the eye can see. The only thing is, he is so high up he can't see the chips and thus has no way of knowing which players are doing well or not. He also has no idea how long this game has been going on. Should Fast Eddie pick a table? Would you?

I hope all of this helps to illustrate why the Martingale System would not work, *even if* there were no real-world limitations on bet-sizing or on bankrolls.

All comments welcome.

[Dov]

I don't think this is true.

I think the infinite bankroll overcomes the house edge in the end.

This is because when you do complete a series you will have an EV of your old BR+1 betting unit.

You are still EV+, you just can't know the EV of a particular wager until the series is completed. You would have to divide your 1 betting unit of profit over all of the bets in the series.

You are correct in that during a losing series you will have experienced -EV bets. However, the metagame conditions allow this to actually be a +EV situation.

Taken from the perspective that there may be a more efficient use of your funds, I can understand that the Martingale would be -EV.

Under these conditions, though, I would expect that your BR would grow with the average # of trials in a series. If you are saying that the longer you play, the wider your variance will get, (which is true), it will still even out in the end when you do actually win, assuming that your winning chances are above 0.

As a matter of fact, I can't think of a situation where you have any chance to win that wouldn't guarantee that you do win except when you are a guaranteed loser. (like drawing dead)

I think you have somehow overlapped 2 concepts that don't, but I'm not completely sure where your error is or if I am the one who is mistaken. (I don't really think that I am, though, because by definition, you WILL win, and when you do, your BR will be larger than when it started.)

[[censored]]

Had lunch with Fast Eddy Felton, he is a cool cat.

[SheetWise]

Assuming the game is not biased, and the wager is close to 50-50, Baccarat or the pass line for example, there does exist a number of series where the probability of failure approaches a theoretical zero. Assume the game was fair, and I could play a progression that lasted for 1000 trials -- I would be quite comfortable playing for a lifetime. Because my confidence limits would be affected by my life expectancy, I would probably be happy, rich, and comfortable with a great many fewer. The casino would certainly have the advantage in the game you describe -- but they would probably want to take their lifetime (or quarterly earnings) into account as well.

SheetWise

[maurile]

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I think the infinite bankroll overcomes the house edge in the end.

This is because when you do complete a series you will have an EV of your old BR+1 betting unit.

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Your old bankroll was infinite. What's infinity plus one?

[Dov]

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Your old bankroll was infinite.

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So why were you playing in the first place?

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What's infinity plus one?

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According to your logic you could never lose anything either. What's infinity minus 1?

[maurile]

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Your old bankroll was infinite.

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So why were you playing in the first place?

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What's infinity plus one?

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According to your logic you could never lose anything either. What's infinity minus 1?

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Infinity plus or minus any finite sum is still just infinity. So an infinite bankroll cannot be increased (or decreased) with the martingale system if you are betting finite sums.

If you have a finite bankroll, using the martingale system is demonstrably -EV in any game like roulette where each trial is -EV.

So either way -- finite bankroll or infinite bankroll -- you cannot turn a -EV game into a +EV game by martingaling it.

[MMMMMM]

[ QUOTE ]
(I don't really think that I am, though, because by definition, you WILL win, and when you do, your BR will be larger than when it started.)

[/ QUOTE ]

Well, that argument meets to its complement, which is: by definition, you WILL lose, and lose many many many times in a row on occasion. And the longer you play, you will set new records for longer and longer losing streaks.

The accumulated small wins you are thinking of are balanced by an extreme theoretical tail or far-out reach of the matrix of possible paths, on the negative side.

The losses compound geometrically more and more, the rarer (longer) they are.

Also, although this is tangential and more complicated, if you had an infinite number of gamblers playing this system, might you have at least one gambler who would never get to make a single winning bet? So he might never complete the first series?

[PairTheBoard]

Correct.

I posted this same idea in the Probabilty Forum for the case where bet sizes remain constant but players with finite bankroll bust out due to variance. The difference here is that average bet sizes increase over time and the Cosmic Casino makes EVEN MORE MONEY on average over time because of it. It makes no difference whether bet sizes increase at random or due to long losing streaks.

PairTheBoard

[MMMMMM]

[ QUOTE ]
[ QUOTE ]
I think the infinite bankroll overcomes the house edge in the end.

This is because when you do complete a series you will have an EV of your old BR+1 betting unit.

[/ QUOTE ]


Your old bankroll was infinite. What's infinity plus one?

[/ QUOTE ]

Just for clarity's sake, I wish I had written "unlimited bankroll" instead of "infinite bankroll".

Not that your question isn't interesting on its own merits.

[Dov]

[ QUOTE ]
Infinity plus or minus any finite sum is still just infinity. So an infinite bankroll cannot be increased (or decreased) with the martingale system if you are betting finite sums.

If you have a finite bankroll, using the martingale system is demonstrably -EV in any game like roulette where each trial is -EV.

So either way -- finite bankroll or infinite bankroll -- you cannot turn a -EV game into a +EV game by martingaling it.

[/ QUOTE ]

Why do you say that you can use a finite bankroll over an infinite time period?

Surely, if the time period is infinitely long, then you can't increase it by another day.

Similarly you can't have an infinite number of trials either, because you would never be able to increment the number of completed trials by 1.

Why do you accept an increasing number of trials, but not an increasing bankroll?

[MMMMMM]

[ QUOTE ]
Assuming the game is not biased, and the wager is close to 50-50, Baccarat or the pass line for example, there does exist a number of series where the probability of failure approaches a theoretical zero.

[/ QUOTE ]

Doesn't your so-called "theoretical zero" correspond to a so-called "theoretically infinite dollar amount (loss)" on the other side of the coin?

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Assume the game was fair, and I could play a progression that lasted for 1000 trials -- I would be quite comfortable playing for a lifetime.

[/ QUOTE ]

Well maybe you should be. However, for all the people who might do it along with you and be successful and happy, wouldn't there be a few who would bear a commensurate loss to your summed profits by hitting that nasty old extreme negative end of the tail? (actually it would be more money because of the house edge on roulette or baccarat or whatever).

[maurile]

I can't make sense of your post.

[Dov]

[ QUOTE ]
What is less obvious is that it is STILL all one long game.

[/ QUOTE ]

I think this is where the problem is.

It IS still one long game, but we adjust the parameters artificially according to short term results.

The Martingale with an infinite bankroll (the way I think you intended it) actually breaks the long run into many short run games which all have a +EV of 1 unit spread across as many wagers as it takes to win 1 game.

As soon as a win is achieved, the system 'resets' itself as if it were actually starting over again for the first time.

What is the difference between bets in the Martingale system when it is the 1st bet of a new series or the next bet following a win?

For all intents and purposes, it is NOT one long game.

[maurile]

[ QUOTE ]
[ QUOTE ]
Assuming the game is not biased, and the wager is close to 50-50, Baccarat or the pass line for example, there does exist a number of series where the probability of failure approaches a theoretical zero.

[/ QUOTE ]

Doesn't your so-called "theoretical zero" correspond to a so-called "theoretically infinite dollar amount (loss)" on the other side of the coin?

[/ QUOTE ]
Yes. Take roulette as an example. We'll bet on red, which pays 1-1. On any particular spin of the wheel, our chance of winning is 18/38 = 0.473684. Our chance of losing is thus 1 - 0.473684 = 0.526316.

I'll do the math for a 15-unit bankroll, and then I'll just post the results for successively higher bankrolls . . . you'll see for yourself that extrapolating up to an infinite bankroll does not produce a positive result.

If we have a 15-unit bankroll, we can fund four losing spins before we go broke in the martingale system. (1 + 2 + 3 + 4 = 15.) We are thus risking 15 units to win 1 unit. Our chance of winning = 1 - (0.526316^4) = 0.9232. We are thus 12-1 favorites to come out ahead. But laying 15-1 odds when we are only 12-1 favorites is a losing proposition. Thus we have a negative expectation with a 15-unit bankroll.

Here are the results for successively higher bankrolls:

3-unit bankroll (can fund 2 spins): we are laying 3-1 odds as a 2.61-1 favorite.
7-unit bankroll (can fund 3 spins): we are laying 7-1 odds as a 5.86-1 favorite.
15-unit bankroll (4 spins): we are laying 15-1 odds as a 12.03-1 favorite.
31-unit bankroll (5 spins): we are laying 31-1 odds as a 23.76-1 favorite.
63-unit bankroll (6 spins): we are laying 63-1 odds as a 46.05-1 favorite.
127-unit bankroll (7 spins): we are laying 127-1 odds as a 88.39-1 favorite.
255-unit bankroll (8 spins): we are laying 255-1 odds as a 168.8-1 favorite.
511-unit bankroll (9 spins): we are laying 511-1 odds as a 321.7-1 favorite.
1023-unit bankroll (10 spins): we are laying 1023-1 odds as a 612.1-1 favorite.
2047-unit bankroll (11 spins): we are laying 2047-1 odds as a 1164-1 favorite.
4095-unit bankroll (12 spins): we are laying 4095-1 odds as a 2212-1 favorite.
8191-unit bankroll (13 spins): we are laying 8191-1odds as a 4204-1 favorite.
16383-unit bankroll (14 spins): we are laying 16383-1 odds as a 7989-1 favorite.
32767-unit bankroll (15 spins): we are laying 32767-1 odds as a 15180-1 favorite.

And so on, up to a bankroll that approaches infinity, and can therefore fund a number of spins that approaches infinity. The trend of laying better odds than our chance of winning warrants never reverses itself: we are never taking the best of it.

As our bankroll approaches infinity, we approach becoming an infinity-to-one favorite to come out ahead. But the odds we are laying approach infinity-to-one even faster. The fact that it is theoretically possible for the roulette wheel to land on black N times in a row for any N means that, even in theory, the martingale never shows an expected profit.

[Dov]

[ QUOTE ]
I can't make sense of your post.

[/ QUOTE ]

It seems to me like you accept the notion that we can have an infinite number of trials, each with its own result, as well as a cumulative result of all trials thus far conducted,

I was just wondering why you dismiss the applicability of this technique to a bankroll that can be considered to be large enough as to be equivallent to infinite.

If you like, then just always assume that more money will be available for the bankroll when you need to place a bet larger than the balance.

[maurile]

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It seems to me like you accept the notion that we can have an infinite number of trials, each with its own result, as well as a cumulative result of all trials thus far conducted.

[/ QUOTE ]
Yes.

[ QUOTE ]
I was just wondering why you dismiss the applicability of this technique to a bankroll that can be considered to be large enough as to be equivallent to infinite.

[/ QUOTE ]
I never said you can't have an infinite bankroll. I just said -- correctly -- that when you add a finite sum to it, you are not increasing it. (Same with an infinite number of trials. If you add a trial to an infinite series of trials, you are not increasing the length of the series.)

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If you like, then just always assume that more money will be available for the bankroll when you need to place a bet larger than the balance.

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Yes, that's what an infinite bankroll is.

[Dov]

[ QUOTE ]
Well, that argument meets to its complement, which is: by definition, you WILL lose, and lose many many many times in a row on occasion. And the longer you play, you will set new records for longer and longer losing streaks.

[/ QUOTE ]

Are you saying that eventually you will encounter a losing streak so long that it will be infinite?

If that's true, then you will also encounter a similar winning streak at some point.

I think that because we are assuming a game where you aren't drawing dead (so to speak) we dismissed this possibility by using an unlimited number of trials.

[ QUOTE ]
The accumulated small wins you are thinking of are balanced by an extreme theoretical tail or far-out reach of the matrix of possible paths, on the negative side.

[/ QUOTE ]

The negative side is completely recovered with a single win. This is only possible because of the unlimited bankroll, but given that unlimited bankroll is a condition of this problem, I still don't see how our gambler can fail to bust the bank 1 unit at a time.

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The losses compound geometrically more and more, the rarer (longer) they are.

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This is why you need an unlimited bankroll. It overcomes this objection.

[maurile]

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If that's true, then you will also encounter a similar winning streak at some point.

[/ QUOTE ]
Possibly tomorrow. :fingerscrossed:

[MMMMMM]

Hi Dov,

I can see how what you are saying appears to make sense, but I think that is an illusion.

Time for sleep now, so later;-)

[Dov]

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It makes no difference whether bet sizes increase at random or due to long losing streaks.

[/ QUOTE ]

If Cosmic Casino has to repay the entire amount lost to them plus 1 unit when the gambler finally wins, then how are they making money here?

[SheetWise]

[ QUOTE ]
Doesn't your so-called "theoretical zero" correspond to a so-called "theoretically infinite dollar amount (loss)" on the other side of the coin?


[/ QUOTE ]

No, because I'm going to use my lifespan as a definition of infinity /images/graemlins/wink.gif . For arguments sake, if infinity is a lifetime -- and you play 24 hours a day until death -- you can find a number of progressions with a finite bankroll that will give you any confidence level you want, and I do believe in a theoretical zero. While a long sequence of losses is required to lose, it isn't to win. I'm saying neither one will happen.

[ QUOTE ]
wouldn't there be a few who would bear a commensurate loss to your summed profits by hitting that nasty old extreme negative end of the tail?

[/ QUOTE ]

I'm saying no. That's why nobody will play this game. I'm not suggesting the house edge ever goes away, it's negative expectation -- but if the player can control the wagering to this degree, that expectation will never be realized.

[maurile]

[ QUOTE ]
This is only possible because of the unlimited bankroll, but given that unlimited bankroll is a condition of this problem, I still don't see how our gambler can fail to bust the bank 1 unit at a time.

[/ QUOTE ]
You have to be careful to distinguish between (a) an extremely large but still finite bankroll, and (b) a literally infinite bankroll.

For any extremely large but finite bankroll, we can do the math directly and see that we are still -EV (in roulette or any other game where each trial is -EV).

If you jump to a literally infinite bankroll, then the casino also has to have a literally infinite bankroll (to cover our biggest possible wagers). So the answer to why you won't eventually bust the casino one unit at a time is that you can't bust an infinite bankroll one unit at a time. You can keep subtracting one from the casino's 'roll all you want, but you will never bust it.

[Dov]

[ QUOTE ]
No, because I'm going to use my lifespan as a definition of infinity

[/ QUOTE ]

This doesn't really work, though.

As an example, 5 years ago it would have been extremely rare for any professional poker player to play more than 125,000 hands in a year.

It is not uncommon for us to be able to play 4, 5, or even 6 times that number of hands now. Does that mean that the lifetime of a pro poker player has been lengthened?

Why put an artificial limit on this?

[Dov]

[ QUOTE ]
If you jump to a literally infinite bankroll, then the casino also has to have a literally infinite bankroll (to cover our biggest possible wagers). So the answer to why you won't eventually bust the casino one unit at a time is that you can't bust an infinite bankroll one unit at a time. You can keep subtracting one from the casino's 'roll all you want, but you will never bust it.

[/ QUOTE ]

Fair enough, but if the game is actually +EV for the casino, then there should be a finite bankroll capable of defeating the unlimited Martingale bankroll, no?

If not, then you need to conclude that the Martingale played with an unlimited bankroll actually does overcome the house edge, regardless of its size.

I know this is true because you will eventually run into a series that overwhelms the finite side.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
It makes no difference whether bet sizes increase at random or due to long losing streaks.

[/ QUOTE ]

If Cosmic Casino has to repay the entire amount lost to them plus 1 unit when the gambler finally wins, then how are they making money here?

[/ QUOTE ]

What you will see as you look over all the tables at the Cosmic Casino is a lot of players who are ahead a little bit, and a few players who are stuck huge amounts. The identities of the players who are ahead and who are stuck keep changing. But as time goes on it becomes more and more likely that the Casino will be ahead overall. What matters is who's rack the money is in at any point in time. Overall, the liklihood is that the Casino will have more money won and its racks than the players. As far as an individual player goes, if you're going to count yourself as being ahead after a win, you also have to count yourself as being behind when on a horrendous losing streak. You may spend more Time being ahead, but if you weight your time spent ahead or behind by the Amount you are ahead or behind, you will spend more Weighted Time being behind.

PairTheBoard

[maurile]

[ QUOTE ]
Fair enough, but if the game is actually +EV for the casino, then there should be a finite bankroll capable of defeating the unlimited Martingale bankroll, no?

[/ QUOTE ]
Yes, any finite bankroll will defeat the Martingale. A finite bankroll on the part of the casino means a finite betting limit -- so the player will be unable to keep doubling his bets past a certain point.

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If not, then you need to conclude that the Martingale played with an unlimited bankroll actually does overcome the house edge, regardless of its size.

[/ QUOTE ]
With literally infinite bankrolls, neither party has an edge. If either side has a finite bankroll, then the house will have an edge (assuming a game like roulette where the house has an edge on each individual trial).

[SheetWise]

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Why put an artificial limit on this?

[/ QUOTE ]

Because there are limits. Using infinity is an artificial abandonment of constraint -- and why you can't define the expected outcome of the game.

My calculation would require a definition of zero (impossibility), and I have seen people define it differently for different purposes. But once that limit is defined, we can apply it to a lifetime of play and achieve that confidence level.

I don't know what to do with infinity /images/graemlins/confused.gif

[Dov]

[ QUOTE ]
Yes, any finite bankroll will defeat the Martingale. A finite bankroll on the part of the casino means a finite betting limit -- so the player will be unable to keep doubling his bets past a certain point.

[/ QUOTE ]

This is not a finite bankroll for the casino. It is a finite bankroll for the player. The assumption was that the player is allowed to wager any amount he likes and chooses to wager 1 unit after a win and double the previous wager after a loss.

A finite bankroll for the casino would be a limited amount of capital that they could pay winning wagers with. (ie: an individual player wins)

[maurile]

[ QUOTE ]
This is not a finite bankroll for the casino. It is a finite bankroll for the player.

