Why doesn't the old roulette trick work?

[Morbo]

I'm sure many has heard of the old roulette strategy of betting, say $1 on red. If you lose, you double your bet and bet again. If you lose again, you double again and bet $4. You just keep doubling until you hit and then you have netted a $1 gain.

I can see that the 0 and 00 toss off the 50/50 chance but no matter how I look at it I can't see why this would not be a legit strategy as long as you have the bankroll to support the huge amount of doubling that you eventually will run into.

Any comments?

[fluff]

Because eventually you'll run up against the house limit and can't double your bet, among other things.

[Thythe]

Sure its somewhat unlikely for awhile that you will hit a limit that is too high...but in the long run you will. $1, 2,4,8,16,32,64,128,256,512....eventually it will hit millions and you still haven't won.

[ZeeJustin]

Don't forget that your EV on each spin is independant of your EV on any other spin. There is no memory in roulette.

Let's forget 0 and 00 to make the math easier.

If your bankroll is 1 bet, you will win 1 50% of the time and lose 1 50% of the time.

If your bankroll is 3 bets, you will win 1 50% of the time (first spin), win 1 25% of the time (from betting 2 on your second bet), and lose 3 25% of the time.

If your bankroll is 7 bets, you will win 1 50% of the time (first spin), win 1 25% of the time (from betting 2 on your second bet), win 1 12.5% of the time (from betting 4 on your third spin), and lose 7 12.5% of the time.

In each scenario, your EV is 0. Sure, you will win far more often than you lose, but when you lose, it's a doozy. This only considers bankrolls of up to 7 bets, but feel free to continue the math, and you will see that in each case, your EV is always 0.

[bugstud]

Add in that #1 your EV is negative, and #2 bad streaks DO happen. Check the archives, one or two people did this, lost, and complained some time ago.

See also: martingale system

[Jim Kuhn]

It WOULD work if there were no house limit and you had an unlimited bankroll. In other words, it won't work!

[Stew]

[ QUOTE ]
It WOULD work if there were no house limit and you had an unlimited bankroll. In other words, it won't work!

[/ QUOTE ]

no it wouldn't b/c theoretically you would never come out ahead b/c of the zero/double zero.

[WarDekar]

Yes it would, it doesn't matter if it's a 50/50 proposition, you will always profit with the martingale system given an unlimited bankroll and no limits. You will just win at a rate relative to the odds of winning. If it's a 25% chance of simply winning 1:1, you will win at that rate.

[Stew]

[ QUOTE ]
Yes it would, it doesn't matter if it's a 50/50 proposition, you will always profit with the martingale system given an unlimited bankroll and no limits. You will just win at a rate relative to the odds of winning. If it's a 25% chance of simply winning 1:1, you will win at that rate.

[/ QUOTE ]

Yep, you're right, I can't recall the specifics of martingale, but isn't your EV exactly zero, unless you stop after winning two spins in a row?

[ZeeJustin]

If your bankroll is infinite, how could you possibly win?

Add $10 to that infinite bankroll. Are you now any richer?

[Stew]

[ QUOTE ]
If your bankroll is infinite, how could you possibly win?

Add $10 to that infinite bankroll. Are you now any richer?

[/ QUOTE ]

you would only win if you stopped when you won two spins in a row.

[ZeeJustin]

You are missing my point. In any gambling situation when you set aside a bankroll, you are considered a winner if your bankroll increases, and a loser if your bankroll decreases. If your bankroll is infinite, it can neither increase nor decrease by definition; therefore it is impossible to either win or lose.

[JTG51]

Yes it would, it doesn't matter if it's a 50/50 proposition, you will always profit with the martingale system given an unlimited bankroll and no limits.

That's completely ridiculous. Without doing any math, just think about what you are saying. Your claim is that if you make enough bets with a negative expectation you are guaranteed a profit in the end. That's pure nonsense.

[JTG51]

If your bankroll is infinite, it can neither increase nor decrease by definition...

