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Here is a part of Bob Anderson's answer in my question (Does variance do any good to gambling game makers?)

Suppose you had two gamblers who were flipping coins against one another with fair odds for 1 dollars a flip. One of the gamblers has 20 dollars and the other has 100 dollars. If someone goes broke the game is over.

what is the formal mathematical formula/distribution that link "the variance and the odds that the gambler who has smaller budget would win" together and what is its proof?

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Victor
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1 Answers1

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I'm not sure what you're looking for. In this particular case you can easily figure out the whole probability distribution and compute the expected value and the variance.

To make this explicit, say $X$ is the random variable that represents the profit of the gambler with the smaller budget. Because the gambler with the smaller budget has a 20/120 probability of winning and a 100/120 probability of losing [this takes some reasoning to see], $P[X=100] = 1/6$ and $P[X=-20]$ = 5/6. So the game is fair, in the sense that $E[X] = 1/6 \times 100 + 5/6 \times -20 = 0$. The variance of $X$ is $\text{Var}(X) = E[X^2] - E[X]^2 = E[X^2] = 1/6 \times 10000 + 5/6 \times 400 = 2000$ and hence the standard deviation is $\sqrt{2000} \approx 44.7$.

Is this what you were looking for?

Magdiragdag
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  • Agreed. However I'd point out that the original question related to "practical gambling", in which case this mathematical definition of "fair" is usually inappropriate, because when you're broke, you don't get to keep coming back for the longshot that gets you even. – Bob Anderson Sep 30 '13 at 22:19