[/ QUOTE ]
Same thing. The betting limit depends on the lesser of the two bankrolls. So if the bank's bankroll is finite, then effectively so is the player's. He can't double up his bets anymore once he gets to the point where the casino can't cover his bet.

[ QUOTE ]
The assumption was that the player is allowed to wager any amount he likes and chooses to wager 1 unit after a win and double the previous wager after a loss.

[/ QUOTE ]
This requires an infinite bankroll for the casino as well. How can the player wager an infinite sum if the casino only has $100?

[Dov]

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What matters is who's rack the money is in at any point in time.

[/ QUOTE ]

I still don't see why if each individual player comes out 1 unit ahead at the end of a series, and you extend them out to infinity, you think the casino will win in the end.

If this were true, then why put a table limit at all? Why not let people keep doubling the bets?

Are you saying that it would be too risky for a real casino to allow someone to do this?

Wouldn't that contradict what was said earlier about them making more money when more money is wagered?

There must be a reason for table limits. Obviously EV isn't everything.

[maurile]

[ QUOTE ]
Are you saying that it would be too risky for a real casino to allow someone to do this?

[/ QUOTE ]
Yes. Binion's used to be famous for being willing to accept huge bets. It let a guy bet $1 million on craps a couple times, I think. (Back when $1 million meant something.)

But there comes a point where it's definitely too risky. If I bet a few billion or whatever on red at Harrah's, I will own the casino if I win. The board of directors at Harrah's doesn't want to take that risk.

[ QUOTE ]
Wouldn't that contradict what was said earlier about them making more money when more money is wagered?

[/ QUOTE ]
No.

[ QUOTE ]
There must be a reason for table limits. Obviously EV isn't everything.

[/ QUOTE ]
EV isn't everything. There's variance, too.

[PairTheBoard]

Dov --
"I still don't see why if each individual player comes out 1 unit ahead at the end of a series, and you extend them out to infinity, you think the casino will win in the end."

I'm not looking at what happens "in the end". We never get to the end. All we can look at is what happens over time as we look at longer and longer measures of time played. Yes, there are times you will be ahead. In fact, most of the time you will be ahead. But there is no point in time where you will be ahead forever. There will always be times in the future when you are behind again. During those periods when you are behind it makes no sense to say, "It doesn't matter that I'm behind now because I know I'm going to be ahead again in the future." If it did make sense then the opposite could be said when you are ahead, "It doesn't matter that I'm ahead now because I know I will be behind again in the future." What does matter is how much time you spend ahead or behind AND BY HOW MUCH. If you weight the amount of time spent ahead and behind by the AMOUNT you are ahead or behind and look at future times T then you can say the following. As T gets larger and larger, the Weighted Time you are behind will exceed the Weighted Time you are ahead by an average amount determined by the House Edge.

PairTheBoard

[felson]

There's no way I am betting my money according to Martingale.

F.E.F.

[Lawrence Ng]

[ QUOTE ]
There's no way I am betting my money according to Martingale.

F.E.F.

[/ QUOTE ]

Do you ever say anything that is even remotely useful or insightful on this board? Seriously, you must be the number 1 rank troll on 2+2.

Lawrence

[jason1990]

Suppose I repeatedly flip a coin for $1 each time; no doubling strategy. If it's a fair coin, then the swings in my bankroll, both up and down, will grow without bound. That is, for any X, there will come a time when I am $X ahead and a time when I am $X behind. So if I have an unlimited bankroll and an unlimited amount of time, I can make a million dollars. I simply wait until I am a million dollars ahead, which will eventually happen, and then I quit, never to play the game again. But the expected value of the number of flips I must make before winning a million dollars is infinity. And I must *never* play the game again after reaching my goal. I'll leave it as food for thought what, if anything, this has to do with the so-called martingale betting system.

[PairTheBoard]

Jason1990 --
"But the expected value of the number of flips I must make before winning a million dollars is infinity."

Actually, this expected value would be a finite number. This is easy to see. After only about 100 Trillion flips, being up a million would only amount to being about 1/10 of a standard deviation over the mean of 50 Trillion heads. After a Billion*Billion flips it's almost like a coin flip for you to be ahead by a million dollars.

With a fair coin you will of course spend just as much time on average being behind a million dollars as ahead. This is similiar in principle to what I'm saying would happen with the martingale backed by unlimited funds.

PairTheBoard

[jason1990]

[ QUOTE ]
Jason1990 --
"But the expected value of the number of flips I must make before winning a million dollars is infinity."

Actually, this expected value would be a finite number. This is easy to see.

[/ QUOTE ]
What you are seeing is an illusion. The expected number of flips it takes to win *one* dollar is infinite.

[Dov]

[ QUOTE ]
During those periods when you are behind it makes no sense to say, "It doesn't matter that I'm behind now because I know I'm going to be ahead again in the future." If it did make sense then the opposite could be said when you are ahead, "It doesn't matter that I'm ahead now because I know I will be behind again in the future."

[/ QUOTE ]

Isn't this exactly how we decide whether or not to continue playing in a poker game?

We figure our Sklansky bucks based on the situation and go from there, right?

If you know that you are going to be ahead in the future, and that the win will overtake the loss, then this is a +EV situation.

I don't see why this is any different than say entering a pot from LP with a small PP hoping to spike a set and recover the PF 'mistake' through implied odds on the flop. You have taken a -EV individual situation and given it a +EV spin by using implied odds.

The Martingale seems to have implied odds that will cover all of your losses + 1 unit.

I don't see how you can get around this. I understand you to be saying that when you finally do win, you should be ahead more than 1 unit if your expectation was positive. I also understand that you want to measure each individual bet. I don't think this is entirely accurate, though.

I posted a 44 hand from EP a few weeks ago where each individual action (bet, call fold) was correct, but each STREET had errors on it. (flop, turn, river). Maybe what we should be calculating is the effective odds of the Martingale. You are using pot odds on every bet.

BTW, I know that this is not a viable gambling technique, but it is interesting to consider.

Meh, I have to go now. Be back later.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
Jason1990 --
"But the expected value of the number of flips I must make before winning a million dollars is infinity."

Actually, this expected value would be a finite number. This is easy to see.

[/ QUOTE ]
What you are seeing is an illusion. The expected number of flips it takes to win *one* dollar is infinite.

[/ QUOTE ]

ok. I think you are working with a definition of the word "win" which you have in mind but are not stating. What I took "win" to mean was to "be ahead" by that amount. Certainly the expected number of flips you would make before finding yourself ahead by one dollar at that point in time is not infinite. If you mean to say that being ahead by a dollar doesn't mean you've won a dollar because you keep playing forever then ok. In that case you never "win" anything.

PairTheBoard

[jason1990]

[ QUOTE ]
Certainly the expected number of flips you would make before finding yourself ahead by one dollar at that point in time is not infinite.

[/ QUOTE ]
It is infinite. I mean exactly that.

[Mayhap]

I agree. Even if you have a stated goal, like 500,000 bets,
you may not ever achieve it.

The Martigale system is perfectly complemented by alcoholic beverages.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
During those periods when you are behind it makes no sense to say, "It doesn't matter that I'm behind now because I know I'm going to be ahead again in the future." If it did make sense then the opposite could be said when you are ahead, "It doesn't matter that I'm ahead now because I know I will be behind again in the future."

[/ QUOTE ]

Isn't this exactly how we decide whether or not to continue playing in a poker game?

We figure our Sklansky bucks based on the situation and go from there, right?

If you know that you are going to be ahead in the future, and that the win will overtake the loss, then this is a +EV situation.

I don't see why this is any different than say entering a pot from LP with a small PP hoping to spike a set and recover the PF 'mistake' through implied odds on the flop. You have taken a -EV individual situation and given it a +EV spin by using implied odds.

The Martingale seems to have implied odds that will cover all of your losses + 1 unit.


[/ QUOTE ]

With Sklansky implied odds, it means you are going to get paid off on future bets where you are a big favorite or even a lock to win the bets. In this game you are a dog on all the bets.

Try looking at it from the Casino's point of view. While you are marking "special" times after wins when you are ahead, the Casino is marking its "special" times when you are way behind and it is way ahead. There's no reason to think that your times for looking at where things are at are any more special than the times the Casino looks at where things are at.

PairTheBoard

[felson]

[ QUOTE ]
[ QUOTE ]
There's no way I am betting my money according to Martingale.

F.E.F.

[/ QUOTE ]

Do you ever say anything that is even remotely useful or insightful on this board? Seriously, you must be the number 1 rank troll on 2+2.

Lawrence

[/ QUOTE ]

You sound quite angry. I'm sorry that you feel that way.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
Certainly the expected number of flips you would make before finding yourself ahead by one dollar at that point in time is not infinite.

[/ QUOTE ]
It is infinite. I mean exactly that.

[/ QUOTE ]

It's true that there is no number of flips, N, such that after N flips you expect to be ahead by one dollar or more. But that is not what I said. And it's not what you said when you used the phrase "expected value of the number of flips".

Let N be the random variable which counts the number of flips at which you first go ahead by one dollar. I don't believe the Expected Value of N is infinite. If that's your assertion, I need to see a proof.

PairTheBoard

[jason1990]

[ QUOTE ]
Let N be the random variable which counts the number of flips at which you first go ahead by one dollar. I don't believe the Expected Value of N is infinite. If that's your assertion, I need to see a proof.

[/ QUOTE ]
It is my assertion. If you really want to see a proof, PM me and I will direct you to a textbook. But here's a pseudo-proof:

Let S(n) be your bankroll after the n-th flip (with S(0)=0). If E[N] is finite, then by Wald's identity, E[S(N)]=E[one flip]*E[N]. But S(N)=1, so E[S(N)]=1; and yet E[one flip]=0, a contradiction.

It's a pseudo-proof because I haven't stated or proved Wald's indentity, but perhaps it seems reasonable without it.

If you don't believe it and think I'm mistaken (which I've been known to be), then perhaps you might want to try and prove it's finite.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
Let N be the random variable which counts the number of flips at which you first go ahead by one dollar. I don't believe the Expected Value of N is infinite. If that's your assertion, I need to see a proof.

[/ QUOTE ]
It is my assertion. If you really want to see a proof, PM me and I will direct you to a textbook. But here's a pseudo-proof:

Let S(n) be your bankroll after the n-th flip (with S(0)=0). If E[N] is finite, then by Wald's identity, E[S(N)]=E[one flip]*E[N]. But S(N)=1, so E[S(N)]=1; and yet E[one flip]=0, a contradiction.

It's a pseudo-proof because I haven't stated or proved Wald's indentity, but perhaps it seems reasonable without it.

If you don't believe it and think I'm mistaken (which I've been known to be), then perhaps you might want to try and prove it's finite.

[/ QUOTE ]

Damn I guess you're right. The "Wald Identity" I have is for stopping times for Brownian Motion. The idea being that you can't change the zero expected value of Brownian Motion using a stopping time with finite mean. I imagine it works like you say in the discrete case as well.

So the expected number of flips to win a dollar is infinite. That's kind of amazing.

PairTheBoard

[slickpoppa]

You clearly misunderstand what it means to have an infinite bankroll. As long as you are always wagering a finite amount, your bankroll after using the martingale will be the same: infinity. Infinity - 6918624892374691286489127469278468927467892346 = infinity

[PairTheBoard]

[ QUOTE ]
You clearly misunderstand what it means to have an infinite bankroll. As long as you are always wagering a finite amount, your bankroll after using the martingale will be the same: infinity. Infinity - 6918624892374691286489127469278468927467892346 = infinity

[/ QUOTE ]

We're measuring success or failure by how much we are ahead or behind.

PairTheBoard

[Dov]

[ QUOTE ]
With Sklansky implied odds, it means you are going to get paid off on future bets where you are a big favorite or even a lock to win the bets. In this game you are a dog on all the bets.

[/ QUOTE ]

Maurile explained in a previous post how the longer you play, the more likely you are to to win a given bet. Your probability approaches infinity over time.

The difference in poker is that you can't recover all of your previous losses when you 'get lucky' and catch your miracle.

Here, when you do catch your miracle, you recover all of your losses plus a profit.

I understand that you are trying to isolate each individual bet as a separate event. I don't think that you can do that here, though, because the system has a defined relationship between the previous bet and the next one. Ignoring this aspect of the system is the part that puzzles me, I guess.

[PairTheBoard]

So the expected number of coin flips to win a dollar betting a dollar at a time is infinite. But the expected number of coin flips to win a dollar using a Martingale looks like:

1/2 + 2/4 + 3/8 + 4/16 + 5/32+ ... =
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ... +
000 + 1/4 + 1/8 + 1/16 + 1/32 + .... +
000000000 + 1/8 + 1/16 + 1/32 + .... +
000000000000000 + 1/16 + 1/32 + .... +
0000000000000000000000 + 1/32 + .... +
.
.
.
= 1 + 1/2 + 1/4 + ... = 2

Can that be right? You mean that using a Martingale reduces the expected value for the number of coin flips to win a dollar from infinite to 2?

PairTheBoard

[PairTheBoard]

I think you should try putting yourself in the Casino's shoes and saying to yourself, I'm going to judge how I'm doing by looking at the totals when the Martingale gambler is way behind after a long string of losses.

PairTheBoard

[SheetWise]

I think the problem in this thread concerns a probability of zero. There is an assumption that if an events probability has any value, that given an infinite number of trials it will eventually occur. I disagree.

[ QUOTE ]
Corollary I: The sum of the probabilities of any event and its complement is unity. That is P = 1.


The complement of an event is, of course, the nonoccurrence of that event. Characterizing an impossible event, we can state;


Corollary II: The probability of an impossible event is zero, or P = 0.


It is worth noting that the converse of Corollary II is not true. Given that the probability of some event equals zero, it does not follow that the event is impossible (there is a zero probability of selecting at random a pre-specified point on a line).


Theory of Gambling and Statistical Logic, Richard Epstein, page 15

[/ QUOTE ]

There are many events that I will easily assign a probability of zero, even though I know the probability is a number above zero.

I believe the probability of playing golf on a full length 18 hole PGA course, and shooting a score of 18 is impossible. I would without thought use this value in any equation.

I believe the probability of having a single selection in the PowerBall lottery, and selecting the winning combination in four consecutive draws is zero.

In the first example, the probability is obviously not zero -- but it's a perfectly good number to use. In the second example I can even assign a numeric value to it. Just because something can happen -- such as a run of 1000 heads or tails from a fair coin -- does not mean it ever will.

The theoretical concept of infinity is not really necessary for this problem, a definition of a theoretical zero is, then you can calculate some very large, yet finite, numbers.

[drudman]

As long as each series can complete (and I think, by definition, it must) you will always win in the long run.
Isn't this correct? I can't see how it couldn't be.

[PairTheBoard]

Saying that it "will eventually occur" is a somewhat loose use of language. What can be said more precisely is that if the probabilty of the event occuring in 1 trial is strictly positive, Then the limit of the probability that it occurs in N trials goes to 1 as N goes to infinity.

Here's an example you might like. Let the random variable X be uniformly distributed on the interval [0,1] - ie. X is equally likely to be in subintervals of the same length. Let x be the outcome of a Trial for the random variable X. ie. x is a specific number in the interval [0,1]. Then x has occurred as a result of the Trial but prior to the trial the probabilty for x occurring was 0. The event {x} had zero probabilty yet it happened anyway.

PairTheBoard

[Dov]

[ QUOTE ]
think you should try putting yourself in the Casino's shoes and saying to yourself, I'm going to judge how I'm doing by looking at the totals when the Martingale gambler is way behind after a long string of losses.

[/ QUOTE ]

That's what I was doing before when I asked why casinos bother with table limits.

The answer I got was that the variance will kill them.

I know this is correct, but I also know that the gambler MUST win if he is allowed to play each Martingale series to completion.

[Dov]

It's funny, but this reminds me of the other thread where they were discussing traversing a distance in steps that are each half as large as the next.

When someone finally says, "I don't get it, just take your first step twice!"

That's what the martingale does. Takes the first step as often as necessary to get to the other side.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
think you should try putting yourself in the Casino's shoes and saying to yourself, I'm going to judge how I'm doing by looking at the totals when the Martingale gambler is way behind after a long string of losses.

[/ QUOTE ]

That's what I was doing before when I asked why casinos bother with table limits.

The answer I got was that the variance will kill them.

I know this is correct, but I also know that the gambler MUST win if he is allowed to play each Martingale series to completion.

[/ QUOTE ]

So the gambler must "win" AND the Casino must "win". There will be times when the gambler is ahead and there will be times when the Casino is ahead. So how do you measure who is doing better?

PairTheBoard

[SheetWise]

It's wrong to look at the outcome of a trial (series of trials) and then assigning it a probablity. What you are calculating now is the probability that it will re-occur.. And that probability may well be zero.

There are many things we can observe -- then assign a probability -- and conclude that they will never happen again. The leap in logic is to assign a probability of zero to an event you just witnessed.

SheetWise

[Dov]

[ QUOTE ]
So the gambler must "win" AND the Casino must "win". There will be times when the gambler is ahead and there will be times when the Casino is ahead. So how do you measure who is doing better?

[/ QUOTE ]

The casino will NOT win if the gambler is allowed to always complete the series. It's impossible.

If there is a precondition that the gambler is always allowed to complete the series, then the casino must account for its reverse implied odds. They are actually paying out more than they are winning provided that the gambler can always complete his series.

Paradoxically, the only way the casino can win here is to put an end to his play or stop the martingale.

[Dov]

[ QUOTE ]
It's wrong to look at the outcome of a trial (series of trials) and then assigning it a probablity. What you are calculating now is the probability that it will re-occur.. And that probability may well be zero.

[/ QUOTE ]

This game has no memory. The probabilities for each trial do not change.

The probability that you will win at least once approaches infinity as long as you have a non zero probability event.

