# Probability of winning a tied game in tennis?

Trying to get this formula has been puzzling me for a bit...

In certain sports, winning a game requires a lead of two points. That is, if the score is tied you have to score two points in a row to win.

In tennis, a point is scored every play. Suppose your probability of scoring the next point is always p. Then, your opponent's probability of scoring the next point is always 1 - p.

Give a formula for your probability w of winning a tied game.

This is a random walk.   The states of the game are $\{\underbrace{-2}_{\textsf{loose}}, -1, 0, 1, \underbrace{~2}_{\textsf{win}}\}$

Assuming the game continues indefinitely until a win, let $P_n$ be the probability of ultimately winning from state $n$.   Since the probability of moving up is a constant $p$ and the probability of moving down is the complement $(1-p)$ then $$P_n=p~P_{n+1}+(1-p)~P_{n-1}$$

Except, of course, for the boundary states, which are $P_2=1, P_{-2}=0$ .

This yield three simultaneous equations: $P_0 = p~P_1+(1-p)~P_{-1} \\ P_1 = \ldots \\ P_{-1}=\ldots$

Complete and solve them (for $P_0$ in terms of $p$)

$\Box$

For you to win, either you win with two straight points ,$\;Pr=p^2$,
or you are still "alive" but back to square one, $Pr = pq+qp = 2pq.$

Thus $w = p^2 + 2pq\cdot w$

$w = \dfrac{p^2}{1-2pq}$