Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 10F / ii / Solution

This is a gambler's ruin hitting problem. For $0<p<1$, let $h_j$ be the probability of reaching $+3$ before $-3$ when the current difference is $j$. Conditioning on the next independent game givesStopping occurs almost surely: conditional on being inside the barriers, three consecutive wins by either player force absorption within the next three games. This event has probability $p^3+q^3>0$, so the chance of surviving $m$ such blocks is at most $(1-p^3-q^3{)}^m$, which tends to zero.

If $p\ne q$, the characteristic roots of the interior recurrence are $1$ and $r=q/p$, so $h_j=A+B{r}^j$. Imposing the two boundary values yieldsIn particular,For $p=q=1/2$ the recurrence instead has affine solutions, and the boundary values give $h_j=(j+3)/6$, hence $h_0=1/2$. The deterministic cases $p=0,1$ give zero and one respectively. The formula covers all of them continuously:The stopping argument and the boundary recurrence show why the answer concerns the first barrier reached rather than the win difference after a predetermined number of games.