If you are looking at a roulette red/black bet, then using your logic, it would be a leap to assume that no trial would produce a win but we can think of series which would do just that. The problem with this thinking, IMO, is that we simultaneously have the opposite possible series as well, ie: infinite wins.

I think that for the sake of this discussion we can let these 2 series cancel each other out despite the fact that the winning series is actually less probable than the losing one. This is because it doesn't really matter as we are saying that both will occur with a 100% probability.

I say that we can do this, just as we can 'take a limit' and cancel the terms there.

When we take the limit of infinite losses, we also take the limit of infinite wins and they cancel each other out. That only leaves you with series that include at least 1 win for the gambler.

And when he wins, he recoups all of his former losses plus 1 unit of profit.

Since there are no more series that do not include a win, he must win and the casino must lose.

[Dov]

[ QUOTE ]
I think that for the sake of this discussion we can let these 2 series cancel each other out despite the fact that the winning series is actually less probable than the losing one. This is because it doesn't really matter as we are saying that both will occur with a 100% probability.

[/ QUOTE ]

Maybe this is where the problem is.

When we encounter a losing streak, it will be many orders of magnitude larger than its equivalent winning streak.

But when you take an infinite limit, it still doesn't really matter because as was pointed out earlier, infinity + x = infinity.

[MMMMMM]

[ QUOTE ]
[ QUOTE ]

So the gambler must "win" AND the Casino must "win". There will be times when the gambler is ahead and there will be times when the Casino is ahead. So how do you measure who is doing better?

[/ QUOTE ]



The casino will NOT win if the gambler is allowed to always complete the series. It's impossible.

If there is a precondition that the gambler is always allowed to complete the series, then the casino must account for its reverse implied odds. They are actually paying out more than they are winning provided that the gambler can always complete his series.

Paradoxically, the only way the casino can win here is to put an end to his play or stop the martingale.

[/ QUOTE ]

It might be better if you put the "must" in quotes when stating that the gambler must win a series (because perhaps that is not completely true)

But even if it is true, it is also true the casino "must" eventually win a streak big enough to eclipse all of the gambler's current accumulated one-unit profits. That horrible bad run is actually statistically expected and is probably "inevitable" if the gambler keeps on playing throughout eternity.

So the gambler "must" have times at which he is ahead of the casino. HOWEVER, the casino also "must" have times when it is ahead of the gambler (on the gambler's very worst streaks), and at these times the casino will often be further ahead of the gambler than the gambler ever was ahead of the casino (and all the more so because of the built-in green zero house edge on roulette).

Let's look at this another way, too.

Say there is a large corporation that will never go bankrupt or out of business, but is free to make or lose money. It's profits or losses are somewhat volatile, too, due to its particular niche and structure.

The accountant says, "We will only do the accounting after the quarters or years which show a profit. Quarters or years showing a loss will simply be carried over into the future, until we are showing a net profit, at which time we will do the accounting."

Would that make any sense? (granted it is not a good analogy, but I'm just trying to illustrate the fallacy of selecting your accounting times based on performance).

[MMMMMM]

[ QUOTE ]
[ QUOTE ]

So the gambler must "win" AND the Casino must "win". There will be times when the gambler is ahead and there will be times when the Casino is ahead. So how do you measure who is doing better?

[/ QUOTE ]



The casino will NOT win if the gambler is allowed to always complete the series. It's impossible.

If there is a precondition that the gambler is always allowed to complete the series, then the casino must account for its reverse implied odds. They are actually paying out more than they are winning provided that the gambler can always complete his series.

Paradoxically, the only way the casino can win here is to put an end to his play or stop the martingale.

[/ QUOTE ]

It might be better if you put the "must" in quotes when stating that the gambler must win a series (because perhaps that is not completely true)

But even if it is true, it is also true the casino "must" eventually win a streak big enough to eclipse all of the gambler's current accumulated one-unit profits. That horrible bad run is actually statistically expected and is probably "inevitable" if the gambler keeps on playing throughout eternity.

So the gambler "must" have times at which he is ahead of the casino. HOWEVER, the casino also "must" have times when it is ahead of the gambler (on the gambler's very worst streaks), and at these times the casino will be further ahead of the gambler than the gambler ever was ahead of the casino (and all the more so because of the built-in green zero house edge on roulette).

These times will become rarer and rarer, but the negative slumps when they do occur will be deeper and deeper for the gambler. So, they both are (nearly) "assured" of being ahead at some points.

Let's look at this another way, too.

Say there is a large corporation that will never go bankrupt or out of business, but is free to make or lose money. It's profits or losses are somewhat volatile, too, due to its particular niche and structure.

The accountant says, "We will only do the accounting after the quarters or years which show a profit. Quarters or years showing a loss will simply be carried over into the future, until we are showing a net profit, at which time we will do the accounting."

Would that make any sense? (granted it is not a good analogy, but I'm just trying to illustrate the fallacy of selecting your accounting times based on performance).

[Dov]

[ QUOTE ]
HOWEVER, the casino also "must" have times when it is ahead of the gambler (on the gambler's very worst streaks), and at these times the casino will often be further ahead of the gambler than the gambler ever was ahead of the casino

[/ QUOTE ]

The only streak that this applies to is one with infinite losses. Any streak that includes a win will be a loss for the casino.

In your business example, it is like offering the clients a full refund on all of their payments to the company if they fulfill a given condition.

You would have to account for the money you would be expected to pay out as a result of clients fulfilling their obligations. This money expected to be paid still needs to be accounted for.

You cannot simply do a balance sheet and get an accurate picture of the business. You would need to do all of the key financial reports and they will include payments expected to be made.

The effect of the zero in the roullette wheel for this situation, as I see it, is that it makes the likelihood of having an infinite losing streak higher than the likelihood of having a winning one.

The problem with this, of course, is that both of those probabilities are 100% and cannot get any higher. This is what I meant when I said that the infinite (unlimited) bankroll would defeat the house edge. It makes both events equally likely.

[SheetWise]

[ QUOTE ]
But even if it is true, it is also true the casino "must" eventually win a streak big enough to eclipse all of the gambler's current accumulated one-unit profits.

[/ QUOTE ]

Disagree - as in a later thread.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
So the gambler must "win" AND the Casino must "win". There will be times when the gambler is ahead and there will be times when the Casino is ahead. So how do you measure who is doing better?

[/ QUOTE ]

The casino will NOT win if the gambler is allowed to always complete the series. It's impossible.

If there is a precondition that the gambler is always allowed to complete the series, then the casino must account for its reverse implied odds. They are actually paying out more than they are winning provided that the gambler can always complete his series.

Paradoxically, the only way the casino can win here is to put an end to his play or stop the martingale.

[/ QUOTE ]

By "win" I meant "be ahead". Sometimes the Casino will be ahead and sometimes the Martingaler will be ahead. If the Martingaler wants to count himself a winner when he's ahead, then he must count himself a loser when he's behind.

PairTheBoard

[PairTheBoard]

Dov --
"The only streak that this applies to is one with infinite losses. Any streak that includes a win will be a loss for the casino."

You are only looking at the end of a streak. The middle of a streak is just as valid a time to look.

PairTheBoard

[MMMMMM]

Well, I do not necessarily agree that the probability of having an endless losing streak is ZERO. Or, you could say that I do not necessarily agree that the gambler is CERTAIN to win a future bet.

Anyway, as the chance of a Long/Longer/SuperLong losing streak decreases geometrically (minus the green zero effect), the penalty for having such a streak increases geometrically.

The merely "very long" losing streaks are expected to negate the gambler's accumulated wins, plus some (due to the house zero(s)).

Also, what is the point of taking accounting only after wins, if the gambler is going to always continue playing?

[PairTheBoard]

[ QUOTE ]
It's wrong to look at the outcome of a trial (series of trials) and then assigning it a probablity. What you are calculating now is the probability that it will re-occur.. And that probability may well be zero.

There are many things we can observe -- then assign a probability -- and conclude that they will never happen again. The leap in logic is to assign a probability of zero to an event you just witnessed.

SheetWise

[/ QUOTE ]

There may be many things that are wrong to do, but I don't think there was anything wrong with what I said.

PairTheBoard

[MMMMMM]

[ QUOTE ]
[ QUOTE ]
But even if it is true, it is also true the casino "must" eventually win a streak big enough to eclipse all of the gambler's current accumulated one-unit profits.

[/ QUOTE ]



Disagree - as in a later thread.

[/ QUOTE ]

I'm not saying the casino ACTUALLY must--I'm saying if the rationale applies that the gambler "must" win a series (of which I am not completely convinced), then why not flip it around and give the house a similar "must"--which may not really be a must.

That said, I haven't followed all your later threads or sub-threads, so until I do I'll back off this part.

[SheetWise]

But remember that for the two extreme outcomes you are talking about, one is necessary for me to lose -- the other is not necessary for me to win.

For me to lose, my opponent needs a long series of consecutive wins. For me to win, all that is necessary is that my opponent doesn't get that sequence. I don't care how the winning occurs, it occurs every time, as long as my opponent doesn't accomplish the long series.

I am not even remotely suggesting that the EV of the game changes, or that a Martingale will work in any realistic situation. But even without the limits of this hypothetical, it can (hypothetically) work. A series of events can be compounded to a point where its probability of ocurrence is zero -- simply observing that it has a positive value does not indicate it will 'eventually' happen (in any world that has real limits).

[Dov]

[ QUOTE ]
simply observing that it has a positive value does not indicate it will 'eventually' happen (in any world that has real limits).

[/ QUOTE ]

This is the whole point. When you take the variables to infinity, the Martingale defeats the house edge assuming that it's edge is not 100%.

Those variables are Bankroll and Trials.

I don't think anyone is actually saying that they believe this to be practical in the real world.

[Dov]

[ QUOTE ]
You are only looking at the end of a streak. The middle of a streak is just as valid a time to look.

[/ QUOTE ]

No, it's not.

The Martingale series has a defined beginning and end point. You will be losing BY DEFINITION in the middle of a series.

You must take the whole thing into account if you are to accurately assess its EV.

If the casino didn't know what the gambler was doing, then you would be correct. But every once in a while, they would have massive losing days which coincided with the days that a series actually ended.

When they started correlate the data, they would find that for every completed series, they were down 1 bet. So their EV+ game was actually losing 1 bet / avg_num_trials_per series during that time period despite the fact that they took the numbers after every single bet.

The gambler really doesn't care about the losing trials. He simply endures them as a waste of time until his winning trial where he makes 1 bet. So, he doesn't take stock of the losses, only the wins. For him, the only thing that matters is the end of a series.

I guess what I'm saying is that if you look at the middle of the martingale series, your conclusions will be automatically wrong because the sample size will be too small to contain the entire system.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
You are only looking at the end of a streak. The middle of a streak is just as valid a time to look.

[/ QUOTE ]

No, it's not.

The Martingale series has a defined beginning and end point. You will be losing BY DEFINITION in the middle of a series.

You must take the whole thing into account if you are to accurately assess its EV.



[/ QUOTE ]

Exactly. I am taking the whole thing into account to assess its EV. I'm taking into account the beginning the middle and the end. You are only taking the beginning and the end into account. That's why your assessment of the EV is wrong.

You said later,
"The gambler really doesn't care about the losing trials. He simply endures them as a waste of time until his winning trial where he makes 1 bet. So, he doesn't take stock of the losses, only the wins. For him, the only thing that matters is the end of a series."

The gambler may not care but the EV does. From the Casino's point of view the exact opposite could be said:

The Casino really doesn't care about the player's winning trials. It simply endures them as a waste of time until the Casino's winning streak where it makes huge gains. So, it doesn't take stock of the losses, only the wins. For it, the only thing that matters is the streaks when it is way ahead.

The EV cares about both when the Casino is ahead and when the gambler is ahead. As I've mentioned before, it measures the Time ahead or behind WEIGHTED BY THE AMOUNT ahead or behind.

I'm afraid I'm starting to repeat myself. I don't think there's much I can add to what I've already said.

PairTheBoard

[Siegmund]

[ QUOTE ]
[ QUOTE ]
Let N be the random variable which counts the number of flips at which you first go ahead by one dollar. I don't believe the Expected Value of N is infinite. If that's your assertion, I need to see a proof.

[/ QUOTE ]
Let S(n) be your bankroll after the n-th flip (with S(0)=0). If E[N] is finite, then by Wald's identity, E[S(N)]=E[one flip]*E[N].


[/ QUOTE ]

Two correct, and completely unrelated, points.

N is a garden variety random variable; it has a geometric distribution with p=9/19 (on the double-zero wheel). E[N]=19/9 and Var[N] = 190/81.

E[S(N)] is 0 for all N for a fair bet, and negative and decreasing monotonically with N for a Martingale on a real roulette wheel. But P(S(N)>0) remains positive, decaying slowly enough that a Martingaler with an infinite bankroll will find himself ahead of the game infinitely many times (and behind infinitely many times too of course.)

---

Also, to point out one other error in this thread: no, the house doesn't need an infinite bankroll. All it needs to do is start with one chip on the table before each progression, along with whatever money the Martingaler has already lost on his current progression. Only someone who *more than* doubles his wager can threaten the house with being broken.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Let N be the random variable which counts the number of flips at which you first go ahead by one dollar. I don't believe the Expected Value of N is infinite. If that's your assertion, I need to see a proof.

[/ QUOTE ]
Let S(n) be your bankroll after the n-th flip (with S(0)=0). If E[N] is finite, then by Wald's identity, E[S(N)]=E[one flip]*E[N].


[/ QUOTE ]

Two correct, and completely unrelated, points.

N is a garden variety random variable; it has a geometric distribution with p=9/19 (on the double-zero wheel). E[N]=19/9 and Var[N] = 190/81.



[/ QUOTE ]

I don't think you can be talking about the same Random Variable N as I define above - even if converted to Roullette from fair coin flips. Maybe you didn't realize we weren't talking Martingale at that point but fixed betting amounts. Or maybe I'm not realizing you're talking Martingale throughout your post.

As defined N amounts to the stopping time at which S(n) first equals 1. So by N's definition, S(N)=1 identically, and E[S(N)]=1. If E[N] were finite as you claim, then by Wald, E[S(N)]=E[S(1)]E[N]. But for the coin flip, E[S(1)]=0 and for Roulette, E[S(1)]<0. A contradiction, therefore E[N] cannot be finite as you claim.

Your latter comments use N as fixed rather than a random variable and look ok.

PairTheBoard

[drudman]

At this point, is do we have consensus that OP was wrong?

[PairTheBoard]

[ QUOTE ]
At this point, is do we have consensus that OP was wrong?

[/ QUOTE ]

Only among those without the math background to understand that he is basically right.

PairTheBoard

[Dov]

[ QUOTE ]
Only among those without the math background to understand that he is basically right.

[/ QUOTE ]

I'm afraid that I may have to agree with you here. I'm actually just going over this thread again and again trying to understand why you guys are right.

Something tells me that you probably are, but I still can't quite see it yet.

I'm sure another week or so of scrutinizing this will help.

I am really just trying to understand why this is correct or not. I appreciate everyone's patience, by the way.

[PairTheBoard]

I think the key to understanding it is to realize that the Casino's viewpoint that I've talked about several times is just as valid as the Martingaler's viewpoint.

Best Wishes

PairTheBoard

[Grivan]

Dov you are assuming the gambler can end whenever he chooses. This is not fair if you are saying he must play for an infinite amount of time. A more reasonable way to look at it would be to allow time to run, but have it stop at some random point in the future (the casino and the gambler having no affect on when it stops). Once it stops look at who is ahead and by how much. Now do this trial a ton of times and you will clearly see that the casino is the winner in this circumstance.

Back to the scenario where it never ends. The gambler cannot choose to stop when he is ahead, he must always keep going until there is a point where he is behind again. On average the gambler will always be behind assuming the trial has run long enough.

[drudman]

[ QUOTE ]
[ QUOTE ]
At this point, is do we have consensus that OP was wrong?

[/ QUOTE ]

Only among those without the math background to understand that he is basically right.

PairTheBoard

[/ QUOTE ]

Well, that's me, but I'll play Devil's Advocate.

1) Each completed series adds +1 to Martingaler.
2) Each series completes.
3) Martingaler is guaranteed to be +x, where x=number of series completed.

Which of the above is false, and why?

[MMMMMM]

[ QUOTE ]
1) Each completed series adds +1 to Martingaler.
2) Each series completes.
3) Martingaler is guaranteed to be +x, where x=number of series completed.

[/ QUOTE ]

Well for one thing, each series hasn't completed yet (you are looking into the future, not the past). So how do you KNOW that each series completes?

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
At this point, is do we have consensus that OP was wrong?

[/ QUOTE ]

Only among those without the math background to understand that he is basically right.

PairTheBoard

[/ QUOTE ]

Well, that's me, but I'll play Devil's Advocate.

1) Each completed series adds +1 to Martingaler.
2) Each series completes.
3) Martingaler is guaranteed to be +x, where x=number of series completed.

Which of the above is false, and why?

[/ QUOTE ]

I think it's true that the event, Martingale Series Completes, occurs infinitely many times with probabilty 1.

But when the Casino goes ahead of the player by a record amount let's call that the completion of the Casino's Anti-Maringale Series.

Then it's also true that the event, Anti-Martingale Series Completes, occurs infinitely many times with probabilty 1.

So just saying that the player's series or the Casino's series completes infinitely many times doesn't tell us who is really winning because they both happen. We need a more precise way of seeing who is really winning. Grivan has the right idea when he says,

Grivan --
"A more reasonable way to look at it would be to allow time to run, but have it stop at some random point in the future (the casino and the gambler having no affect on when it stops). Once it stops look at who is ahead and by how much. Now do this trial a ton of times and you will clearly see that the casino is the winner in this circumstance."

The "random point in the future" needs to be nailed down a little better, but Grivan's basic point is exactly right.

