# Wimbledon's final

Tom and Jack are playing the final of Wimbledon and they are 6:6 at the last set. They play to the bitter end until one of them is ahead by two games. For Tom the probability to win the next games is $$p$$, and for Jack $$1-p$$. Every games is independent from the others.

1. Find the probability that the match end 9 to 7 for one of them.

For $$A=($$Tom wins 9 to 7$$)$$ and $$B=($$Jack wins 9 to 7$$)$$, we have

$$\rightarrow \mathbb{P}(A\cup B)=2p(1-p)[p^2+(1-p)^2]$$

1. Find the probability that it needs more than 4 games to end the match.

For $$X=($$# games to the end$$)$$, we have

$$\rightarrow \mathbb{P}(X>4)=1-\mathbb{P}(X\leq 4)=1-\mathbb{P}(X\leq 4|A\cup B)=1-\frac{\mathbb{P}(X\leq 4 \cap A)+\mathbb{P}(X\leq4\cap B)}{\mathbb{P}(A)+\mathbb{P}(B)}=1-\frac{2[p^2+(1-p)^2]}{[p^2-(1-p)^2]}$$

1. Find the probability that Tom wins.

Hoping 1) and 2) are right, do you have any ideas for point 3)? Thanks in advance.

• The game can't end in a draw, right? – Alex Jul 14 at 13:52
• @Alex That's right. – Francesco Totti Jul 14 at 13:53
• For question (1) you might want to check the signs: $2p(1-p)[p^2+(1-p)^2]$ could look better to me – Henry Jul 14 at 13:56
• @Henry Right, I edited. So the second answer is wrong. – Francesco Totti Jul 14 at 13:58

I think the simplest way is as follows. Almost surely, they eventually finish, and this must be after an even number of games $$2k$$.

If Tom wins after $$2k$$ extra games, than means they won one game each in all of the first $$k-1$$ pairs, then Tom won twice, which has probability $$q_kp^2$$ for some $$q_k>0$$; similarly Jack wins after $$2k$$ games with probability $$q_k(1-p)^2$$ for the same $$q_k$$. This means that, conditional on the number of games being $$2k$$, the probability of Tom winning is $$\frac{P(T\text{ wins after }2k)}{P(\text{game ends after }2k)}=\frac{q_kp^2}{q_kp^2+q_k(1-p)^2}=\frac{p^2}{p^2+(1-p)^2},$$ which does not depend on $$k$$. So this must also be the unconditional probability that Tom wins.

• Thanks for your answer. Would you please be more specific how you obtain $\frac{p^2}{p^2+(1-p)^2}$? – Francesco Totti Jul 14 at 15:32
• @FrancescoTotti sure, have edited. – Especially Lime Jul 15 at 7:41

A possible approach is via computing series. Say that Tom wins after $$2n+2$$ games: then he either won the last two games, which happens with probability $$p^{n+2}(1-p)^n$$, or he won the last three games, which again happens with probability $$p^{n+2}(1-p)^n$$ but can only happen if $$n>0$$. Since there are no other possibilities, it is enough to compute the value of the series above and add up the result.

• You are right, I read the problem the wrong way. – Leo163 Jul 14 at 14:23

A more systematic approach is using Markov chains with difference equations. You have two absorbent states: $$T_2, J_2$$ ( I use $$\{T, J\}$$ for players and $$T_i, J_j$$ for $$i$$ games ahead. You start with a '0' state/noone ahead, and the probability of winning the game for either gamer is $$h_{0,T_2}, h_{0,T_2}$$: $$h_{0, T_2} = ph_{T_1, T_2} + (1-p)h_{J_1,T_2}\\ h_{0, J_2} = ph_{T_1, J_2} + (1-p)h_{J_1,J_2}\\ h_{T_1, T_2} = p \times 1 + (1-p)h_{0,J_2}\\ h_{J_1, J_2} = (1-p)\times1 + (1-p)h_{0,J_2}$$ Obvisouly $$h_{T_2,T_2} = 1=h_{J_2,J_2}, h_{T_2, J_2}=0=h_{J_2, T_2}$$.

From here, you need to construct 2 more difference equations.

EDIT, for 2), consider $$H$$=Tom wins, $$T$$=Jack wins. You have the following probabilities ($$S$$: game finishes in 4 or less trials): $$P(S) = P(HH)+P(TT)+ \\ + P(HTHH) + P(THTT) + P(HTTT)+P(THHH)$$

• Sure, but the exercise falls within the first section of book. I have to solve it with the probability rules. – Francesco Totti Jul 14 at 14:02
• pls see the edit – Alex Jul 14 at 14:17
• @Alex Why is $P(HTT)$ or $P(THH)$ valid? – Learning Mathematics Jul 14 at 14:24
• @Alex The winning condition in the question is to be "ahead of two games" but not winning two games in a row – Learning Mathematics Jul 14 at 14:31
• @Alex I've understood the same thing. – Francesco Totti Jul 14 at 14:53