Am I stupid? I can't fit these two concepts into any type of harmony. - Page 4

PairTheBoard · #31 08-25-2005, 06:00 AM

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if u have edge than u should use kelly criteria and u will never bust out.

the idea is to bet and precentage amount from ur bankroll (and not a fixed amount).

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Good point!

PairTheBoard

MtDon · #32 08-26-2005, 01:34 AM

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"If you play for an infinite lenght of time, all possible finite runs of cards will be dealt an infinite number of times. Since you will always have an finite number of chips, you will have to run into a run of cards which will break you."

Wrong

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Why?

BillC · #33 08-26-2005, 12:22 PM

"If you have an edge, then your probability of going bust is always < 1, no matter how low your win rate, or how small your bankroll. This probability, called the risk of ruin, depends on your win rate, your standard deviation, and your bankroll, and it can be computed by the formula in this thread for games like blackjack and poker. "

This assumes that the bet size is "small" relative to the bankroll. If i bet my entire bankroll on each trial, my ROR is 1. So what happens if you bet say 10% of your starting bank?

PairTheBoard · #34 08-26-2005, 12:54 PM

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"If you play for an infinite lenght of time, all possible finite runs of cards will be dealt an infinite number of times. Since you will always have an finite number of chips, you will have to run into a run of cards which will break you."

Wrong

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Why?

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What will happen with non-zero probability is that for any long bad streak you want to identify, it will always happen when your bankroll has grown large enough to sustain it.

PairTheBoard

BillC · #35 08-26-2005, 01:22 PM

What I meant was that the usual ROR formula assumes small bet sizes and a relatively unskewed payoff distribution.

For highly skewed games such as video poker (and i think most poker tournaments) the ROR calculation is not so nice
see this article

pokerplayer28 · #36 08-26-2005, 02:32 PM

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"If you play for an infinite lenght of time, all possible finite runs of cards will be dealt an infinite number of times. Since you will always have an finite number of chips, you will have to run into a run of cards which will break you."

Wrong

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Why?

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when do your finite run of cards end?

#37 08-26-2005, 04:42 PM

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And when you bring down the variance to "normal" levels, then the random walk model can be used with high precision even in NLH.

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Well, it depends on our assumptions about how the game is set up. If we assume that you are playing against opponents with infinitely large bankrolls, then the random walk model no longer holds. On any hand, there is a very small chance that you will lose your entire stack. Your winnings at any rate of BB/100 will not eliminate that chance, and therefore you will bust out eventually.

If you play against opponents who add or subtract from their stacks so they're at 1 trillion BB at the beginning of each new hand, then you're correct, assuming both you and your opponents play to maximize EV, rather than to maximize or minimize the chance you will bust.

David Sklansky · #38 08-26-2005, 08:04 PM

"What will happen with non-zero probability is that for any long bad streak you want to identify, it will always happen when your bankroll has grown large enough to sustain it."

PairTheBoard

Who wrote this? I thought we had software preventing more than one person from using the same name.

BruceZ · #39 08-27-2005, 02:47 AM

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What I meant was that the usual ROR formula assumes small bet sizes and a relatively unskewed payoff distribution.

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The initial bankroll can be a single bet. In my derivation of the ROR formula, which is essentially Sileo’s derivation, the ROR for a bankroll of size B is derived as the ROR for a 1 bet bankroll raised to the B power. This assumes that the winnings are reinvested in the bankroll, and that your win rate and standard deviation do not change. My comments above assumed that we are maintining a constant win rate.

If your increase your betting limits as your bankroll grows, then you will go broke with probability 1 if you continue to bet more than twice the Kelly fraction of your bankroll, where the Kelly fraction is approximately EV/sigma^2.

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For highly skewed games such as video poker (and i think most poker tournaments) the ROR calculation is not so nice
see this article

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The ROR formula that I linked to is derived by assuming that the game with skewed payoffs can be modeled as a coin flip game with the same win rate and variance via the central limit theorem. There may be games for which this model breaks down, but it works well for blackjack and poker, and any game for which the risk of ruin depends primarily on the win rate and standard deviation, and very little on the higher moments.

PairTheBoard · #40 08-27-2005, 04:00 AM

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"What will happen with non-zero probability is that for any long bad streak you want to identify, it will always happen when your bankroll has grown large enough to sustain it."

PairTheBoard

Who wrote this? I thought we had software preventing more than one person from using the same name.

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Actually, I let my 6 year old niece make that post. She got it from her 1st grade arithmetic book. She tells me all the kids are into this stuff these days because of all the poker shows on TV.

PairTheBoard