PairTheBoard

[drudman]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
At this point, is do we have consensus that OP was wrong?

[/ QUOTE ]

Only among those without the math background to understand that he is basically right.

PairTheBoard

[/ QUOTE ]

Well, that's me, but I'll play Devil's Advocate.

1) Each completed series adds +1 to Martingaler.
2) Each series completes.
3) Martingaler is guaranteed to be +x, where x=number of series completed.

Which of the above is false, and why?

[/ QUOTE ]

I think it's true that the event, Martingale Series Completes, occurs infinitely many times with probabilty 1.

But when the Casino goes ahead of the player by a record amount let's call that the completion of the Casino's Anti-Maringale Series.

Then it's also true that the event, Anti-Martingale Series Completes, occurs infinitely many times with probabilty 1.

So just saying that the player's series or the Casino's series completes infinitely many times doesn't tell us who is really winning because they both happen. We need a more precise way of seeing who is really winning. Grivan has the right idea when he says,

Grivan --
"A more reasonable way to look at it would be to allow time to run, but have it stop at some random point in the future (the casino and the gambler having no affect on when it stops). Once it stops look at who is ahead and by how much. Now do this trial a ton of times and you will clearly see that the casino is the winner in this circumstance."

The "random point in the future" needs to be nailed down a little better, but Grivan's basic point is exactly right.

PairTheBoard

[/ QUOTE ]

I disagree. It's obvious that the casino will be a large winner if the betting must randomly stop at some point. But that wasn't what OP said. OP said that it is a fallacy that with an infinite bankroll Martingaling is successful. That is wrong, because the player chooses when to stop, not the casino.

As for the anti-Martingaler series objection, it seems pretty weak. It doesn't even make sense. There is no chain of bets that ends with the casino winning a unit back from the player. Obviously the casino will win some ridiculous bets, but the player will complete his series by definition, and all of those ridiculous record bets will become immaterial. The player will be +1 unit.

The player isn't even really betting on roulette. He's betting that he won't have an infinitely long losing streak. It's a lock, Jerry!

[MMMMMM]

[ QUOTE ]
I disagree. It's obvious that the casino will be a large winner if the betting must randomly stop at some point. But that wasn't what OP said. OP said that it is a fallacy that with an infinite bankroll Martingaling is successful. That is wrong, because the player chooses when to stop, not the casino.


[/ QUOTE ]

Actually, I said this:

"If you intend to keep repeating the system, you cannot logically "take accounting" at only those times when you fancy to take accounting."

But even if you do NOT intend to play forever, or repeatedly, the system does not work. Yes, it will almost surely work for you if you do it just once or a few times. But for every time you try it, there exists the possibility of indefinitely prolonged losing streaks. Though unlikely, the penalty of a horrid streak may be so severe as to obviate all gains from prior runs.

You cannot be sure that will not occur. And if you and a countless number of gamblers were all playing the same system, from the same bottomless bankroll, on countless tables simultaneously, you guys would be expected to be showing a net loss all along the way. In other words, even thoughg you and many of the gamblers might be winning as you complete as series, therew would exist your horrendously unlucky counterparts who would be losing more than you guys are winning.

And you have no way of knowing in advance if you are going to be one of the ones for whom the luck seems "normal", or "good"--or absolutely unbelievably, never-endingly horrible. Just as you have no way of knowing if the next streak you embark on will get worse....and worse...and worse...

For all the small wins "earned" by completing the series, there exists a theoretical counterpart or series which is losing that much and more. And you can't know in advance what your next lot will be.

Yes, it seems like a lock to win a small amount (perhaps again and again). But theoretically speaking, it isn't.

[PairTheBoard]

drudman --
"As for the anti-Martingaler series objection, it seems pretty weak. It doesn't even make sense. There is no chain of bets that ends with the casino winning a unit back from the player. Obviously the casino will win some ridiculous bets, but the player will complete his series by definition, and all of those ridiculous record bets will become immaterial. The player will be +1 unit."

It's not weak. After the end of every Martingale series that puts the player ahead by amount A, the Casino is looking for a series of bets LLLLLLL...L consisting of n straight losses for the player where 2^n > A. That is a series of bets that ends with the Casino taking one or more chips from the player's pocket. The fact that they play on from that point is no more valid than the fact that they play on after the end of a Martingale Series. What's weak is your refusal to look at it from the Casino's perspective.

PairTheBoard

[Dov]

I thnk the part of this that's so hard for us non-math people is the fact that there is only 1 series that ends with a L, and that is not considered to be a complete series.

Losing forever is the only way that the Martingaler loses. Every other sequence results in a gain. And he simply continues until he wins.

I understand the point about having multiple people playing at the same time and I think it is a good one.

The problem with the accounting is this, as I see it:

The casino's edge is ok because they are offering the game. This is the 'price' they are charging you for the privelege (option) to play.

The Martingaler is not required to play. However, we have stipulated in our problem that he must continue playing forever.

Now, since the Martingaler plays as he sees fit, he only takes account of when he completes a series. This is because he knows that the rest of it is just waiting.

The casino takes account after every bet. They have no way of knowing if he will, in fact, place another bet or not.

I'm starting to think that they may both be right. Still, it seems that the casino isn't taking everything into account.

They should be able to understand that if the Martingaler continues to play, and never quits when he loses, then they are almost certain not to win.

This is not against the rules of the game, as someone else was saying that it would be taking future bets into account. I think this is imperative.

Otherwise, it would be kind of like saying that as long as your preflop game is perfect, your postflop skills don't matter because you already made your money in the beginning. If something happens later in the hand to change the EV of your situation, and you don't recognize it, you will lose, even though you were originally the favorite.

I think that this is what happens here. The Martingaler is not playing the same game as the casino.

The casino plays one bet and wins every time. The Martingaler plays one series and wins every time. They are actually playing different games.

They both win.

How did I do?

[drudman]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
At this point, is do we have consensus that OP was wrong?

[/ QUOTE ]

Only among those without the math background to understand that he is basically right.

PairTheBoard

[/ QUOTE ]

Well, that's me, but I'll play Devil's Advocate.

1) Each completed series adds +1 to Martingaler.
2) Each series completes.
3) Martingaler is guaranteed to be +x, where x=number of series completed.

Which of the above is false, and why?

[/ QUOTE ]

Pairtheboard,

Like I say above, I'm playing Devil's Advocate. Which is false, 1 or 2? If both are true, then 3 is true. You say 3 is false, so please make your objection.

Bet 1. Win. +1

Bet 1. Lose. -1
Bet 2. Win. +1

Bet 1. Lose. -1
Bet 2. Lose. -3
Bet 4. Win. +1

...and so on. Are you trying to say that at some point:

Bet 1. Lose. -1
Bet 2. Lose. -3
...
Bet x. Win. <not +1>

[MMMMMM]

[ QUOTE ]
They should be able to understand that if the Martingaler continues to play, and never quits when he loses, then they are almost certain not to win.


[/ QUOTE ]

"Almost" certain isn't certain.

The less likely the sequence of straight losses, the greater the penalty, in proportion (plus a bit).

If your single hypothetical gamber were replaced by countless simultaneous gamblers on independent tables, that's pretty much the same thing as one gambler gambling for endless duration, isn't it? But at no time could you expect the sum total of those gamblers' results to be showing a net profit.

When you start a sequence, you don't know for certain in advance that it won't be the worst sequence ever--or even that it will eventually win. Just as you don't know that there won't be one player in that countless field of players who won't lose every single bet. And if he does, he will lose far, far more than his positive counterpart who might be winning every single bet.

[Dov]

[ QUOTE ]
...and so on. Are you trying to say that at some point:

Bet 1. Lose. -1
Bet 2. Lose. -3
...
Bet x. Win. <not +1>

[/ QUOTE ]

No, what they are saying is that there will be a series which does not end with a W.

In addition, this series will occur with a 100% probability.

I was saying that the opposite will occur as well. That is you will have a series WWW...W through infinite.

The problem is, that if you ignore the aspect of infinity +1 = infinity and pretend that they are finite. What will happen is that the losing streak will be infinitely larger than the winning streak.

Not only that, but because of the house edge, you will be "more likely" to experience the losing streak than the winning streak, even though they will both actually occur with 100% frequency.

LOL

Looking at this kind of makes me think of trying to use Newtonian physics on Quantum particles.

hmm...

[Dov]

[ QUOTE ]
And if he does, he will lose far, far more than his positive counterpart who might be winning every single bet.

[/ QUOTE ]

Yes, I get it now. I just explained it to my 'partner' here.

I think I was trying to have it both ways in terms of infinity. Still, the question remains, I guess, that since the math concepts say that infinity +/- 1 = infinity, then the whole problem is pointless, because no one can actually have any edge at all, even on a 0 probability event.

One quick question I guess should answer it for me.

Solve for x:

infinity - infinity = x

[drudman]

Whoa, you're saying that in an infinite series of rolls, he will necessarily have an infinitely long losing streak?

I'm pretty sure that's not true.

The Martingaler's bankroll will never change. But we can still measure his net gain/loss. Every single series that isn't an infinitely long losing streak ends with +1. The money he bets is immaterial. It might as well not even be money - it is Monopoly money, or bananas, or whatever. What the Martingaler is interested in is his net gain/loss, which 1 unit per series completed.

EDIT: Basically, the whole thing comes down to whether or not there can be a series of infinite losses. I need a probability theorist to answer that. Common sense and my once somewhat-formidable knowledge of math both say that in an infinite series of bets of any kind, there is a 0% chance that there will be an infinitely long chain of only one result.

[jason1990]

If the casino had an infinite number of patrons and each one left as soon as he won one series, then each leaving patron would leave with $1. But at any point in time, the casino will be ahead. This shouldn't be too surprising. After all, making something out of nothing is easy with infinities.

A bus carrying an infinite number of passengers arrives at a hotel with an infinite number of rooms. Problem is, every room in the hotel is occupied. So the hotel manager announces over the PA that if you are in room n, you should immediately move to room 2n. Suddenly, all the odd rooms are empty and he can accomodate the passengers on the bus.

[Dov]

[ QUOTE ]
Whoa, you're saying that in an infinite series of rolls, he will necessarily have an infinitely long losing streak?

[/ QUOTE ]

Exactly

[ QUOTE ]
I'm pretty sure that's not true.

[/ QUOTE ]

It is true.

Maurile started a table earlier in this thread that showed that your probability of winning each successive bet in the series grows geometrically, but never reaches 100%.

Of course, there is also a series of infinite wins.

How can this not be true?

[ QUOTE ]
Basically, the whole thing comes down to whether or not there can be a series of infinite losses. I need a probability theorist to answer that. Common sense and my once somewhat-formidable knowledge of math both say that in an infinite series of bets of any kind, there is a 0% chance that there will be an infinitely long chain of only one result.

[/ QUOTE ]

Where's Pzhon when you need him/her?

[hoyaboy1]

[ QUOTE ]
What's weak is your refusal to look at it from the Casino's perspective.

PairTheBoard

[/ QUOTE ]

The thing is that I can choose to stop after I win my series - but the casino can't make me stop at a time ideal for them (at least we are assuming this, I think).

[Dov]

[ QUOTE ]
The thing is that I can choose to stop after I win my series

[/ QUOTE ]

No you can't. In this game the Martingaler plays forever,

[hoyaboy1]

[ QUOTE ]
[ QUOTE ]
The thing is that I can choose to stop after I win my series

[/ QUOTE ]

No you can't. In this game the Martingaler plays forever,

[/ QUOTE ]

Ok, but isn't the whole point of the theory that it is broken into series and you can stop after a win? I'm not sure it is supposed to be valid if you aren't given that ability.

[kpux]

I'm pretty sure that an infinitely long losing streak doesn't exist. Wouldn't that imply that the EV of an even money bet is negative? That doesn't seem to make sense.

While I haven't thought this through completely, it would seem that the Martingaler can expect a win just through even money bets, but only because he can stop when he chooses. And that is the edge he has over the house, who has to play on demand.

[MMMMMM]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The thing is that I can choose to stop after I win my series

[/ QUOTE ]



No you can't. In this game the Martingaler plays forever,

[/ QUOTE ]



Ok, but isn't the whole point of the theory that it is broken into series and you can stop after a win? I'm not sure it is supposed to be valid if you aren't given that ability.

[/ QUOTE ]

Even if you plan to stop, you must realize that you are gambling that you will eventually win a bet. You aren't absolutely 100% sure you will, though. And the penalty for not winning a bet increases geometrically after each losing spin.

Even if you have lost 15 bets in a row, or 15,000 bets in a row, the odds of your winning the *next* spin are still less than 50%. So while you may look into the future, saying "I have to win a bet *eventually*", the wheel doesn't know that, and each new spin is a new chance because the wheel has no memory. Meanwhile your losses are multiplying geometrically.

It may seem like it is too farfetched to imagine that you might NEVER win another spin. But if you had essentially unlimited players each playing on independent wheels, there could be one player who just loses every single bet he makes. This is parallel to the chances of your sitting down and just losing every bet--it's an infinitesimally small chance, but given unlimited time, you could fall into that slot. And the finacial penalty would be unlimited, too.

For every winning player or winning sequence, there is a theoretical counterpart which is losing the same amount or more. So theoretically the system isn't going to produce a profit even with no restrictions on table limits or bankroll.

[MMMMMM]

[ QUOTE ]
Whoa, you're saying that in an infinite series of rolls, he will necessarily have an infinitely long losing streak?

I'm pretty sure that's not true.

[/ QUOTE ]

In an infinite series of rolls, which is akin to an infinite number of players playing independently, there could well be one player who just never wins. That doesn't mean his losing streak will necessarily continue to extend to infinity. It just means that no matter where you are on the timeline when you ask him, that's the result he will be having at that time.

At least, that's the way I see it.

[bobman0330]

I don't think you understand MMMMMM... If you strip the problem of all meaning and make a lot of shaky assumptions, you can WIN ONE UNIT!!

The casino shouldn't be too concerned though, because in any real-time setting, they can put all the money the Martingaler is currently down, keep it in a money-market account, and make a ton of money in interest.

[Dov]

[ QUOTE ]
I don't think you understand MMMMMM... If you strip the problem of all meaning and make a lot of shaky assumptions, you can WIN ONE UNIT!!

[/ QUOTE ]

LOL, I know you're just being sarcastic, but...

What if we make each unit infinite too? Then would it be worth winning JUST ONE?

[ QUOTE ]
The casino shouldn't be too concerned though, because in any real-time setting, they can put all the money the Martingaler is currently down, keep it in a money-market account, and make a ton of money in interest.

[/ QUOTE ]

Until they lose a ton of money in early withdrawal fees when he wins a series earlier than they expected.

[pzhon]

Here is a proof I think is more intuitive, with more details in white:

Let p(n) be the probability you have never been ahead before the nth wager.

The expected first time to be ahead is the sum of the p(n).
<font color="white">In the usual sum for the expected value, change i to the sum from n=1 to i of 1. Then change the order of summation.</font>
p(2k-1)=p(2k)= (2k C k)/2^(2k) ~ 1/sqrt(pi k)
<font color="white">Use the reflection principle to write the number of lattice paths to (a,b) staying above the diagonal as a difference between binomial coefficients. The sum telescopes. Then use Stirling's approximation for n!.</font>
The sum of c/sqrt(n) diverges, so the expected number of turns before your first win is infinite.

[MMMMMM]

[ QUOTE ]
[ QUOTE ]
Whoa, you're saying that in an infinite series of rolls, he will necessarily have an infinitely long losing streak?

I'm pretty sure that's not true.

[/ QUOTE ]



In an infinite series of rolls, which is akin to an infinite number of players playing independently, there could well be one player who just never wins. That doesn't mean his losing streak will necessarily continue to extend to infinity. It just means that no matter where you are on the timeline when you ask him, that's the result he will be having at that time.

At least, that's the way I see it.

[/ QUOTE ]

Let me be clear about something here, because I realize this could quickly get a lot more complicated. I'm not saying it's guaranteed that at any time you ask, there will be someone who has never won a bet in their current series. I'm saying it could well happen. And I'm saying that no matter when or whom you ask, you can't be absolutely sure it won't happen in the future.

[SomethingClever]

Bottom line... if you can always bet 2x after a loss, you can choose to be a winner as long as you stop on a win.

[ QUOTE ]
If you were truly to play this game eternally, your chips would slowly go down, on average, due to the house edge.

[/ QUOTE ]

The "average" amount of your bankroll is meaningless, though, when you can recoup all your losses with a single bet.

Sure, if you play eternally, you will experience bigger and bigger losses, on average, but if you quit on a win, you have won.

Edit: Oh, and the probability of being a losing player under the system with an infinite bankroll, infinite betting limits and an infinite amount of time appears to be infinity - 1. Do you see why? /images/graemlins/tongue.gif

[hoyaboy1]

[ QUOTE ]
For every winning player or winning sequence, there is a theoretical counterpart which is losing the same amount or more. So theoretically the system isn't going to produce a profit even with no restrictions on table limits or bankroll.

[/ QUOTE ]

For this to work, the odds of never winning x potential losses would have to be greater than the odds of eventually winning x 1 unit, right? Is this necessarily true?

Edit - I'm sure this can be figured out, but I haven't done much stats stuff in a while and forget how

[bobman0330]

Here's the killer proof:

Eventually, the house will have so many of your checks stacked up on their racks that their mutual gravitation will overwhelm the electrostatic and other quantum mechanical forces that hold atoms apart, causing the checks to collapse into a black hole. Special relativity and quantum mechanics as we understand them today state that it is impossible to retrieve any matter from a black hole. Ergo, your chips will be lost forever.

[Grivan]

[ QUOTE ]
Bottom line... if you can always bet 2x after a loss, you can choose to be a winner as long as you stop on a win.