I'm really nitpicking here for no good reason, but that's not really true. Some infinities are larger than others. For example, the set of real numbers is larger than the set of natural numbers. Uncountably infinite is larger than countably infinite.

Of course that's all irrelevant since even if your bankroll was infinite it could never be uncountably infinite, which probably makes this a good place for me to stop babbling.

[Jim Kuhn]

You are thinking that you can flip a fair coin ten million times and it will always be the same heads/tails? What about 20 million?

[JTG51]

You are thinking that you can add a series of negative numbers and end up with a positive number? How many does it take?

[Thythe]

[ QUOTE ]
You are thinking that you can add a series of negative numbers and end up with a positive number? How many does it take?

[/ QUOTE ]

We aren't talking expected value here. No matter how bad the odds are, if you can keep increasing your bet to infinitum and there are no limits you will make money. For example say the odds are 4:1 against. You bet 25cents and lose...now you bet 2 dollars and lose.....now you bet 10 dollars and lose.....you keep increasing your bet until you do win....and since you can increase your bet continuously on your first win you will show a profit. Having an infinite amount of money and seeing no table limits, you must win eventually.

[WarDekar]

Well you have to STOP. That's the key. As someone else pointed out if you have an infinite bankroll you can't exceed infinite so I guess it's a moot point, but whatever.

[Nottom]

Well you don't really have to win eventually. If you have infinite money and infinite time, you will eventually hit an infinatly long losing streak.

[WarDekar]

Thank you, someone realizes my point. It is moot though since if you have an infinite BR you don't need anymore, but still. You will always profit unless your last "roll" or whatever it is is a loser. You have to, that's the definition of the Martingale system. Of course you do have to stop sometime though in order to profit.

If you don't believe it, try it. If you know how to program, program it for millions of trials. You will win at the rate of your chances to win the first roll or spin or whatever, as long as you stop at a win.

[WarDekar]

True, so I guess that's when the old EV comes from haha. But in practice that won't happen, but also in practice you have limits and you don't have infinite BR. So let's just all agree to stop the argument because in some ways we're all right and in some ways we're all wrong. You have to stop to win, but if we're talking about infinite values I think we need to assume you play infinitely long. So, if it's less than a 50/50 proposition, the chances of an infinitely long losing streak are higher than an infinitely long winning streak, therefore you will lose.

[Ed Miller]

Your claim is that if you make enough bets with a negative expectation you are guaranteed a profit in the end.

Why does dropping this bomb never end the discussion immediately? I don't understand. What else is there to say?

[daryn]

this comes up so often /images/graemlins/frown.gif /images/graemlins/frown.gif /images/graemlins/frown.gif /images/graemlins/frown.gif /images/graemlins/frown.gif

nobody understands math anymore!

[fluff]

People not understanding math should get /images/graemlins/laugh.gif /images/graemlins/laugh.gif /images/graemlins/laugh.gif /images/graemlins/laugh.gif /images/graemlins/laugh.gif
not /images/graemlins/frown.gif /images/graemlins/frown.gif /images/graemlins/frown.gif /images/graemlins/frown.gif /images/graemlins/frown.gif

[SossMan]

[ QUOTE ]
Having an infinite amount of money and seeing no table limits, you must win eventually.

[/ QUOTE ]

Why? Because, well hell, infinity is a long time? Nobody can lose that long, can they? Wouldn't they be "due"?
No George Bush "fuzzy math" please.

[Wake up CALL]

[ QUOTE ]
Your claim is that if you make enough bets with a negative expectation you are guaranteed a profit in the end.

Why does dropping this bomb never end the discussion immediately? I don't understand. What else is there to say?

[/ QUOTE ]

Answer: Because this little bomb was incorrect in the context in which it was offered. Does this answer your question Ed? Do I get a chapter in your new book? /images/graemlins/smile.gif

[Stew]

[ QUOTE ]
Well you don't really have to win eventually. If you have infinite money and infinite time, you will eventually hit an infinatly long losing streak.

