A Probability Model for Blackjack
February 21st, 2014 at 6:13:47 AM
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In the thread "Don Johnson's Blackjack Win" I posted some conclusions about the game of blackjack using a binomial probability model. While some found these postings useful, they raised objections that the results had to be inaccurate because I was not considering the variance of the game of blackjack, which is considered to be 1.1. I think a number of useful conclusions can be made about the game using the binomial probability model, but I hesitate to post them anticipating a reception of this sort.
To calculate the probability of reaching a certain win stopping point b before reaching a loss stopping point x I used the classical gambler's ruin formula, which is for a coin-toss game and hence is based on a binomial distribution. Teliot uses a formula based on the normal distribution and includes the variance of the game. The variance of a coin-toss game is considered to be 1 because one always wins or loses the same amount.
Both the classical formula and Teliot's have the same form: the numerators and the denominators consisting of an exponential minus 1. In Teliot's formula the exponential is e-2μx/σ2 where μ is the negative of the house edge, x is the loss goal and σ2 is the variance of the game. The exponential in the denominator is the same except that x is replaced by b, the win goal.
The exponentional in the classical formula is (q/p)x where p is the player's probability of winning and q is his probability of losing. The house edge is generally considered to be q - p, so that we may express q/p in terms of μ as (1 + μ)/(1 - μ).
It is evident that if σ2 ≠ 1 for a particular game, the exponential is unchanged if we instead set σ2 = 1 and change μ to μ/σ2. Since p = (1 - μ)/2 the new p would be p' = (1 - μ/σ2)/2.
Would there be any objections to my using the binomial probability model if I make this adjustment? For the game of blackjack instead of 0.0029 as the house edge I would use 0.00264. With this adjustment the classical formula should give the same results as Teliot's.
To calculate the probability of reaching a certain win stopping point b before reaching a loss stopping point x I used the classical gambler's ruin formula, which is for a coin-toss game and hence is based on a binomial distribution. Teliot uses a formula based on the normal distribution and includes the variance of the game. The variance of a coin-toss game is considered to be 1 because one always wins or loses the same amount.
Both the classical formula and Teliot's have the same form: the numerators and the denominators consisting of an exponential minus 1. In Teliot's formula the exponential is e-2μx/σ2 where μ is the negative of the house edge, x is the loss goal and σ2 is the variance of the game. The exponential in the denominator is the same except that x is replaced by b, the win goal.
The exponentional in the classical formula is (q/p)x where p is the player's probability of winning and q is his probability of losing. The house edge is generally considered to be q - p, so that we may express q/p in terms of μ as (1 + μ)/(1 - μ).
It is evident that if σ2 ≠ 1 for a particular game, the exponential is unchanged if we instead set σ2 = 1 and change μ to μ/σ2. Since p = (1 - μ)/2 the new p would be p' = (1 - μ/σ2)/2.
Would there be any objections to my using the binomial probability model if I make this adjustment? For the game of blackjack instead of 0.0029 as the house edge I would use 0.00264. With this adjustment the classical formula should give the same results as Teliot's.
February 21st, 2014 at 10:05:47 AM
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Drink deep, or taste not the Pierian spring.
February 21st, 2014 at 1:58:51 PM
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Obviously neither the binomial nor the normal distribution will give completely accurate results.
That's because the actual distribution of the BJ results is neither of them.
The normal distribution takes variance into account but it has also the problem that it is a continuous function (bell curve in graphic form).
The binomial does not take into account variance but on the other hand is a discrete function (step by step in graphic form).
BJ has variance but is also a discrete function.
When the number of trials is big then the continuous aspect of the normal is not a problem.
When the number of trials is small then the step by step is more appropriate.
You are proposing a revised formula using binomial and a a variance element as well.
You might be on to something.
I have not really looked deep into this formula you are proposing.
The important thing is, does it give the right result (ie a good approximation) and under what circumstances?
Please tell us what kind of results does your formula give and how accurate it is.
That's because the actual distribution of the BJ results is neither of them.
The normal distribution takes variance into account but it has also the problem that it is a continuous function (bell curve in graphic form).
The binomial does not take into account variance but on the other hand is a discrete function (step by step in graphic form).
BJ has variance but is also a discrete function.
When the number of trials is big then the continuous aspect of the normal is not a problem.
When the number of trials is small then the step by step is more appropriate.
You are proposing a revised formula using binomial and a a variance element as well.
You might be on to something.
I have not really looked deep into this formula you are proposing.
The important thing is, does it give the right result (ie a good approximation) and under what circumstances?
Please tell us what kind of results does your formula give and how accurate it is.