[/ QUOTE ]

You cannot have it both ways. You can't have the game go on for infinite and have the ability to stop on a win. Since if you stop it is no longer an infinitly long game. So stopping at any point is not valid. What people are doing is coming up with a way to measure who is winning, since you cannot reasonably use a single point in time to do this.

[SomethingClever]

[ QUOTE ]
[ QUOTE ]
Bottom line... if you can always bet 2x after a loss, you can choose to be a winner as long as you stop on a win.

[/ QUOTE ]

You cannot have it both ways. You can't have the game go on for infinite and have the ability to stop on a win. Since if you stop it is no longer an infinitly long game. So stopping at any point is not valid. What people are doing is coming up with a way to measure who is winning, since you cannot reasonably use a single point in time to do this.

[/ QUOTE ]

Agreed. If you play infinitely, you will lose money on average.

However, unless you are the one case out of infinity that literally never wins a bet, you will win if you stop.

Also, you can "virtually" assure yourself a win even if you do multiple stop wins over a career of indeterminate length.

This is because assuming an infinite bankroll, etc... the odds that you will ever hit the infinity-long losing streak are infinitely improbable.

Yes, I'm a big HH2G fan.

[drudman]

Okay, so the pro-Martingalers are arguing that there cannot be an infinite series of losses.

The anti-Martingalers are arguing that there can be.

I really have a hard time believing the anti-Martingalers. It keeps being brought up that if an infinite number of players use this strategy, there will always be players who have not won yet. I'm not sure if I buy this, because you can only check to see how many players have yet to win at finite numbers of trials... isn't this right?

Also, doesn't 1/2^n=0 when n goes to infinity? Because this represents the odds of losing n bets in a row. Didn't we establish this in the "Hawking Infinite Series" thread?

[MMMMMM]

[ QUOTE ]
Okay, so the pro-Martingalers are arguing that there cannot be an infinite series of losses.

The anti-Martingalers are arguing that there can be.

I really have a hard time believing the anti-Martingalers. It keeps being brought up that if an infinite number of players use this strategy, there will always be players who have not won yet. I'm not sure if I buy this, because you can only check to see how many players have yet to win at finite numbers of trials... isn't this right?

[/ QUOTE ]

This can get complicated and over my head at some point, so just let me say that I'm NOT saying there WILL ALWAYS be players who have not won yet--just that there CAN be. And you have no way of knowing FOR SURE that there WON'T be.


[ QUOTE ]
Also, doesn't 1/2^n=0 when n goes to infinity? Because this represents the odds of losing n bets in a row.

[/ QUOTE ]

I don't know, but if it does, couldn't the same argument be applied at that point by saying you have already lost infinite money (since your losses double at the same rate that 1/2^n goes to infinity).

[SomethingClever]

If you play perpetually, you will lose money on average in a game where the house has a slight edge.

Of course it won't matter with your infinite bankroll.

If you decide that you will have 1 stopping point and you have an infinite bankroll, it is almost infinitely probable that you will win.

Adding several other stopping points along the way will very slightly increase the chance that you will go on an infinitely long losing streak, but it's still almost infinitely improbable.

[PairTheBoard]

drudman --
"Okay, so the pro-Martingalers are arguing that there cannot be an infinite series of losses.

The anti-Martingalers are arguing that there can be."

No. I think MMMMMMMMMMMM is getting off base with that argument. SomethingClever has it right. In the beginning MMMMMMMMM was talking about the case where the Martingaler continues playing forever. In that case the pro-Martingalers argue that he will just chalk up 1 chip after another as each Martingale series completes. The con-Martingalers argue that the Casino will chalk up stacks of chips after stacks of chips as one long streak after another happens, and that on average the Casino will be Ahead by more chips per unit time than the Martingaler.

But if you change this scenario to one where the Martingaler quits after he wins say 1 chip then it's true that he will Quit a Winner with Probabilty One. However, QUITING A WINNER DOES NOT NECESSARILY MAKE HIM A WINNER. If he has to go into Debt for a Huge Amount of Money for a Long Period of time, you cannot count him a winner.

Consider this result of a Bet and tell me who do you think lost the Bet. As a result of the bet Mr. M loans Mr. C $1 million for 1 year. At the end of the year Mr. C must pay Mr. M back $1 million and $1. Did Mr. M win the bet because he gets an extra dollar back at the end of the year? Or did Mr. C win the bet?

PairTheBoard

[jason1990]

If you have infinitely many players, each repeatedly playing some game, then after the n-th trial (with probability one, no matter what n is), there will be infinitely many players who have not won yet.

But...if you look at any fixed player, then (with probability one) he will eventually win.

Sorry that I'm not explaining why, but I just wanted to point out something subtle which can cause confusion and apparent paradoxes.

[drudman]

[ QUOTE ]
drudman --
"Okay, so the pro-Martingalers are arguing that there cannot be an infinite series of losses.

The anti-Martingalers are arguing that there can be."

No. I think MMMMMMMMMMMM is getting off base with that argument. SomethingClever has it right. In the beginning MMMMMMMMM was talking about the case where the Martingaler continues playing forever. In that case the pro-Martingalers argue that he will just chalk up 1 chip after another as each Martingale series completes. The con-Martingalers argue that the Casino will chalk up stacks of chips after stacks of chips as one long streak after another happens, and that on average the Casino will be Ahead by more chips per unit time than the Martingaler.

But if you change this scenario to one where the Martingaler quits after he wins say 1 chip then it's true that he will Quit a Winner with Probabilty One. However, QUITING A WINNER DOES NOT NECESSARILY MAKE HIM A WINNER. If he has to go into Debt for a Huge Amount of Money for a Long Period of time, you cannot count him a winner.

Consider this result of a Bet and tell me who do you think lost the Bet. As a result of the bet Mr. M loans Mr. C $1 million for 1 year. At the end of the year Mr. C must pay Mr. M back $1 million and $1. Did Mr. M win the bet because he gets an extra dollar back at the end of the year? Or did Mr. C win the bet?

PairTheBoard

[/ QUOTE ]

Mr. M has net gain of $1, and Mr. C has a net loss of $1, and as such I would say that Mr. M is the winner.

What if I put it this way:

You put $1 mil in a savings account, and one year later withdraw it, plus interest. Have you not made (won) money?

[MMMMMM]

[ QUOTE ]
If you decide that you will have 1 stopping point and you have an infinite bankroll, it is almost infinitely probable that you will win.

Adding several other stopping points along the way will very slightly increase the chance that you will go on an infinitely long losing streak, but it's still almost infinitely improbable.

[/ QUOTE ]

Agreed. However that "almost infinitely improbable" corresponds to an "almost infinitely large" loss.

Likewise if you round it off as "zero chance" then you must round off the loss as "infinitely large". So at the hypothetical "zero chance point" (whether rounded or otherwise derived) you have ALREADY lost infinite money.

[MMMMMM]

[ QUOTE ]
If you have infinitely many players, each repeatedly playing some game, then after the n-th trial (with probability one, no matter what n is), there will be infinitely many players who have not won yet.

But...if you look at any fixed player, then (with probability one) he will eventually win.

[/ QUOTE ]

True one, or just rounded-off or assumed or otherwise derived one?

[Dov]

[ QUOTE ]
Special relativity and quantum mechanics as we understand them today state that it is impossible to retrieve any matter from a black hole. Ergo, your chips will be lost forever.

[/ QUOTE ]

I don't think this is true. I saw something a couple of weeks ago about how Steven Hawking is advancing a theory that black holes actually emit some kind of particle.

[MMMMMM]

[ QUOTE ]
drudman --
"Okay, so the pro-Martingalers are arguing that there cannot be an infinite series of losses.

The anti-Martingalers are arguing that there can be."

No. I think MMMMMMMMMMMM is getting off base with that argument.

[/ QUOTE ]

Maybe so; I worried about that; off main point at least. Yet can anyone tell me WHY (without just meaning 100% "for all practical intents and purposes")

[MMMMMM]

[ QUOTE ]
[ QUOTE ]

Special relativity and quantum mechanics as we understand them today state that it is impossible to retrieve any matter from a black hole. Ergo, your chips will be lost forever.



[/ QUOTE ]

I don't think this is true. I saw something a couple of weeks ago about how Steven Hawking is advancing a theory that black holes actually emit some kind of particle.


[/ QUOTE ]

Yes, but not chips.

[maurile]

[ QUOTE ]
[ QUOTE ]
Special relativity and quantum mechanics as we understand them today state that it is impossible to retrieve any matter from a black hole. Ergo, your chips will be lost forever.

[/ QUOTE ]

I don't think this is true. I saw something a couple of weeks ago about how Steven Hawking is advancing a theory that black holes actually emit some kind of particle.

[/ QUOTE ]
http://en.wikipedia.org/wiki/Black_hole_information_paradox

[Dov]

Thanks Maurile.

That was it.

[jason1990]

It is a mathematical fact (which can be proven) that those probabilities are exactly equal to the number 1.

So I guess that would be "true one."

[EliteNinja]

If I had an infinite bankroll, couldn't I just buy the casino and play for free?

The more important question is:
Why bother playing Roulette if one has an infinite bankroll?



If both the player and casino have infinite bankrolls, each finite win/loss no matter how large would be equivalent to zero. That means, neither the player nor the casino has an edge over each other because the bankrolls of either player will not change.

However..... if you could lay infinite wagers in an infinite time, the finite wins/losses would become infinite so that at the end of infinite time, the summation of the infinite finite losses by the player would be larger than the summation of infinite finite wins by the player. i.e. the player would have lost infinity.

Therefore, the final question is:

Does infinity (player bankroll) - infinity (player's losses at the end of infinity if the end of infinity exists) = 0 ?


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[Dov]

[ QUOTE ]
If both the player and casino have infinite bankrolls, each finite win/loss no matter how large would be equivalent to zero. That means, neither the player nor the casino has an edge over each other because the bankrolls of either player will not change.

[/ QUOTE ]

I mentioned this before, but got no response. What I had said was that in this case, an infinite bankroll defeats all edges.

[ QUOTE ]
Does infinity (player bankroll) - infinity (player's losses at the end of infinity if the end of infinity exists) = 0 ?

[/ QUOTE ]

I asked this too. In the form of:

Solve for x:

infinty - infinity = x

x = ?

[drudman]

[ QUOTE ]
If I had an infinite bankroll, couldn't I just buy the casino and play for free?

The more important question is:
Why bother playing Roulette if one has an infinite bankroll?



If both the player and casino have infinite bankrolls, each finite win/loss no matter how large would be equivalent to zero. That means, neither the player nor the casino has an edge over each other because the bankrolls of either player will not change.

However..... if you could lay infinite wagers in an infinite time, the finite wins/losses would become infinite so that at the end of infinite time, the summation of the infinite finite losses by the player would be larger than the summation of infinite finite wins by the player. i.e. the player would have lost infinity.

Therefore, the final question is:

Does infinity (player bankroll) - infinity (player's losses at the end of infinity if the end of infinity exists) = 0 ?


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[/ QUOTE ]

This is not the question at hand. It would be a very simple problem if it were simply which value approaches infinity faster, wins or losses. The question is, is it possible for the player to begin a series that will never end in a win. And the answer is no, it is not.

I'm officially no longer just playing Devil's Advocate.

[MMMMMM]

[ QUOTE ]
It is a mathematical fact (which can be proven) that those probabilities are exactly equal to the number 1.

So I guess that would be "true one."

[/ QUOTE ]

OK, well then: wouldn't the loss at the far convergence point also be infinite dollars?

As the probability of never getting a win is halved each time as you project the trials into the future, so too is the loss doubled with each projected trial. So if you extend the sequence out far enough to equate the probability of a win to 1.00, you are also enlarging the corresponding loss at that point to infinite money.

And that's with a fair coin, not a roulette wheel.

[drudman]

I really think everyone is starting to totally misunderstand the point.

Forget the betting. The betting doesn't matter. You are betting with funny money. It is just an inconvenience... irrelevant to your goal, which is to complete series. You always win when you complete a series, and every series completes NECESSARILY.

Stop thinking about each individual bet, and the terrifically mind-boggling exponentially large amounts of money that are lost on many of them. Your probability of eventually winning a bet, and therefore recouping ALL of that lost money is 1.

You can increase the house edge of the bet even. Play Pai Gow Poker if you like. Sic Bo. Let it Ride. Whatever your poison. So long as the house edge is less than 100%, your probability of completing a series is 1. You always complete a series. And you always win when you complete a series. Now that you are looking at it in terms of series, and not in terms of whatever individual bets comprise the series, you realize that you can never ever lose.

[MMMMMM]

[ QUOTE ]
The question is, is it possible for the player to begin a series that will never end in a win. And the answer is no, it is not.

[/ QUOTE ]

Firstly, that is not the main point of the thread.

Secondly, even if that is true, and even if it is a hinging point, the following is then also true (with a fair coin not a roulette wheel):

P= probability of no win by N trials

1 Loss in a row----&gt; P=1/2, financial loss = $1

3 Losses in a row--&gt; P=1/8, financial loss = $7

5 Losses in a row--&gt; P=1/32, financial loss = $31

Infinite losses in a row--&gt;P=Zero, financial loss = $Infinity

So the certainty point of which you speak also corresponds to losing infinite dollars.

With a fair coin.

Not a roulette wheel.

[MMMMMM]

[ QUOTE ]
I really think everyone is starting to totally misunderstand the point.

Forget the betting. The betting doesn't matter. You are betting with funny money. It is just an inconvenience... irrelevant to your goal, which is to complete series. You always win when you complete a series, and every series completes NECESSARILY.

Stop thinking about each individual bet, and the terrifically mind-boggling exponentially large amounts of money that are lost on many of them. Your probability of eventually winning a bet, and therefore recouping ALL of that lost money is 1.

You can increase the house edge of the bet even. Play Pai Gow Poker if you like. Sic Bo. Let it Ride. Whatever your poison. So long as the house edge is less than 100%, your probability of completing a series is 1. You always complete a series. And you always win when you complete a series. Now that you are looking at it in terms of series, and not in terms of whatever individual bets comprise the series, you realize that you can never ever lose.

[/ QUOTE ]

But to reach that certainty, you have to be willing to risk INFINITE money.

In other words, even if you have infinite money, you can't reach that CERTAINTY of which you speak, unless you are willing to risk losing it ALL.

Whether that is practically possible is not the issue. You are forced by the system to must a commensurate amount to the chance of failure or success of the entire series. And if you project enough trials in to make that chance of failure ZERO, the corresponding risk is your ENTIRE infinite bankroll.

You can't bet for eternity at ever-increasing amounts without risking an infinite bankroll. The one infinity is not greater than the other (and in this case they are highly correlated).

[drudman]

[ QUOTE ]
[ QUOTE ]
The question is, is it possible for the player to begin a series that will never end in a win. And the answer is no, it is not.

[/ QUOTE ]

Firstly, that is not the main point of the thread.

Secondly, even if that is true, and even if it is a hinging point, the following is then also true (with a fair coin not a roulette wheel):

P= probability of no win by N trials

1 Loss in a row----&gt; P=1/2, financial loss = $1

3 Losses in a row--&gt; P=1/8, financial loss = $7

5 Losses in a row--&gt; P=1/32, financial loss = $31

Infinite losses in a row--&gt;P=Zero, financial loss = $Infinity

So the certainty point of which you speak also corresponds to losing infinite dollars.

With a fair coin.

Not a roulette wheel.

[/ QUOTE ]

But when you reach the point of certainty, your loss is infinity dollars, and your win is infinity + 1 dollars.

[MMMMMM]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

The question is, is it possible for the player to begin a series that will never end in a win. And the answer is no, it is not.

[/ QUOTE ]



Firstly, that is not the main point of the thread.

Secondly, even if that is true, and even if it is a hinging point, the following is then also true (with a fair coin not a roulette wheel):

P= probability of no win by N trials

1 Loss in a row----&gt; P=1/2, financial loss = $1

3 Losses in a row--&gt; P=1/8, financial loss = $7

5 Losses in a row--&gt; P=1/32, financial loss = $31

Infinite losses in a row--&gt;P=Zero, financial loss = $Infinity

So the certainty point of which you speak also corresponds to losing infinite dollars.

With a fair coin.

Not a roulette wheel.

[/ QUOTE ]



But when you reach the point of certainty, your loss is infinity dollars, and your win is infinity + 1 dollars.

[/ QUOTE ]

Um, I don't think so. Because you haven't won it back yet. And you just bet infinity dollars AND LOST.

[maurile]

[ QUOTE ]
And you just bet infinity dollars AND LOST.

[/ QUOTE ]
That's got to hurt.

[PairTheBoard]

drudman --
"Mr. M has net gain of $1, and Mr. C has a net loss of $1, and as such I would say that Mr. M is the winner.

What if I put it this way:

You put $1 mil in a savings account, and one year later withdraw it, plus interest. Have you not made (won) money? "

You say Mr. M has a net gain and Mr. C has a net loss of a dollar. Yet you ignore your exact next point in that Mr. M has lost the 1 year's savings account interest on that $1 million and Mr. C has gained it, which amounts to much more than $1.

So now, who won and who lost that bet, and which side of it would you take; the gained $1 or the gained year's interest on the $1 million.

And the relation to the Martingaler is that although there is zero probabilty the Quiting Martingaler will not "win" $1, there is a non-zero Positive probability that he will run so bad doing it that it will amount to the scenario of giving up a year's interest on $1 million or worse.

PairTheBoard

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
drudman --
"Okay, so the pro-Martingalers are arguing that there cannot be an infinite series of losses.

The anti-Martingalers are arguing that there can be."

No. I think MMMMMMMMMMMM is getting off base with that argument.