[/ QUOTE ]

No, b/c you can never infinitely lose what you infinitely have.

[Ed Miller]

Why do I never agree with anything you say? Maybe I'm an idiot...

[Homer]

Listen to JTG, Ed and daryn, they know what they're talking about.

-- Homer

[ZeeJustin]

[ QUOTE ]
Your claim is that if you make enough bets with a negative expectation you are guaranteed a profit in the end.

Why does dropping this bomb never end the discussion immediately? I don't understand. What else is there to say?

[/ QUOTE ]

This bombshell doesn't disprove the theory that the system works with an infinite bankroll, although it does disprove it with any finite bankroll.

I've already stated the reason for this. If your bankroll is infinite, a finite amount of wins / losses cannot possibly alter the size of your bankroll; therefore you can neither win nor lose.

If you were to say with a finite bankroll, the system is always negative EV, I would obviously agree with you.

If you were to say with an infinite bankroll, the system is always negative EV, I have to disagree.

[Thythe]

Let us say that a person plays this game with an infinite bankroll and infinite time with infinite table maxes. Let us further suppose that this person starts betting $1 and increases the bet at a rate where at the first win they will quit with a profit. That is any person successfully winning a bet will quit the game with a profit. Any one who continually loses will stay in the cycle.

Introducing time into the equation we will say that the person completes one bet every minute. So that this person has a 49% chance of winning money in the first minute and a 51% chance of losing and staying in the cycle. We will label time as 't.'

In 2 minutes this person has a higher chance of quitting this game a winner. Further we can say that as
t-->infinity this person has a probability of 1.00 of earning a profit. With infinite time it seems this type of problem can only be solved using limits and thus any person involved in this ludacris infinite example should expect a profit.

[Nottom]

[ QUOTE ]
No, b/c you can never infinitely lose what you infinitely have.

[/ QUOTE ]

I never claimed you would lose everything ... just something.

[slavic]

Why does dropping this bomb never end the discussion immediately? I don't understand. What else is there to say?

Denial of reality, the whole freakin industry lives on it.

[Ed Miller]

If you were to say with an infinite bankroll, the system is always negative EV, I have to disagree.

I'm sorry, but you are simply wrong. A negative EV bet is a negative EV bet. It doesn't matter how large your bankroll is. Each bet is INDEPENDENT. INDEPENDENT. I repeated it (and put it in caps) because that is the important part.

The roulette wheel doesn't give a flying f... that you just bet and lost (or bet and won). Each bet is INDEPENDENT, and each bet is -EV. Why on earth would it matter how big your bankroll is?

Maybe this is what confuses you... it is true that with an infinite bankroll, you would never go broke. But each bet would still be a net money loser. You cannot add a bunch of INDEPENDENT negative numbers and expect to get a positive one. Even if it's an infinite number of negative numbers. /images/graemlins/grin.gif

I am correct without question. I'm not arguing this point any further, because to me it's like arguing with someone who is trying to tell me that 2+2=5.

[Wake up CALL]

[ QUOTE ]
Why do I never agree with anything you say? Maybe I'm an idiot...

[/ QUOTE ]

I couldn't say why Ed. Perhaps you will agree with this: You are obviously not an idiot.

Here is the post by JTG to which you referred and the post to which he was referring:

[ QUOTE ]
by Wardecker Yes it would, it doesn't matter if it's a 50/50 proposition, you will always profit with the martingale system given an unlimited bankroll and no limits.

[/ QUOTE ]

[ QUOTE ]
by JTG That's completely ridiculous. Without doing any math, just think about what you are saying. Your claim is that if you make enough bets with a negative expectation you are guaranteed a profit in the end. That's pure nonsense.

[/ QUOTE ]

Now as I wrote earlier JTG is correct in general about negative expectation bets. The problem arose when he (and you) applied it to a context in which you progressively adjust your bet beyond the negative expectation each time until the bet finally wins. As long as the bet is not impossible to win and the stipulations in Wardeckers post are properly adhered to then eventually using the Martingale system you will show a profit of one unit. You may deny this as often and as long as you wish but that does not make you correct.

