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Old 10-29-2005, 12:05 AM
jason1990 jason1990 is offline
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Join Date: Sep 2004
Posts: 205
Default Re: Can one overcome a -EV game?

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O.K. Jason, I agree with your summation of infinite sequences. Since you know more about these things than I do it is my belief that I do not need to use an infinite sequence to prove that I am correct. I am confident that you can provide proof that the Casino will go broke using finite numbers for each (the casino and the opponent) given no cap on bettings. Correct?

Vince

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The first step toward making a proof is formulating a provable (or disprovable) statement. I don't know exactly what it is you want to prove, but I can try to work with you to create a mathematical formulation of your claim. You said, "the Casino will go broke using finite numbers for each (the casino and the opponent) given no cap on bettings." If I take this literally, it seems to mean

A. Suppose the "Casino" and the "Player" play a game that pays n to 1, the Player's chance of winning is p<1/(n+1), and there are no betting limits. Then there exist finite numbers A and B such that if A and B are the Player's and Casino's bankrolls, respectively, and the Player uses the "martingale strategy," then the probability the Casino goes broke is 1.

Is this what you mean? Somehow, I think it might not be. I'm guessing you mean

B. Let q be an arbitrary number strictly less than 1 (such as 0.99999, for example). Suppose the "Casino" and the "Player" play a game that pays n to 1, the Player's chance of winning is p<1/(n+1), and there are no betting limits. Then there exist finite numbers A and B such that if A and B are the Player's and Casino's bankrolls, respectively, and the Player uses the "martingale strategy," then the probability the Casino goes broke is greater than q.

Is this it? Or is it something else entirely?
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