[/ QUOTE ]

Maybe so; I worried about that; off main point at least. Yet can anyone tell me WHY (without just meaning 100% "for all practical intents and purposes")

[/ QUOTE ]

You know, MMMMMM, as I recall my studies it's kind of tricky handling infinite trials type scenarios. But I'm pretty sure you can say that when flipping coins, Heads will come up infinitely often. If a heads is required for the Martingaler's win, that should mean he will win with probabilty 1. I'm real fuzzy on the details for probabilties with infinite trials though. It's a lot easier to look at probability statements for finite trials.

PairTheBoard

[PairTheBoard]

EliteNinja --
"The more important question is:
Why bother playing Roulette if one has an infinite bankroll?"

I guess we're having fun doing it. We don't seem to care that it doesn't really impact our bankroll. We're just trying to figure out whether we're winning or losing if we do it.

PairTheBoard

[SheetWise]

[ QUOTE ]
In other words, even if you have infinite money, you can't reach that CERTAINTY of which you speak, unless you are willing to risk losing it ALL.

[/ QUOTE ]

I was just curious, how long of a losing streak would that take? /images/graemlins/wink.gif

[mosta]

you know they invented something called "math" for questions like this.

[Dov]

[ QUOTE ]
you know they invented something called "math" for questions like this.

[/ QUOTE ]

Why don't you enlighten us as to how you solve this problem mathematically?

[drudman]

[ QUOTE ]
drudman --
"Mr. M has net gain of $1, and Mr. C has a net loss of $1, and as such I would say that Mr. M is the winner.

What if I put it this way:

You put $1 mil in a savings account, and one year later withdraw it, plus interest. Have you not made (won) money? "

You say Mr. M has a net gain and Mr. C has a net loss of a dollar. Yet you ignore your exact next point in that Mr. M has lost the 1 year's savings account interest on that $1 million and Mr. C has gained it, which amounts to much more than $1.

So now, who won and who lost that bet, and which side of it would you take; the gained $1 or the gained year's interest on the $1 million.

And the relation to the Martingaler is that although there is zero probabilty the Quiting Martingaler will not "win" $1, there is a non-zero Positive probability that he will run so bad doing it that it will amount to the scenario of giving up a year's interest on $1 million or worse.

PairTheBoard

[/ QUOTE ]

What are you talking about? The situations are analagous. In both, the guy who puts his million on hold gets it and extra back at the end of the year. I think you have confused the analogies.

[drudman]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]

The question is, is it possible for the player to begin a series that will never end in a win. And the answer is no, it is not.

[/ QUOTE ]



Firstly, that is not the main point of the thread.

Secondly, even if that is true, and even if it is a hinging point, the following is then also true (with a fair coin not a roulette wheel):

P= probability of no win by N trials

1 Loss in a row----&gt; P=1/2, financial loss = $1

3 Losses in a row--&gt; P=1/8, financial loss = $7

5 Losses in a row--&gt; P=1/32, financial loss = $31

Infinite losses in a row--&gt;P=Zero, financial loss = $Infinity

So the certainty point of which you speak also corresponds to losing infinite dollars.

With a fair coin.

Not a roulette wheel.

[/ QUOTE ]



But when you reach the point of certainty, your loss is infinity dollars, and your win is infinity + 1 dollars.

[/ QUOTE ]

Um, I don't think so. Because you haven't won it back yet. And you just bet infinity dollars AND LOST.

[/ QUOTE ]

I'm at my limit of mathematical knowledge, so if you say that you can go bust with an infinite bankroll, I can't really argue.

[SheetWise]

I find Knights and Knaves logic to be a lot easier than Infinity-vs-Infinity.

Earlier I hoped to get a bite in looking at this as a finite situation, even if we had to use somewhat large numbers. While I know the value of running simulations, and I know the value of calculations -- I'm interested in what value you all think using 'infinity' in your hypothetical has.

I have seen the concept of infinity disprove Euclidian geometry -- and while I thought it was interesting, to me Euclidian geometry has bested infinity.

Do all of you believe that because something can occur that it eventually will?

[SomethingClever]

[ QUOTE ]
Infinite losses in a row--&gt;P=Zero, financial loss = $Infinity

[/ QUOTE ]

As my losses approached infinity, I would begin to believe that the casino is rigged.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
drudman --
"Mr. M has net gain of $1, and Mr. C has a net loss of $1, and as such I would say that Mr. M is the winner.

What if I put it this way:

You put $1 mil in a savings account, and one year later withdraw it, plus interest. Have you not made (won) money? "

You say Mr. M has a net gain and Mr. C has a net loss of a dollar. Yet you ignore your exact next point in that Mr. M has lost the 1 year's savings account interest on that $1 million and Mr. C has gained it, which amounts to much more than $1.

So now, who won and who lost that bet, and which side of it would you take; the gained $1 or the gained year's interest on the $1 million.

And the relation to the Martingaler is that although there is zero probabilty the Quiting Martingaler will not "win" $1, there is a non-zero Positive probability that he will run so bad doing it that it will amount to the scenario of giving up a year's interest on $1 million or worse.

PairTheBoard

[/ QUOTE ]

What are you talking about? The situations are analagous. In both, the guy who puts his million on hold gets it and extra back at the end of the year. I think you have confused the analogies.

[/ QUOTE ]

I think you're confused.

Case 1. M has $1 million. He could put it in a savings account for a year and earn say $20,000 interest. But instead, because he lost a bet, he has to let C use the money for a year. C pays M back $1 million plus $1. But C earns the $20,000 interest on the money. M gets back an extra dollar but loses $20,000 in interest. Clearly, M lost on that bet.

Case 2. M gambles coin flips with C. One flip every 10 days. M plays a martingale system, doubling up after each loss. M goes on a bad run and goes down over $1 billion before he hits a winner after about a years's play, and goes up by $1. In the mean time, C has put all M's losses during the year into short term interest bearing notes and has earned over $1 million in interest. M is up $1 from his "winning" Martingale system but is OUT over $1 million in interest he would have earned had he not gambled and "won".

While there is zero probabilty that M will not "win" his $1, there is a strictly postitive, non-zero probabilty that he will run as bad as described in Case 2 doing it. There is also a non-zero probabilty he will run a Trillion Times Worse.

If you insist on such a scenario really being a "win" for M then I don't see what more can be said to convince you. You just won't be convinced.

PairTheBoard

[MMMMMM]

[ QUOTE ]
I'm at my limit of mathematical knowledge, so if you say that you can go bust with an infinite bankroll, I can't really argue.

[/ QUOTE ]

If you can say that a non-zero probability (that of never winning a future bet) is actually a zero probability (by rounding-off, limits, or whatever) then you can't object to a limited but enormous bet being considered an "infinite" bet.

The two infinite amounts are directly correlated.

One is the non-zero probability of never winning another bet, which is doubled with every future spin.

The other is the bet size for that spin, which doubles with every roll.

If you consider the first to be, effectively, ZERO, you have to consider the second to be, effectively, INFINITE.

The two are exactly correlated (with a fair coin).

So advanced knowledge of math has nothing to do with it. If you assign a ZERO probability to an infinitesimally small probability, you have to assign an INFINITE size to the corresponding humungously large bet.

[drudman]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
drudman --
"Mr. M has net gain of $1, and Mr. C has a net loss of $1, and as such I would say that Mr. M is the winner.

What if I put it this way:

You put $1 mil in a savings account, and one year later withdraw it, plus interest. Have you not made (won) money? "

You say Mr. M has a net gain and Mr. C has a net loss of a dollar. Yet you ignore your exact next point in that Mr. M has lost the 1 year's savings account interest on that $1 million and Mr. C has gained it, which amounts to much more than $1.

So now, who won and who lost that bet, and which side of it would you take; the gained $1 or the gained year's interest on the $1 million.

And the relation to the Martingaler is that although there is zero probabilty the Quiting Martingaler will not "win" $1, there is a non-zero Positive probability that he will run so bad doing it that it will amount to the scenario of giving up a year's interest on $1 million or worse.

PairTheBoard

[/ QUOTE ]

What are you talking about? The situations are analagous. In both, the guy who puts his million on hold gets it and extra back at the end of the year. I think you have confused the analogies.

[/ QUOTE ]

I think you're confused.

Case 1. M has $1 million. He could put it in a savings account for a year and earn say $20,000 interest. But instead, because he lost a bet, he has to let C use the money for a year. C pays M back $1 million plus $1. But C earns the $20,000 interest on the money. M gets back an extra dollar but loses $20,000 in interest. Clearly, M lost on that bet.

Case 2. M gambles coin flips with C. One flip every 10 days. M plays a martingale system, doubling up after each loss. M goes on a bad run and goes down over $1 billion before he hits a winner after about a years's play, and goes up by $1. In the mean time, C has put all M's losses during the year into short term interest bearing notes and has earned over $1 million in interest. M is up $1 from his "winning" Martingale system but is OUT over $1 million in interest he would have earned had he not gambled and "won".

While there is zero probabilty that M will not "win" his $1, there is a strictly postitive, non-zero probabilty that he will run as bad as described in Case 2 doing it. There is also a non-zero probabilty he will run a Trillion Times Worse.

If you insist on such a scenario really being a "win" for M then I don't see what more can be said to convince you. You just won't be convinced.

PairTheBoard

[/ QUOTE ]

You are wholly ridiculous. At what point did this argument become a question of whether the Martingaler is losing money that he could have made by doing something else?

[MMMMMM]

[ QUOTE ]
Do all of you believe that because something can occur that it eventually will?

[/ QUOTE ]

No, because infinity does not imply infinite diversity.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
drudman --
"Mr. M has net gain of $1, and Mr. C has a net loss of $1, and as such I would say that Mr. M is the winner.

What if I put it this way:

You put $1 mil in a savings account, and one year later withdraw it, plus interest. Have you not made (won) money? "

You say Mr. M has a net gain and Mr. C has a net loss of a dollar. Yet you ignore your exact next point in that Mr. M has lost the 1 year's savings account interest on that $1 million and Mr. C has gained it, which amounts to much more than $1.

So now, who won and who lost that bet, and which side of it would you take; the gained $1 or the gained year's interest on the $1 million.

And the relation to the Martingaler is that although there is zero probabilty the Quiting Martingaler will not "win" $1, there is a non-zero Positive probability that he will run so bad doing it that it will amount to the scenario of giving up a year's interest on $1 million or worse.

PairTheBoard

[/ QUOTE ]

What are you talking about? The situations are analagous. In both, the guy who puts his million on hold gets it and extra back at the end of the year. I think you have confused the analogies.

[/ QUOTE ]

I think you're confused.

Case 1. M has $1 million. He could put it in a savings account for a year and earn say $20,000 interest. But instead, because he lost a bet, he has to let C use the money for a year. C pays M back $1 million plus $1. But C earns the $20,000 interest on the money. M gets back an extra dollar but loses $20,000 in interest. Clearly, M lost on that bet.

Case 2. M gambles coin flips with C. One flip every 10 days. M plays a martingale system, doubling up after each loss. M goes on a bad run and goes down over $1 billion before he hits a winner after about a years's play, and goes up by $1. In the mean time, C has put all M's losses during the year into short term interest bearing notes and has earned over $1 million in interest. M is up $1 from his "winning" Martingale system but is OUT over $1 million in interest he would have earned had he not gambled and "won".

While there is zero probabilty that M will not "win" his $1, there is a strictly postitive, non-zero probabilty that he will run as bad as described in Case 2 doing it. There is also a non-zero probabilty he will run a Trillion Times Worse.

If you insist on such a scenario really being a "win" for M then I don't see what more can be said to convince you. You just won't be convinced.

PairTheBoard

[/ QUOTE ]

You are wholly ridiculous. At what point did this argument become a question of whether the Martingaler is losing money that he could have made by doing something else?

[/ QUOTE ]

At what point? At the point of the first post of this thread actually. Do you agree or disagree with the original point MMMMMM was making IN THE CASE WHERE THE MARTINGALER DOESN'T QUIT? You've never expressed an opinion on that. The reason the non-Quiting Martingaler loses is that time spent being behind is just as important as time spent being ahead. That's what makes the Cosmic Casino a winner and the non-Quiting Martingalers as a whole, losers. The same principle applies to the quiting Martingaler. Time spent being behind matters.

PairTheBoard

[drudman]

The average of all of the non-quitting Martingaler's wins and losses is in the black.

His time spent down is unimportant. Like I have said, he is betting with funny money, because he can never run out of it, and because he will eventually win. Funny. Money.

[MMMMMM]

[ QUOTE ]
The average of all of the non-quitting Martingaler's wins and losses is in the black.

[/ QUOTE ]

This is most definitely wrong.

[PairTheBoard]

[ QUOTE ]
The average of all of the non-quitting Martingaler's wins and losses is in the black.

His time spent down is unimportant. Like I have said, he is betting with funny money, because he can never run out of it, and because he will eventually win. Funny. Money.

[/ QUOTE ]


We're now talking about the non-quiting Martingalers playing against the Casino's Roullette edge. The Casino also has unlimited funds.

Funny Money? If the money the Martingaler goes behind when on a losing streak is Funny Money, then the money he goes ahead after a win is Funny Money too, since he keeps on gambling and will eventually go behind once again.

Your viewpoint is Martingaler-Centric. The Casino's viewpoint is just as valid. When the Martingaler goes ahead the Casino doesn't care because it knows it will eventually go back ahead of the Martingaler again - and it will do so more often in #dollars than the Martingaler.

The key measure is (Time Spent Ahead)*(Dollars Ahead).

Run a simulation of 1,000,000 Martingalers playing the Casino and see what happens. You will see a net flow of funds from the combined bankrolls of the Martingalers into the Casino's coffers over time at a rate about equal to the House edge times the action. Do it one roullette spin per player at a time and check the flow of funds after each set of 1,000,000 spins.

PairTheBoard

[drudman]

Why? Every one win he has wipes out every n losses he has sustained, and then some. Every single win puts him back in the black, and the black gets bigger by one every time. He may have terrifically large funny money losses, but he will have one win even larger than all of the previous losses combined in every series. Every series has an average win/loss &gt; 0. The longer the series, the closer the average of that series approaches zero. But all of those positive values adds up. The average of them is positive and non-zero.

[drudman]

The casino's viewpoint is negative. They don't want the Martingaler to play. They know that no matter how bad they beat up on the Martingaler, they will always have to pay back every dime, plus 1 unit bet. They cannot avoid always having to pay up.

Look at it this way. The casino stacks up all of the money they take from the Martingaler on the left of the table. The Martingaler stacks up all of his unit bets won on the right. The stack on the right side of the table never ever decreases. It always increases. The stack on the left does decrease... in fact it decreases to zero frequently! The casino slides all of the cash back to the Martingaler's bankroll every time.

It works even if the casino doesn't pay off the wins using their stack of winnings, so long as you let the Martingaler stack up his huge payouts along with his unit bets.

Maybe you'd like to be the casino, Pairtheboard? I'll give you $10,000 at 9AM every day, and you can stack it up on your kitchen table. Then at 5PM every night, you give it back, plus a dollar. Would you want this deal? (don't try to say "well I can invest it and make more than $1 with it" because the casino does not invest money)

[PairTheBoard]

drudman --
"Look at it this way. The casino stacks up all of the money they take from the Martingaler on the left of the table. The Martingaler stacks up all of his unit bets won on the right. The stack on the right side of the table never ever decreases."

The stack on the right decreases every time the Martingaler loses a bet. If he loses a long streak of bets the stack goes into the red.

Run the simulation and see what happens.

PairTheBoard

[SheetWise]

So, 842 views and 155 posts later -- we have a starting point?

[MMMMMM]

If the house edge is zero (fair coin) then their joint expected loss or win is zero. If the house edge is 5.26% then they are losing more and more the longer they play. If you want to find the avcerage loss per gambler, multiply the total volume bet times the house edge and divide by the number of gamblers.

[drudman]

[ QUOTE ]
drudman --
"Look at it this way. The casino stacks up all of the money they take from the Martingaler on the left of the table. The Martingaler stacks up all of his unit bets won on the right. The stack on the right side of the table never ever decreases."

The stack on the right decreases every time the Martingaler loses a bet. If he loses a long streak of bets the stack goes into the red.

Run the simulation and see what happens.

PairTheBoard

[/ QUOTE ]

The Martingaler doesn't use the stack on the right to make his bets.

[drudman]

I'm all done here. You enjoy looking at those piles of chips you've won from the Martingaler's bankroll pile while they last... you will always push them all back, +1. I'll stack my +1s all the way to the ceiling and on into space.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
drudman --
"Look at it this way. The casino stacks up all of the money they take from the Martingaler on the left of the table. The Martingaler stacks up all of his unit bets won on the right. The stack on the right side of the table never ever decreases."

The stack on the right decreases every time the Martingaler loses a bet. If he loses a long streak of bets the stack goes into the red.

Run the simulation and see what happens.

PairTheBoard

[/ QUOTE ]

The Martingaler doesn't use the stack on the right to make his bets.

[/ QUOTE ]

I'm doubting you really believe what you're saying drudman. You must be getting a little embarressed when you start talking about M's loses not counting because they're "funny money", the Casino's viewpoint doesn't matter because it's "negative" and putting your winning chips on the right and not touching them while you lose with funny money chips on your left that don't matter. Any reasonable Jurist hearing arguments like that is going to throw the devil's case out of court.

Just run the simulation and see what actually happens.

PairTheBoard

[Aces McGee]

I think a big problem with this thread is that there's no logical time to "take accounting," because the game is endless. If accounting is done at the end of a series, the player will always be up at least one unit. If it's done in the middle of the series, the player may still be up overall, he may be down a couple of units, or he may be down hundreds or thousands or millions of units, depending on how long his losing streak lasts.

The problem, as I see it, is that there's no valid time to stop and evaluate who's ahead. The player is likely to be down more money and for much longer than he is ahead, but a game with no endpoint cannot have a winner or loser.