Allow me to put this in a poker context. Suppose you run into another player who has an infinite bankroll as well as yourself. Now he offers you a game of holdem where he is dealt AA every hand and you get 7/2 offsuit every hand. There will be an equal ante from each player preflop with no additional betting. The cards will be dealt out randomly and whoever has the best poker hand after the river wins the pot. As an equalizer you are allowed to determine the ante amount before each hand is dealt, he must play until you decide to end the match.

As long as you double the ante amount each hand (ie: Martingale) you will eventually win whatever your initial ante was on the first hand. Now you may have played many hands, each hand with a negative expectation yet as long as you do not die before you choose to end the contest you are guaranteed to win.

This example uses the same principle as in the roulette example.

Wake

[Abagadro]

But you have it wrong.

EV has nothing to do with it. It is about doubling progression.

If you double your wager every time, this is what happens.

X = 1 unit. Doubling it each time means 2X, 4X, 8X.

Using shorthand, X(1...n) is the sum total amount bet and lost, including your fist wager. If you double the amount each time, the last bet (Y) will be the sum of all previous bets, plus 1 unit (X) (this is inherent in doubling).

Thus, Y = the last amount bet and will always be X(1...n)+X.

Your payout on the last bet on an even money wager will be 2Y.

You net winnings on the last bet will be 2Y-[Y+X(1...n)] (i.e. you will win the final payout less the last bet, less the sum total of the amount lost already).

Since Y=[X(1...n)+X], the winnings are

2[X(1...n)+X]-[X(1...n)+X+X(1...n)].


So, your net gain is:

[2X(1...n)+2X]-[2X(1...n)+X]

which equals (drumroll) X!

Therefore you have won (net gain) one unit (X) as long as n is an infinite number and you stop when you win a bet.

Plug in any numbers you want.



(BTW, I am open to the possibility that this proof is wrong, but I ran over it several times and couldn't find a problem with it. Please review it and see if I goofed up somewhere).

[pudley4]

[ QUOTE ]
Therefore you have won (net gain) one unit (X) as long as n is an infinite number and you stop when you win a bet

[/ QUOTE ]

Where is the proof that you will ever win that one necessary bet?

[mosta]

If the pay-off is 1-to-1, it doesn't matter what the true probability is--doesn't matter how fair or unfair the bet is. Each time you lose, you double your bet. If you have ANY chance of winning (and infinite bankroll, etc), eventually you WILL win once and get ALL your money back. EV irrelevant. (if the pay-off is less than one-to-one, you increase your bet by a greater factor accordingly--actual probability still irrelevant.)

[mosta]

Here's a game for you. I bet on green 00, for even money. I bet one dollar, you pay me one if I hit, take my one if I miss--exceedingly unfair, exceedingly "minus EV", yes? If I can keep doubling until eventually 00 hits, that is a break-even game. (no, you can't simulate it on a computer. but you can prove it with highschool-level math (just need the infinite series someone showed above.))

[Homer]

If you have ANY chance of winning (and infinite bankroll, etc), eventually you WILL win once and get ALL your money back.

Are you sure about that? It seems to me that there will be an infintesimally small chance (but still a chance) of losing infinity, which, when combined with all the single unit wins will produce an EV equal to the EV of a single wager.

-- Homer

[Homer]

I'm thinking about this some more. To solve this, don't you have to do some sort of divergence/convergence test of an infinite series? I have no idea what I'm talking about, so maybe I should learn some math and then come back.

[mosta]

[ QUOTE ]
[ QUOTE ]
Therefore you have won (net gain) one unit (X) as long as n is an infinite number and you stop when you win a bet

[/ QUOTE ]

Where is the proof that you will ever win that one necessary bet?

[/ QUOTE ]

the proof is that the probability of hitting is greater than zero and you have an infinite number of trials.

