Can one overcome a -EV game?

[jason1990]

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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?

[Alex/Mugaaz]

Vince. Why don't you understand that someone can be the most likely to get all the money on the table but have the worst of it? These ideas aren't contradictory.

Here is a horrible example. You vs 100 ten year olds in a fight to the death. You are definitely not a favorite to be alive at the end of the battle, but none of the ten year olds are the favorite to beat you, each one who fights you is most likely going to lose.


Think about this until you get it. If you can understand this then you can understand why you can't beat a -ev game.

[skiier04]

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If you toss a fair coin 2 times the probability distribution will be as follows.

2 heads: .25
2 tails: .25
head:tail:.25
tail:head:.25

If you toss the coin an infinite number of times the probabiliy distribution will be similar. One of the distributions will be an infinite series of heads and one an infinite series of tails, etc. No? Or are you through has mathematics able to eliminate a potential result? I'm sure common sense plays no part in this discussion but even common sense tells us that one possibility is to continuously toss a head.

http://www.netnam.vn/unescocourse/statistics/52.htm

Vince

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You cant flip a coin an infinate number of times.

[Vincent Lepore]

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Is this it? Or is it something else entirely?


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Hmmm...Yes with one minor exception. Martingale might be interesting but a simple bet the whole bankroll on each toss of the coin (%1 bias for the casino) will do.

I'm pretty sure the following applies, especially when one uses bet the whole bankroll on each toss.

Oh, one other thing. Thank you for taking the time.

Vince

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Gambler's Ruin



Let two players each have a finite number of pennies (say, for player one and for player two). Now, flip one of the pennies (from either player), with each player having 50% probability of winning, and transfer a penny from the loser to the winner. Now repeat the process until one player has all the pennies.

If the process is repeated indefinitely, the probability that one of the two player will eventually lose all his pennies must be 100%. In fact, the chances and that players one and two, respectively, will be rendered penniless are

(1) P1 = N2/N1 + N2
(2) P2 = N1/N1 + N2

i.e., your chances of going bankrupt are equal to the ratio of pennies your opponent starts out to the total number of pennies.

Therefore, the player starting out with the smallest number of pennies has the greatest chance of going bankrupt. Even with equal odds, the longer you gamble, the greater the chance that the player starting out with the most pennies wins. Since casinos have more pennies than their individual patrons, this principle allows casinos to always come out ahead in the long run. And the common practice of playing games with odds skewed in favor of the house makes this outcome just that much quicker.




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Vince

[Vincent Lepore]

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You cant flip a coin an infinate number of times.

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Who can't?

Vince

[Vincent Lepore]

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Why don't you understand that someone can be the most likely to get all the money on the table but have the worst of it?

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I do understand this very well. What's your point. I've never said anything different.

Vince

[TomCollins]

Vince,

Are you just trying to prove that if you have a large enough bankroll, you can bust a casino x% or more of the time if you bet their entire bankroll every hand? No one has ever debated that. In fact, with a bankroll of three times as much as the casino, you can bust the casino nearly 75% of the time. But the other 25+% of the time, they have quadrupled up.

[Tater10]

If you break the casino, you lose your comp buffet, -EV.