-McGee

[MMMMMM]

[ QUOTE ]
I'm doubting you really believe what you're saying drudman. You must be getting a little embarressed when you start talking about M's loses not counting because they're "funny money", the Casino's viewpoint doesn't matter because it's "negative" and putting your winning chips on the right and not touching them while you lose with funny money chips on your left that don't matter. Any reasonable Jurist hearing arguments like that is going to throw the devil's case out of court.

[/ QUOTE ]

Yeah, his viewpoint is COMPLETELY player-centric.

[ QUOTE ]
Just run the simulation and see what actually happens.


[/ QUOTE ]

I doubt he will. Sounds like he doesn't want to risk losing belief in the Martingale.

[PairTheBoard]

[ QUOTE ]
I think a big problem with this thread is that there's no logical time to "take accounting," because the game is endless. If accounting is done at the end of a series, the player will always be up at least one unit. If it's done in the middle of the series, the player may still be up overall, he may be down a couple of units, or he may be down hundreds or thousands or millions of units, depending on how long his losing streak lasts.

The problem, as I see it, is that there's no valid time to stop and evaluate who's ahead. The player is likely to be down more money and for much longer than he is ahead, but a game with no endpoint cannot have a winner or loser.

-McGee

[/ QUOTE ]

Take accounting the way it's always done. At the end of every roullette spin. With just one Martingaler it's hard to see how to measure all the ups and downs. But with a million Martingalers each getting one spin at a time, you can easily measure the net flow of funds after every set of million spins - one for each martingaler - to see the funds flowing out of the Martingaler's collective pockets and into the Casino Coffers. That's the original point of MMMMMM and it can be easily simulated for proof.

PairTheBoard

[Aces McGee]

Okay. But I believe I provided for that in my original response. Even though the casino is likely to be ahead more often than not, there's no way to say that the either the casino or the player has "beaten" the other, because the game is endless.

-McGee

[jason1990]

[ QUOTE ]
Take accounting the way it's always done. At the end of every roullette spin. With just one Martingaler it's hard to see how to measure all the ups and downs. But with a million Martingalers each getting one spin at a time, you can easily measure the net flow of funds after every set of million spins - one for each martingaler - to see the funds flowing out of the Martingaler's collective pockets and into the Casino Coffers. That's the original point of MMMMMM and it can be easily simulated for proof.

PairTheBoard

[/ QUOTE ]
A million martingalers is not enough. Even with a million, the casino's cashflow will continue to oscillate positive and negative and you will still have the issue of when to take accounting. With a million martingalers, there is a positive probability on each spin that all million will win. When that happens for the n-th time (and it will happen infinitely often if the game is played forever), then the casino will be down at least a million * n dollars.

By the way, as I said, there will be infinitely many spins in which all million martingalers win simultaneously. But I doubt you would ever see one of those in any simulation. The expected number of spins before you see that event is about 10^325,000.

[drudman]

The sim is flawed because you can't take accounting at any finite point, as per the OP. Martingaling works because a win is inevitable. In an infinite series, you will see all of the wins, and they will cancel out all of the losses and then some. The "then some" increases one by one into infinity.

[superleeds]

[ QUOTE ]
No, because infinity does not imply infinite diversity

[/ QUOTE ]

Yes it does. If there are infinate universes there are an infinte amount of superleeds's and an infinate amount of MMMMMM's and in some of those universes they are communicating via a poker chat site and in some of these discussions MMMMMM is arguing against superleeds's assertion that infinity does not imply infinte diversity and indeed everything inbetween. /images/graemlins/grin.gif

[MMMMMM]

[ QUOTE ]
[ QUOTE ]

No, because infinity does not imply infinite diversity

[/ QUOTE ]



Yes it does. If there are infinate universes there are an infinte amount of superleeds's and an infinate amount of MMMMMM's and in some of those universes they are communicating via a poker chat site and in some of these discussions MMMMMM is arguing against superleeds's assertion that infinity does not imply infinte diversity and indeed everything inbetween.

[/ QUOTE ]

Well I see you put a grin after the post, but I do not know whether you made it wholly, or just partially, in jest.

Anyway, infinity does not imply infinite diversity: a simple example is an infinity of rational numbers.

The series of rational numbers goes on forever, but there are also no irrational numbers anywhere in it to be found. It is an infinity of numbers, but it is not an infinitely diverse infinity of numbers.

If there exist infinite universes, there might be many more with MMMMMMes and Superleedses in them...or there might be just a few...or just one more such universe...or no other such universes. It cannot be presumed that MMMMMMes and Superleedses are prerequisite for universes.

[jason1990]

[ QUOTE ]
In an infinite series, you will see all of the wins, and they will cancel out all of the losses and then some.

[/ QUOTE ]
Or you could say,

[ QUOTE ]
In an infinite series, you will see all of the losses, and they will cancel out all of the wins and then some.

[/ QUOTE ]
Imagine you're flipping a coin for $1 a flip. It may not be as obvious here, but there will definitely come a time when you are $1 ahead. If you call that time the end of a "sequence," then you can watch your money grow $1 at a time and consider all your losses to be "funny money."

But that's just one perspective. The casino is free to regard the first time you're down $1 as the end of their "sequence." If they do that, then they're watching your money shrink $1 at a time and they consider all your wins to be "funny money."

[MMMMMM]

[ QUOTE ]
The sim is flawed because you can't take accounting at any finite point, as per the OP. Martingaling works because a win is inevitable. In an infinite series, you will see all of the wins, and they will cancel out all of the losses and then some. The "then some" increases one by one into infinity.

[/ QUOTE ]

It is a strange form of accounting which counts only the peaks and not the valleys. More especially so because the valleys gradually get ever deeper and deeper.

Picture a wave oscillator chart, fluctuating above and below a fixed line of zero. The first win puts the wave 1 above the zero line. A dip will put it below.

As time goes on, you would observe some higher highs and lower lows. This is virtually inevitable, and this trend would continue. But the lower lows would slowly get greater and greater.

From the casino's perspecive, those lower lows (for the gambler) are mathematically expected.

So let's say one way of accounting would be: score the higher highs and lower lows ONLY. They both should occur. Then average the higher highs and lower lows, because you really can't fairly count ONLY the highs, any more than the casino can fairly count ONLY the lows.

I'm not saying the above is the best way of scoring. It's not. But it is better than your completely lopsided way of scoring by counting only the highs which is 100% player-centric. And it would demonstrate that the player on average is getting further and further behind.

Look at it this way too: If you can count only the highs, the casino can count only the lows. So by that method, whichever side has the Cosmic Accountant in its employ is the side that is winning. Or better yet, give them both an Accountant trained in this manner of accounting, so they can both be winning! Does that really make any sense at all?

[MMMMMM]

haha, you said what I said, but in a third as many words or less;-)

[superleeds]

[ QUOTE ]
Well I see you put a grin after the post, but I do not know whether you made it wholly, or just partially, in jest.

[/ QUOTE ]

Partially I guess, because I do think your wrong.

[ QUOTE ]
Anyway, infinity does not imply infinite diversity: a simple example is an infinity of rational numbers.

The series of rational numbers goes on forever, but there are also no irrational numbers anywhere in it to be found. It is an infinity of numbers, but it is not an infinitely diverse infinity of numbers.

[/ QUOTE ]

But exceptions do not mean implications do not exist. If you take the sq. root of the series of infinate rational numbers you have a infinate series of rational and irrational numbers so using your example I can make irrational numbers appear using a simple step.

[ QUOTE ]
If there exist infinite universes, there might be many more with MMMMMMes and Superleedses in them...or there might be just a few...or just one more such universe...or no other such universes. It cannot be presumed that MMMMMMes and Superleedses are prerequisite for universes

[/ QUOTE ]

There is no might about it. If there are infinate universes there are infinate us'es (sp?) simply because I would never be able to search every universe without an infinate number still to be screened.

[SheetWise]

MMMMMM-

You have to remeber that the Martingaler never has any losses. Losing hands are simply an investment that gets you one step closer to a win. I would carry them on the balance sheet as an asset -- perhaps as a prepaid expense, or maybe an account receivable. /images/graemlins/wink.gif

[MMMMMM]

My limited and basic recall:

Two infinities are of the same order or magnitude if their elements can be matched in a one-to-one correspondence.

For instance, the set of whole numbers is an infinity of the same magnitude as the set of all even numbers, because you can match the first, second, third, and fourth elements as follows: 1 matches 2, 2 matches 4, 3 matches 6, 4 matches 8...999 matches 1998, etc.....and in fact you can match all elements of each set in that fashion. Mathematically speaking, the two infinities are therefore of the same magnitude.

The set of all integers also is of the same magnitude as the set of all positive integers. See above; same reason.

However there are infinities that cannot be matched on a one-to-one basis with lower infinities. An example is the set of real numbers to the set of integers. The set of real numbers includes irrational numbers and therefore has elements that cannot be matched on a one-to-one basis with any corresponding integers. Hence the set of real numbers is of a higher magnitude than the set of integers. In a very true mathematical sense, it is a "greater" sort of infinity.

There are sets of infinities that go beyond this. I seem to recall that the next higher infinity is the set of all possible curves (including all crazy squiggly lines, etc.)

The possible magnitudes of infinities are also infinite, and are ordered with a symbol called the Aleph..

A good book on this which I only glanced at years ago is: Infinity And The Mind, by Rudy Rucker.

Anyway, infinite universes may suggest the *possibility* of infinite diversity, but infinite universes wouldn't necessarily mean that everything that COULD POSSIBLY exist, DOES exist.

Heck, you could even have infinite universes all identical to each other...in which case they would each contain exactly one MMMMMM and one Superleeds;-))

Someone more versed in the mathematics of infinities could undoubtedly explain all of this better than I can.

[MMMMMM]

[ QUOTE ]
You have to remeber that the Martingaler never has any losses. Losing hands are simply an investment that gets you one step closer to a win.

[/ QUOTE ]

Ah, but that's just it: they DON'T get you one step closer to a win. The roulette wheel has no memory.

/images/graemlins/grin.gif

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
Take accounting the way it's always done. At the end of every roullette spin. With just one Martingaler it's hard to see how to measure all the ups and downs. But with a million Martingalers each getting one spin at a time, you can easily measure the net flow of funds after every set of million spins - one for each martingaler - to see the funds flowing out of the Martingaler's collective pockets and into the Casino Coffers. That's the original point of MMMMMM and it can be easily simulated for proof.

PairTheBoard

[/ QUOTE ]
A million martingalers is not enough. Even with a million, the casino's cashflow will continue to oscillate positive and negative and you will still have the issue of when to take accounting. With a million martingalers, there is a positive probability on each spin that all million will win. When that happens for the n-th time (and it will happen infinitely often if the game is played forever), then the casino will be down at least a million * n dollars.

By the way, as I said, there will be infinitely many spins in which all million martingalers win simultaneously. But I doubt you would ever see one of those in any simulation. The expected number of spins before you see that event is about 10^325,000.

[/ QUOTE ]

This is true. Although I don't think the question of when to take an accounting should be an issue. Take an accounting after every spin.

With ever increasing bet sizes over time the 1,000,000 Martingaler Results get to look more like a single Martingaler's Results with Huge Singleton Bets swamping the smoothing out effect of the numerous Martingalers. You might keep the results smoother by increasing the pool of Martingalers at play by a certain percent after each spin. Maybe this would help in looking for some kind of rigourous statement for what happens as N goes to infinity.

If bet sizes were fixed we could say that after N trials there is a certain probabilty a gambler will be ahead playing against a certain house edge. It would be nice if we could formulate a similiar statement for a properly expanding pool of Martingalers.


PairTheBoard

[mosta]

[ QUOTE ]
[ QUOTE ]
you know they invented something called "math" for questions like this.

[/ QUOTE ]

Why don't you enlighten us as to how you solve this problem mathematically?

[/ QUOTE ]

this topic comes up regularly and in the past advanced students of math [not me] have discussed (and proved!) various ambiguous issues. this thread is so far from anything logically coherent that it's best hope for any kind of value is only as amusement--particularly considering that the OP has in the past retorted to his critics that he has a near 200 IQ! here, I'll give you a little taste: if 1&gt;=P(x)&gt;0 then lim as n-&gt;inf of (1-P)^n= ZERO. I'll leave it to the high school students to explain that. if someone ever formulates a logical statement (about means or expected values) in this thread, the high schoolers can work on that too...

[mosta]

here's an eg. I post at the end just to cite/paste an email from a mathematician friend. some of the discussion gets highly techincal, but if you focus you can keep track of the central issues:

archive thread (http://archiveserver.twoplustwo.com/showflat.php?Cat=&amp;Number=556731&amp;page=&amp;view=&amp;sb=5&amp;o =&amp;fpart=all&amp;vc=1)

edit: you can make various statments that intuitively sound equivalent (eg: 1. P of eventual succes = 1 (yes) 2. EV &gt; 0 (no--it is undefined!)), but when you do some math instead of talking out your a$$, you get some results, some surprising, some confusing, some not.

[Dov]

What was wrong with this post by Pzhon?

[ QUOTE ]
Here is a proof I think is more intuitive, with more details in white:

Let p(n) be the probability you have never been ahead before the nth wager.

The expected first time to be ahead is the sum of the p(n).
In the usual sum for the expected value, change i to the sum from n=1 to i of 1. Then change the order of summation.
p(2k-1)=p(2k)= (2k C k)/2^(2k) ~ 1/sqrt(pi k)
Use the reflection principle to write the number of lattice paths to (a,b) staying above the diagonal as a difference between binomial coefficients. The sum telescopes. Then use Stirling's approximation for n!.
The sum of c/sqrt(n) diverges, so the expected number of turns before your first win is infinite.

[/ QUOTE ]

[mosta]

I shouldn't have sounded like I was criticizing every post in the thread. it's certain posters in particular that always impress me in a certain way...

[mosta]

is your claim that EV is &lt; 0? skimming the thread I didn't notice any mathematical statements from you. I'd LOVE to see you try to _prove_ this (ie, not with analogies and word pictures). do you do any proofs on near 200 score IQ tests?? do you know what a proof is??

[Dov]

Thanks Mosta,

This is a good thread.

I wish you had linked it a couple of days ago.

[PairTheBoard]

[ QUOTE ]
here's an eg. I post at the end just to cite/paste an email from a mathematician friend. some of the discussion gets highly techincal, but if you focus you can keep track of the central issues:

archive thread (http://archiveserver.twoplustwo.com/showflat.php?Cat=&amp;Number=556731&amp;page=&amp;view=&amp;sb=5&amp;o =&amp;fpart=all&amp;vc=1)

edit: you can make various statments that intuitively sound equivalent (eg: 1. P of eventual succes = 1 (yes) 2. EV &gt; 0 (no--it is undefined!)), but when you do some math instead of talking out your a$$, you get some results, some surprising, some confusing, some not.

[/ QUOTE ]

Here is the last post in that archive thread. As it happens, the post is by mosta:

mosta --
"after consulting with my resident expert, I'm going to vote--Henke 1, bigpooch 0, on split decision: I think it's incorrect that EV=1, but no one says why saying that roulette trick works has to mean that EV&gt;0. good bout all around!.

comments quoted/ lifted from personal email:

Q: Does there exist a sequence of random variables Y_n such that
i) Y_n is almost surely eventually 1, and
ii) E(Y_n) is negative for all n?

["Almost surely" is a technical term meaning "with probability 1".]

The answer is yes, and winnings from the doubling strategy (p individual win
&lt; 1/2) is an example. The first post in the string proved i) and Henke's
first post proved ii). [Well, both of them close enough for government work,
anyway -- they're right in spirit at least].


----------


bigpooch started off on the wrong foot when he tried to prove i) implies
E(Y_n) -&gt; 1 as n gets large:
"Then one can see that
E(F(tau)) = 1 or lim (n --&gt; infinity) E(F(n)) --&gt; 1. "
This is just false --- he's confusing the process at the stopping time
F(tau) [about which he has made a true statement] with the unstopped process
F(n).

-------


There were several side issues, but nobody asked the real question. Why is
ii) so bad for gamblers? Note Henke's first comment:
"You only proved that the probability of at least one win approaches one as
the number of trials approaches infinity. However, to state that the
roulette trick works, I think you would have to prove that it's +EV which
isn't the same thing."
He's absolutely right "[it] isn't the same thing", but why should "roulette
trick works" mean "+EV"? Unfortunately, all subsequent debate is focused on
attacking the part he was right about.

------


Side issues: Henke's distinction between being zero and approaching zero is
not meaningful, but he is right when he says you need further assumptions to
make sense of 0*infinity. bigpooch was pretty careless to let the 0*infinity
genie out of the bottle.

------


Haha, I just read some more of the thread, where bigpooch makes this
seemingly ironclad claim:

E = +$1 x P(tau=1)
+$1 x P(tau=2)
+$1 x P(tau=3)
+...

and then one post later accuses Henke of the mistake _he_ just made:

"Okay I understand your mistake in thinking: going to the
infinite case is not necessarily the same as taking the
case for finite n and taking the limit as n approaches
infinity."

Elaboration: If Y_n is our net change in wealth after playing the nth game,
then bigpooch has implicitly assumed that there is a limiting random variable
Y = lim Y_n as n gets large and that E(Y) = lim E(Y_n). [You can check (as
Henke did) that E(Y_n) is getting large and negative with n (if p&lt;1/2), so
this equality of expectations is definitely violated.] If bigpooch were able
to make sense of the Y = lim Y_n statement (he would say "pointwise a.e."),
then indeed Y would be the constant 1, but he would not be able to obtain
the limit in a way necessary to justify the E(Y) = lim E(Y_n) statement (he
would have to say "in L^1")."
=============================
=============================



It seems to me that drudman's assertion that he will just pile up win after win as each martingale series ends, is countered by the simple observation that the expected value of his net gain/loss after n spins, En, is negative and the limit of En as n goes to infinity is negative infinity. It looks like mosta is saying this is true Even if the Martingaler stops after one win. I was not sure about this, although I was convinced it was true in the MMMMM model where the Martingaler plays on forever.