P(-Hn) [not hitting in n trials] is (1-P(H))^n.
1 >= P(H) [hitting] >0
then 1 > (1-P(H)) >= 0
and (1-P(H))^n goes to zero as n goes to infinity.
ie, P(-Hn) as n goes to infinity = 0

what this says is that any fraction less than one and greater or equal to zero multiplied by itself an infinite number of times gets arbitrarily close to, ie indistinguishable from, ie essentially the same as, zero. if your chance to win is only .01 and your chance of losing is .99, .99 times itself infinitely many times is eventually, vitually zero. you can't lose forever.

[pudley4]

That's exactly it. Since the probability of winning is >0 and <1, the geometric series will converge. See my proof in a new post.

[bernie]

you're not getting 1-1 on a roulette wheel. remember the initial question?

there is no system to beat a roulette wheel in the long run. NONE. the presence of the green number(s) make it unbeatable and non-breakevenable. (new word?)

same with craps.

what next? a heavy study into the possible lifestyle of a big wheel professional?

b

[mosta]

the claim is not to beat the wheel but to break even, and it is premised on an infinite bankroll, infinite time, and no house limit (which conditions don't exist). and I didn't ask for one-to-one on red or black. I asked for one-to-one on the single green 00. get it? read the posts that you are replying to, several times, slowly.

[mosta]

Now here is a computation I think would be interesting. Suppose you bring one million bucks to the wheel. You bet on red or black starting with one dollar and doubling when you lose. The house limit is one million. And there is one spin every two minutes. What is the mean time to bust? What is the probability of busting out in one weekend. What about for 10 million dollars? I'm at work, so I'll just put it out and hope that Buzz will come along and crank it out.

[bernie]

you will not break even on a roulette wheel using a martingale system. no matter how big the bankroll, it is a losing proposition longterm.

that would mean you could play both red and black, and anytime the green hits, you double your bet on the next red and black bet. thereby, winning and getting back to even.

eventually, you will lose. all of it.

b

[ZeeJustin]

[ QUOTE ]
Maybe this is what confuses you... it is true that with an infinite bankroll, you would never go broke. But each bet would still be a net money loser. You cannot add a bunch of INDEPENDENT negative numbers and expect to get a positive one. Even if it's an infinite number of negative numbers.


[/ QUOTE ]

I never stated or even implied otherwise. I have already stated that this system is never positive EV. The reason you are getting frustrated from debating this is because you are apparently ignoring everything I say.

This is what I'm saying: If you take an infinitely large bankroll, and subtract (yes, subtract, negative ev, losing money, whatever you wish to call it) any finite amount, you cannot end up with anything less than the infinitely large amount you started with.

[Ed Miller]

Are you sure about that? It seems to me that there will be an infintesimally small chance (but still a chance) of losing infinity, which, when combined with all the single unit wins will produce an EV equal to the EV of a single wager.

Yes Homer, that's what's going on. I didn't bother mentioning it because these people that don't understand my point about a independent bets CERTAINTLY ain't gonna understand an argument like that one. /images/graemlins/smile.gif

[pudley4]

Right, but in your original post you didn't include the proof that your probability of eventually hitting does in fact equal one.

[pudley4]

[ QUOTE ]
Are you sure about that? It seems to me that there will be an infintesimally small chance (but still a chance) of losing infinity, which, when combined with all the single unit wins will produce an EV equal to the EV of a single wager.

Yes Homer, that's what's going on. I didn't bother mentioning it because these people that don't understand my point about a independent bets CERTAINTLY ain't gonna understand an argument like that one. /images/graemlins/smile.gif

[/ QUOTE ]

See my other thread for proof (based on the sum of a geometric series)

[mosta]

actually with an infinite number of trials there is not a "slight" chance of your number never coming up. there is no chance. and you're right. I don't "get" your analysis. could you please try it with some math.

the flaw with the martingale is the betting limit. that's all. it's really very simple.

