Wimbledon's final

Question

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.

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

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

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

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

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

Find the probability that Tom wins.

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

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 '20 at 13:56

Especially Lime · Accepted Answer · 2020-07-15T07:40:44.900

6

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.

edited Jul 15 '20 at 07:40

answered Jul 14 '20 at 14:43

Especially Lime

40,828

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 '20 at 15:32
@FrancescoTotti sure, have edited. – Especially Lime Jul 15 '20 at 07:41

Alex · Answer 2 · 2020-07-14T14:57:45.067

1

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): $$ P(S) = P(HH)+P(TT)+ \\ + P(HTHH) + P(THTT) + P(HTTT)+P(THHH) $$

edited Jul 14 '20 at 14:57

answered Jul 14 '20 at 13:59

Alex

19,262

Sure, but the exercise falls within the first section of book. I have to solve it with the probability rules. – Francesco Totti Jul 14 '20 at 14:02
pls see the edit – Alex Jul 14 '20 at 14:17
@Alex Why is $P(HTT)$ or $P(THH)$ valid? – Y.T. Jul 14 '20 at 14:24
'4 games or less' includes 3 games; to win, either player has to win two games in a row, hence $HTT$ and $THH$ are admissable events – Alex Jul 14 '20 at 14:29
1

@Alex The winning condition in the question is to be "ahead of two games" but not winning two games in a row – Y.T. Jul 14 '20 at 14:31
OK, my understanding was that the first with two wins wins the game. I fixed the probabilities. – Alex Jul 14 '20 at 14:50
1

@Alex I've understood the same thing. – Francesco Totti Jul 14 '20 at 14:53
OK, I fixed both MC and winning probabiltiy – Alex Jul 14 '20 at 14:58

Leo163 · Answer 3 · 2020-07-14T14:28:43.007

1

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.

edited Jul 14 '20 at 14:28

answered Jul 14 '20 at 14:10

Leo163

2,647

You are right, I read the problem the wrong way. – Leo163 Jul 14 '20 at 14:23

Wimbledon's final

3 Answers3