what is the formal mathematical relationship between the variance and the odds that the gambler who has smaller budget here?

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?

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 $\sqrt{2000} \approx 44.7$.