[WarmonkEd]

This link analyzes betting systems and explains why they DO NOT WORK

http://wizardofodds.com/gambling/scam.html

[Wake up CALL]

If you had read your own link it agrees with what has been posted here by implication. In other words when he writes in reference to the Martingale system "The fallacy is the promise of guaranteed winnings by using the Martingale. Occasionally the player will lose several bets in a row and will reach a point where he doesn't have enough money to double." It is the same as we are stating that it will work if you have no house limit and an unlimited bankroll.

[Wake up CALL]

[ QUOTE ]
Yes Homer, that's what's going on. I didn't bother mentioning it because these people that don't understand my point about a independent bets CERTAINTLY ain't gonna understand an argument like that one.


[/ QUOTE ]

Ed,

Why don't you just ask David to explain it to you. Then either come back to the thread and admit you were mistaken, or not!

[citanul]

i've got to say that bernie's post is by far the most elegant. to top it off, it is so plainly simple that even those who don't want to think about the obvious mathematical flaws in their own arguments can't ignore the point.

kudos,

citanul

[Ed Miller]

Then either come back to the thread and admit you were mistaken, or not!

Please quote me something I said in this thread that is wrong.

[Ed Miller]

Ah. I understand what you are objecting to (and now also your comment about context). You are correct... JTG's comment was not appropriate in context (that is, in response to WarDekar's post). ZeeJustin is incorrect that the expected value of a Martingale progression at roulette is non-negative.

I didn't read what you wrote (i.e. this post) the first time around.

I view all gambling issues from an EV perspective. There is no betting progression that will turn roulette into a +EV game, no matter how big your bankroll is and how many bets you are willing to make.

EDIT: BTW, this argument about Martingale eventually winning a bet is no different than arguing that you can bet in such a way as to make your risk of ruin arbitrarily large in a positive EV game. True, but, to me, uninteresting and misleading.

[ZeeJustin]

[ QUOTE ]
ZeeJustin is incorrect that the expected value of a Martingale progression at roulette is non-negative.


[/ QUOTE ]

If you want to ignore me, that's one thing, but please don't put words in my mouth.

I agree that the EV is negative. What I'm saying is that with an infinitely large bankroll, any finite amount of negative EV will not change the size of the bankroll, and therefore, there are no profits / losses no matter what system you're using or what bets you are making. You can even bet with a 0% chance of winning, and this will still hold true.

My point is simply that analyzing this problem with an infitely large bankroll will not get anyone anywhere. As I've said 10 times before, a basic mathematical analysis of this problem with a limited bankroll will prove that it is indeed negative EV, and you will indeed be a longterm loser.

[fluff]

Are you goofballs still discussing this?? Infinite house limits? Infinite bankroll?

Even if I had such a thing as an infinite bankroll, there is eternal life, and there is a great big casino in heaven with infinite house limits, I can tell you one thing for damn sure: I won't be spending my time playing freaking roulette for all infinity so I can win one effing unit. /images/graemlins/crazy.gif

Although, if gambling is a sin, that would be the perfect punishment: And I say unto thee, if ye are found to commit the sin of wagering, thy punishment shall be an eternity spend at ye infernal roulette wheel, confined for all eternity to wager the Martingale system so that ye might win one effing unit when time comes asunder.

[bigpooch]

Well, most everyone ought to know what happens with a finite
bankroll: as Bob Stupak once said, "Having one-thousandth of
one percent the worst of it, if he plays long enough, that
one-thousandth of one percent will bust the richest man in
the world." (from Theory of Poker : Expectation and Hourly
Rate)

Whenever people think of infinity, someone gets things wrong
for some reason. Perhaps, they didn't take real analysis
in undergraduate school! If some Martian had an infinite
bankroll and lives forever, he could keep increasing his bet
size in such a way to recoup his losses or make a mere
dollar. He WILL EVENTUALLY WIN as long as he is allowed to
bet anything he wants and he is allowed to play as long as
he wants.

