A Less Obvious Martingale Fallacy

[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 [img]/images/graemlins/grin.gif[/img]

[MMMMMM]

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He doesn't really compensate for it

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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.

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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!]

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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 [img]/images/graemlins/grin.gif[/img]

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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]

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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?


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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>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>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->inf (the probability that there exists an M>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]

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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.

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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----> P=1/2, financial loss = $1

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

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

Infinite losses in a row-->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.

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

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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->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)>0 ] goes to zero as n goes to infinity - where P[ T(n)>0 ] is the probabilty that T(n)>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