Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 2 / 10F (Probability)

Put $q=1-p$ and track the difference $S_n$ between the numbers of wins. It is a biased random walk with increments $+1$ and $-1$, stopped on first reaching $3$ or $-3$.

To finish at $+3$ on game five, there must be four wins for $A$ and one for $B$, with the final game won by $A$. The unique loss for $A$ can be in position one, two or three. If it were in position four or five, the first three games would already have stopped the contest. Each of the three allowed paths remains strictly inside the barriers until the last step and has probability $p^{4}q$. By interchanging players, there are also three paths ending at $-3$, each with probability $p{q}^4$. ThereforeFor a fair game this is $3/16$. Counting all five-game sequences with final win difference three would overcount paths on which the contest ended earlier.

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.