It doesn't even matter if he has MUCH THE WORST OF IT. As
long as he has some probability epsilon (strictly greater
than zero) of winning, he will eventually win at some point
(with a probability of 1) and thereby after this winning
wager he will be up, albeit a very small amount (anything
is small compared to infinity!). By the way, every bet
has -EV but that does not deter the Martian: he has an
infinite bankroll!

Even more astounding is this: he will be up money (albeit
only a few dollars) INFINITELY OFTEN if he plays forever.
Of course, forever is a very long time, and our cosmos may
not last that long! If this Martian were to only quit when
he is up, he WILL SURELY BE UP! What a wonderful waste of
time for the Martian though! But what does he care? He
lives forever!

Back to reality: unfortunately, this won't work in this
cosmos because the amount of a bankroll is finite and the
age of the universe may not be infinite! If you were to
try this out in practice, you could win a very small amount
or GET CRUSHED for your entire bankroll! /images/graemlins/smile.gif

[daryn]

you're on the right trail here homer. here's the "hand-waving" explanation.

sometimes in mathematics you have two functions.. say f(x) = x^2 and f(x) = x^3.

now, we both know that as x goes to infinity in both of these functions, then f(x) goes to infinity as well. but some functions go to infinity faster than others.. this is an abstract concept but a mathematically valid one.

regarding the infinite bankroll and infinite number of bets, i believe that you lose money faster than you make money, as time goes to infinity. i'm not a math expert but as a physics expert i think i'm pretty math savvy,.. but maybe one of the true math experts that hang around these parts can come to my aid here and either bolster my argument or dash it to bits!

[Thythe]

I think bigpooch above has it pretty good

[daryn]

by the way, can someone please rehash the "monty hall paradox" so we can talk about THAT for another 3 weeks? thanks.

[Ulysses]

[ QUOTE ]
Listen to JTG, Ed and daryn, they know what they're talking about.

-- Homer

[/ QUOTE ]

Listen to Thythe, Jim, WarDekar, WakeUp Call, mosta, Abagadro, pudley4, and bigpooch, they know what they're talking about.

[Chris Daddy Cool]

A whole lot of infinite talk going on. Maybe somebody already posted this, but...
If you have an infinite amount of trials, you could run into an infinitely long losing streak before you ever win one.

[Homer]

[ QUOTE ]
[ QUOTE ]
Listen to JTG, Ed and daryn, they know what they're talking about.

-- Homer

[/ QUOTE ]

Listen to Thythe, Jim, WarDekar, WakeUp Call, mosta, Abagadro, pudley4, and bigpooch, they know what they're talking about.

[/ QUOTE ]

I don't get it.

[bigpooch]

No you couldn't! If your chances of winning each individual
trial is epsilon > 0, then your chances of not winning EVER
is ZERO. The probability of an infinitely long losing
streak is zero. As I have stated before, infinity is quite
difficult for most humans to grasp!

[bigpooch]

That's such a poorly explained paradox and so many people
are very badly mistaken about that! I believe that ages
ago I posted something about that as an aside to the
envelope paradox (which, in my view, is quite interesting!).
The key is get Monty to like you! No joke! /images/graemlins/smile.gif

[Ulysses]

[ QUOTE ]
I don't get it.

[/ QUOTE ]

I just listed all the people who correctly explained why doubling your bet each time will result in you eventually winning at roulette with a net positive outcome.

Ed is correct that if you add a bunch of negative numbers together, they will never add to a positive. But, that's not what you are doing here.

Here's what you're doing. Each time prior to a win, your result is some negative number. To that negative number you are adding a number that will either be:

a) another negative number
b) a positive number greater than the initial negative number

Rinse and repeat.

With any chance of winning, you will eventually win. And when you win, you are now a winner at roulette. That's all.

So, yes, a series of negative expectation bets can result in a guaranteed positive outcome given this betting sequence.

[Ulysses]

[ QUOTE ]
I don't understand. What else is there to say?