I also think that En can be seen at work in the model of a properly expanding pool of Martingalers at play.

I think in order to really find satisfaction with this thing we need more statements about the probabilty space of infinite sequences of wins-losses when coupled with the varying bet amounts. People want to know what happens "in the end" which involves the infinite sequences in this case. I understand identifying the sequences with binary expansions of real numbers and using the lesbegue measure on the space, although I'm not exactly sure why that should be the obviously correct way to do it. Maybe that yields nothing of particualar interest though, because the sequences almost surely cannot be identified with specific dollar results. Or can they?



PairTheBoard

[MMMMMM]

[ QUOTE ]
is your claim that EV is &lt; 0? skimming the thread I didn't notice any mathematical statements from you. I'd LOVE to see you try to _prove_ this (ie, not with analogies and word pictures). do you do any proofs on near 200 score IQ tests?? do you know what a proof is??

[/ QUOTE ]

Are you addressing me?

Of course the gambler's EV is less than zero on a roulette wheel, using the Martingale or any other system.

EV = total volume bet * advantage (or disadvantage). In this case the advantage for the gambler is negative on every spin (positive for the house), so no matter how many spins he might make (&gt;1) his EV is &lt;0. The house edge on a roulette wheel is fixed somewhere around 5.25% as I recall. So his EV would be -5.25% * total amounts bet.

Also, I never claimed an IQ near 200.

I do sense a hostile tone in your post. Why?

[MMMMMM]

So, Mosta, how does this point that impress you:

As the projected chance of eventual success doubles with each projected future trial, the amount that must be wagered at that point doubles also. So if you are going to use limits (or any other principle) to assert that the chance the gambler will lose a future bet is zero, you must also use the same principle to assert that the gambler has to be able to wager his entire infinite bankroll on one spin.

You can't claim that halving forever is equivalent to zero, unless you agree that doubling forever is equivalent to infinity.

[MMMMMM]

It is also a mathematical fact that NO betting system can overcome the house advantage when utilized against an independent-trials-based game with a built-in house advantage on every trial. The house extracts an EV of the house edge times the volume bet.

I will not attempt to PROVE it, but any mathematician will tell you that that is so.

It should also be obvious that it is so: because the EV for the gambler is negative on every single roll.

Adding a string of negative numbers cannot produce a positive number. Not a formal proof, but pretty basic, and clearly true.

[MMMMMM]

All gambling games have an EV of: Advantage * TotalBet.

The gambler has a disadvantage against the roulette wheel ON EVERY SINGLE WAGER.

You can't sum a string of negative expectations and obtain a positive expectation. You can't sum a string of negative numbers and obtain a positive result.

Hence those who think an attempted Martingale series has an EV of 1 are wrong.

This has nothing to do with bankroll limits. It has everything to do with the fact that no amount of negatives, summed, produces a positive.

Very simple when looked at in this way.

[Dov]

[ QUOTE ]
All gambling games have an EV of: Advantage * TotalBet.

[/ QUOTE ]

Neither of these numbers are negative. Therefore this will not be a negative number, ever. In Martin's case, (Advantage * Bet) - Bet will be negative, but that's not how we calculate EV as you pointed out.

[ QUOTE ]
Hence those who think an attempted Martingale series has an EV of 1 are wrong.

This has nothing to do with bankroll limits. It has everything to do with the fact that no amount of negatives, summed, produces a positive.

Very simple when looked at in this way.

[/ QUOTE ]

I like the way you tried to simplify it, but I don't think it works. You are not, in fact, adding negative numbers, you are adding numbers that are smaller than your initial wager until the win (I know, I'm agreeing with you) which is greater than the sum of the losses by 1. The net effect is that you are adding 1/n trials where n is the number of trials in the series.

That gives me an idea. How's this for a rudimentary proof?

Since the house advantage is fixed, it is impossible for it to increase or decrease on any given spin of the wheel.

Since the Martingaler's 'advantage' per spin = 1/n where n is the number of trials to complete a series, his advantage fluctuates with each series.

If the Martingaler's advantage changes, then the house edge cannot be fixed.


Example:

Lose - Win

If he wagers $1 on the first bet and loses, then wagers $2 on the second bet and wins, then Martin will be up $1 for 2 spins or $.50 / spin.

Lose - Lose - Win

-$1, -$2, +$4 = +$1 over 3 spins = .33 / spin

It feels like there's something wrong with this, but I'm not sure what it is.

I should go back to sleep.

Any good?

[MMMMMM]

Hmm, Dov, not sure if I made myself clear or if I followed your point entirely, but quickly:

The house's advantage is fixed on every spin of the roulette wheel. So too is the Martingaler's disadvantage.

If the house has an positive expectation of 5.25% on every spin, then the Martingaler has a negative expectation of 5.25% on every spin. Neither party's advantage or disadvantage on each spin fluctuates according to the bet sizing or patterns or anything else. It is simply fixed for each trial.

A string of negatives cannot be summed into a positive.

[Dov]

I followed you pretty well.

I was just trying to see if you can prove that since Martin's advantage per spin seems to change and we know that it can't that you can't say that his EV +1/n where n is the number of trials in the series.

Just taking a shot at my first ever proof.

LOL

[superleeds]

[ QUOTE ]
The house's advantage is fixed on every spin of the roulette wheel. So too is the Martingaler's disadvantage.

[/ QUOTE ]

The Martingaler's changes every spin given that with an infinate number of rolls he will win. His expected win every time he starts a series is at least 1/infinity per roll

[MMMMMM]

[ QUOTE ]
[ QUOTE ]
The house's advantage is fixed on every spin of the roulette wheel. So too is the Martingaler's disadvantage.

[/ QUOTE ]

The Martingaler's changes every spin given that with an infinate number of rolls he will win. His expected win every time he starts a series is at least 1/infinity per roll

[/ QUOTE ]

His mathematical disadvantage does NOT change every spin...or on any spin.

This is the most basic and incontrovertible part of the whole scenario.

[SheetWise]

In this hypothetical, I think you are wrong to look at the EV. One of the things you have to remember is that the casino realizes its expectation when you win.

Consider a roulette wheel that pays 35:1 for a win, while the true payoff should be 37:1. Thet make no money when you lose. They could have been paying you 40:1 -- it doesn't matter -- you lost. It's when you win, and they pay you at less than true odds that they realize their advantage.

If the player wins every time, they will never realize their advantage, because they are taking it out of your winnings.

How's that for a twist?

[superleeds]

[ QUOTE ]
His mathematical disadvantage does NOT change every spin...or on any spin

[/ QUOTE ]

Agreed. But he compansates for this.

[MMMMMM]

[ QUOTE ]
[ QUOTE ]
His mathematical disadvantage does NOT change every spin...or on any spin

[/ QUOTE ]



Agreed. But he compansates for this.

[/ QUOTE ]

So if what you believe is true, we have a paradox.

If he truly compensates for it, he is essentially causing a series of negative values to sum to a positive value.

But...even an infinite number of negative values cannot sum to a positive value.

So, which is wrong? For reasons outlined elsewhere, I believe he actually doesn't "compensate for this."

[superleeds]

[ QUOTE ]
So if what you believe is true, we have a paradox

[/ QUOTE ]

I don't believe so. Whilst each individual roll has a negative expected value, if at some point a win will be acheived (which infinity allows) and a compenasating factor (the martingale system) has been employed
each roll now has a positive expected value.

[Dov]

[ QUOTE ]
If he truly compensates for it, he is essentially causing a series of negative values to sum to a positive value.

[/ QUOTE ]

You can multiply by -1 to change the negative to a positive.

Maybe that's what happens when he wins. Maybe a win = (Loss incurred so far(-1)) + 1

Nah, that doesn't make any sense.

Oh shoot, I just got sucked out on in my game. That decides it then, Martin is definitely -EV. /images/graemlins/ooo.gif

[Dov]

[ QUOTE ]
Agreed. But he compansates for this.

[/ QUOTE ]

He doesn't really compensate for it.

The logic would be that he is not getting paid enough when he wins, therefore he is losing. Instead of being up 1 unit, he should be up much more than that.

[superleeds]

[ QUOTE ]
He doesn't really compensate for it

[/ QUOTE ]

Yes he does. He knows that eventually he will win 1 unit if he doubles his bet everytime he loses. He has an infinate amount of money so that is no problem, he has an infinate amount of times he is able to make this bet, therefore he is compensating for the times he loses.

[superleeds]

I think we are just looking at it from opposite but equally valid view points. Mine is that in any series he will eventually win if he can go on indefinately, and yours is that allowed an infinate number of series he will eventually hit a series where he never wins. I think both are true. Maybe thats the paradox /images/graemlins/grin.gif

[MMMMMM]

[ QUOTE ]
He doesn't really compensate for it

[/ QUOTE ] [ QUOTE ]

Yes he does. He knows that eventually he will win 1 unit if he doubles his bet everytime he loses. He has an infinate amount of money so that is no problem, he has an infinate amount of times he is able to make this bet, therefore he is compensating for the times he loses.

[/ QUOTE ]

OK, let's try an example which I believe should show the fallacy of this notion.

Instead of doubling up, and only after losses, let's say the gambler increases his bet by a factor of ten times ON EVERY SINGLE BET. Now, his net result will fluctuate wildly, both in the plus and in the minus columns. In fact, whether he is ahead or behind will be entirely dependent on his last result ONLY.

So, you may argue that whenever he falls behind he is sure to pull ahead in the future--as others have argued in the Martingale example. By the same argument, however, our ever-increasing-ten-times bettor is sure to not only pull ahead at some future point, but also to fall behind at some point. Yet by your logic he is "compensating" for having a negative advantage on every spin, by virtue of sometimes pulling ahead. But this cannot be true, because in this example, you can flip it perfectly upside down and say the same thing from the casino's standpoint. And now it IS the exact same thing (since the bet pattern does not alter based on wins or losses)--except for one little detail: the house edge.

So in the above example your premises and argument are used but the conclusion is clearly false. The bettor is sure to jump ahead AND behind on future spins. That however in no way allows you to take accounting only when he is ahead, or to claim that such a practice compensates for the house advantage.

By this betting pattern, whenever he falls ahead or behind it about is nine times greater than the sum of all his previous results. And he is expected to do so frequently--ON BOTH SIDES.

So merely claiming that because he has to eventually win (or lose) is irrelevant to what his expected value is, or whether he is "compensating" for the house edge. As above, the casino could simply employ the mirror image bookkeeping system. It's purely a fallacy.

[superleeds]

If at some point in the series he wins and that puts in the positive moneywise he his ahead. Yes?

This will always be true regardless of odds. As long as he bets enough to cover previous losses and make a profit should he win the current roll. Yes?

[BluffTHIS!]

[ QUOTE ]
I think we are just looking at it from opposite but equally valid view points. Mine is that in any series he will eventually win if he can go on indefinately, and yours is that allowed an infinate number of series he will eventually hit a series where he never wins. I think both are true. Maybe thats the paradox /images/graemlins/grin.gif

[/ QUOTE ]

The paradox is how this stupid thread can go on for so long. Since it's only purpose is to analyze gambling situations, it can only have a point if gambling situations can be found that accomodate the the assumptions made here, namely allowing such infinite series, an infinite bankroll and no house limit (and also possibly that the player would not quit at some predetermined level of profit). Even if you assume Bill Gates would want to play a martingale, find someone willing to fade his action without limit or discussing this serves no purpose at all.

[MMMMMM]

[ QUOTE ]
If at some point in the series he wins and that puts in the positive moneywise he his ahead. Yes?

This will always be true regardless of odds. As long as he bets enough to cover previous losses and make a profit should he win the current roll. Yes?


[/ QUOTE ]

What the heck does that have to do with his Expected Value?

He isn't COMPENSATING for anything; you're just applying selective accounting.

And why don't you just flip it, as I suggested? As long as he bets enough on every roll to make that roll bigger than all his prior rolls, he will continue toroll both wins and losses, and keep having higher highs and lower lows. You can't take accounting only at the positive peaks, just as the casino can't take accounting only at the negative valleys.

So the fact that he is expected to have higher highs is IRRELEVANT, because he is ALSO expected to have lower lows.

[PairTheBoard]

BluffTHIS --
"discussing this serves no purpose at all."

Hell, we could probably spend another 20 or 30 pages just talking about THAT.

PairTheBoard

[GrekeHaus]

I'm jumping in here a bit late, and I'm not going to read the whole thing, but from looking at page 1 and the last page, it seems that really no progress has been made. So I will offer a more mathematical analysis of the problem.

If that gambler has a finite ammount of time, or a finite bankroll, then it's trivial to show that the game is a losing game. The problem people seem to be having here is that they're still thinking about the problem in finite terms.

Supposed we've concluded n trials. We know that probability of losing every trial n+1,n+2,... is strictly 0. Therefore, we know with probability 1 that there exists some N&gt;n such that our player is ahead after N trials!

Similarly, since the game is losing we know with probability 1 that there is an M&gt;n such that our player is behind after M trials.

The problem that we run into is that in order for this to be a losing system, we'd have to know that

lim n-&gt;inf (the probability that there exists an M&gt;n such that our player is ahead after n trials) = 0

But this loosely defined function doesn't converge to anything so this is not the case. Similarly, you can't show that this is a winning system in this way.

However, if you define 1 trial to mean betting until you win, then the probability that you'll be ahead after any number of trials is 1, so it is a winning system. But you could also have the player keep spinning until they were down one unit and count that as a trial. Then it's a winning system.

It's kind of like this series:

(1-1) + (2-2) + (3-3) = 0+0+0+..=0

but we could also rewrite it as

1 + (-1+2) + (-2+3)+... = 1+1+1+....=inf

Depending on how you look at the problem, you can view it in a number of different ways, but what it really comes down to is that the series is non-convergent, which is why there is so much disagreement.

[GFunk911]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The question is, is it possible for the player to begin a series that will never end in a win. And the answer is no, it is not.

[/ QUOTE ]

Firstly, that is not the main point of the thread.

Secondly, even if that is true, and even if it is a hinging point, the following is then also true (with a fair coin not a roulette wheel):

P= probability of no win by N trials

1 Loss in a row----&gt; P=1/2, financial loss = $1

3 Losses in a row--&gt; P=1/8, financial loss = $7

5 Losses in a row--&gt; P=1/32, financial loss = $31

Infinite losses in a row--&gt;P=Zero, financial loss = $Infinity

So the certainty point of which you speak also corresponds to losing infinite dollars.

With a fair coin.

Not a roulette wheel.

[/ QUOTE ]

But when you reach the point of certainty, your loss is infinity dollars, and your win is infinity + 1 dollars.

[/ QUOTE ]

No it's not. Your loss is infinity and your win is $1, not infinity+$1

[GFunk911]

First off, let me state for the record that I hate this thread and everyone in it, no exceptions, and that includes myself.

The problem is looking at a situation with certain elements infinite and other elements finite. It is nonsensical.

Let's look at the avg casino win on any iteration (i.e. one roulette spin) for an infinite room of infinitely bankrolled martindalers, on a per person basis, for spin N (N is zero based)

sum{x:0-&gt;N}[ (W%*(1-W%)^x) * (2^x) ] * (1-2W%)
(W%*(1-W%)^x) = % of players who have lost x spins in a row
(2^x) = amount wagered by player who has lost x spins in a row
(1-2W%) = house edge

An infinite room of martindalers will lose money on every spin.

This won't convince anyone cause the premise is nonsensical and the actors are zealots, but there it is. I was gonna type more but I can't devote any more time to this swamp or horror.

[PairTheBoard]

I think everyone is in agreement that the Martingaler goes ahead and goes behind infinitely often with probabilty 1. I think it's also true that the martingaler spends more time being ahead than being behind. In fact, I think it can be shown that if A(n)== The number of spins the Martingaler was ahead after n spins, and B(n)== The number of spins the Martingaler was behind after n spins; Then B(n)/A(n) goes to zero as n goes to infinity with probabilty 1. However, this does not take into account the amount by which the Martingaler sometimes falls behind. I think that if we define wB(n)== The sum over all spins where he's behind of the Amounts he's behind at each spin, and wA(n) similiarly for the spins where he's ahead over n spins; Then wB(n)/wA(n) goes to -infinity as n goes to infinity with probabilty 1. Because of the House Edge.

Furthermore, if the Cosmic Casino is set up with a properly expanding pool of Martingalers, and we let T(n)== The amount the total pool of Martingalers is ahead after n spins, then P[ T(n)&gt;0 ] goes to zero as n goes to infinity - where P[ T(n)&gt;0 ] is the probabilty that T(n)&gt;0.

I think these things can all be rigourously proven. I leave that to the working mathematicians here. However, even if they are proven there will still be endless debate as to what to make of it.

PairTheBoard

[Komodo]

</font><blockquote><font class="small">Svar till:</font><hr />
The sim is flawed because you can't take accounting at any finite point, as per the OP. Martingaling works because a win is inevitable. In an infinite series, you will see all of the wins, and they will cancel out all of the losses and then some. The "then some" increases one by one into infinity.

[/ QUOTE ]

That is exactly here where many peoples thinking is flawed. The win isnt inevitable. In an infinite serie, the probability will be closer and closer to 100%, but never 100. The assumption that every martingaler will be able to complete his serie is matchematically impossible. It will be 99,99999999999999999999999999 etc etc but never 100%.

[ThinkQuick]

[ QUOTE ]
[ QUOTE ]
Your old bankroll was infinite.

[/ QUOTE ]

So why were you playing in the first place?


[/ QUOTE ]

Just funny again