Gamblers ruin formula

Question

Hello , I have been reading about gamblers ruin and I found this formula
can anyone confirm its accuracy ? I assume they only bet one chip a time

Answer 1

To derive that formula, let $A(n)$ be the probability that Anne wins if she started with $n$ coins; clearly $A(32) =1$ since Carol has nothing to bet, and $A(0) = 0$ since Anne has nothing to bet. We are looking to determine $A(23)$.

Equally clearly, if $0 < n < 32$, then after one betting outcome the players are playing the same game but with a starting $n$ which is one different from the original $n$, so for $0 < n < 32$, $$ A(n) = \frac{5}{12} A(n+1) + \frac{7}{12} A(n-1) = pA(n+1)+qA(n-1) $$ which we can re-arrange (by solving for $A(n+1)$) to $$ A(n+1) = \frac{1}{p}A(n) - \frac{q}{p}A(n-1) = \frac{p+q}{p}A(n) + \frac{q}{p}A(n-1) \\ A(n+1) = A(n)+\frac{q}{p} \left[ A(n) - A(n-1) \right] $$ Now let's look in particular at the difference $A(2)-A(1)$ (this is the clever step in the proof): $$ A(n+1) - A(n)=\frac{q}{p} \left[ A(n) - A(n-1) \right] $$ $$ A(2) - A(1) = pA(2)+qA(0) = \frac{q}{p} \left[ A(1) - A(0) \right] = \frac{q}{p}A(1) $$ and then
$$ A(3) - A(2) = \frac{q}{p} \left[ A(2) - A(1) \right] = \left( \frac{q}{p} \right)^2 A(1) \\A(4) - A(3) = \left( \frac{q}{p} \right)^3 A(1) $$ and eventually $$ A(32) - A(31) = \left( \frac{q}{p} \right)^{31} A(1)$$ But that means (by telescoping the differences) that $$A(32) = \sum_{i=1}^{31}\left( \frac{q}{p} \right)^{31} A(1)$$ and of of course $A(32) = 1$ from above.

The geometric sum for the expression for $A(32)$ can be expressed in closed form, and this gives an equation which can be solved for $A(1)$. That in turn leads to the formula for all the $A(n)$, which is the formula you were concerned with.