[/ QUOTE ]

Do you understand now, Ed? /images/graemlins/grin.gif

[Ulysses]

[ QUOTE ]
That's pure nonsense.

[/ QUOTE ]

Haven't seen you in this thread for a while either, champ. /images/graemlins/grin.gif

[Homer]

Yeah, I get it now and am more than willing to admit that I was wrong. Actually, no, not more than willing, just regular old willing. I mean, it's not like I'm eager to admit it. Anyway, seeing as how this is the first time I've ever been wrong about anything, I think I should be given a mulligan.

-- Homer

[Ulysses]

[ QUOTE ]
Anyway, seeing as how this is the first time I've ever been wrong about anything, I think I should be given a mulligan.


[/ QUOTE ]

Fair enough. Just don't let it happen again.

[JTG51]

I didn't add anything worthwhile to the discussion the first time around, so I figure I'm better off not saying anything. /images/graemlins/smile.gif

The whole problem comes down to the fact that the probability of losing an infinite series of bets approaches zero. I obviously accept that. What I have a hard time accepting is that saying the probability approaches zero is the same thing as saying it is zero. It seems to me that since there is some non-zero chance of losing every bet, given an infinite series of trials you should eventually experience an infinite losing streak. A lot of people a lot smarter than me are saying that's wrong so I suppose it is, I'm just not sure I understand why.

I should have paid better attention in my calculus classes.

[JTrout]

I suck at calculus. Formulas give me a headache. But I don't need these skills to understand that the roulette system is doomed.

It reminds me of when I was 10 and the fair came to town. There was a dice game that you bet on whether you could roll over 7, or under 7. (you got 3 to 1 on seven).
There were
some folks playing and doing this system, and winning a few quarters.
"See!" they would say, " It's foolproof!"
I was struck with the urge to run home a break my piggy bank. But a few minutes of observing, and thinking, allowed common sense to prevail.
I was fortunate enough to have been blessed with a little common sense.
With each additional bet (larger bankroll), you can get closer to a sure thing (100%), but you can NEVER reach it.

It is similar to the question of:
If I am 100 yds from the finish line, and each day I cut the distance from the finish line in half, how many days will it take me to reach the finish line?
I don't need calculus or a formula to know I'll never get there.

Same deal.
JTrout.

[WarDekar]

At least you understand what we were saying. We've wasted 50 posts talking about something that is a moot point anyway because there is no such thing as an infinite bankroll and infinite limits, so look at the time we just wasted.

Also, what I said a long time ago also holds true. Over a long period of time and a long period of bets I believe the limit will approach the chances of winning any given bet times the amount you've wagered so far, in fact I know that's the case and if anyone REALLY wants to know the math behind it I'm sure I could put it together (this is assuming the last roll wasn't the end of an incredibly long losing streak).

I just read what I wrote and it might not make sense if you don't know what I mean to begin with. I'm saying your PROFITS will approach the percentage*(amount wagered initially). So if you "roll" 100,000 times and your chances of winning any particular time is 40%, and your unit wagered to start is 1, you'll end up with a profit of 40,000. If it's 70% it will be 70,000, etc. Although if it's a value greater than 50 it will approach from the right more often than not, if it's lower than 50 from the left more often than not.

[Chris Daddy Cool]

Maybe I'm belaboring the point here and further proving my own stupidity in calculus, but with an infinite number of trials with the possible outcomes of win or loss, wouldn't there be at some point a string of infinite losses, which could theoritically begin at the the very first "roll"?
I guess I'm saying that within infinity trials, wouldn't there be an infinite streak of wins, an infinite streak of losses, and an infinite streak of random results like WLWLWLLWW?

[WarDekar]

It's basic statistics, look up a page on stats and you'll see that as the number of trials approaches infinity the probability of only losses approaches zero.

[daryn]

just joking about monty hall by the way.. it seems like it's one of those things that just keeps coming up over and over again, like dutch boyd and is moneymaker a fish?