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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.

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

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

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  • $\begingroup$ You are right, I read the problem the wrong way. $\endgroup$ – Leo163 Jul 14 at 14:23
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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): $$ P(S) = P(HH)+P(TT)+ \\ + P(HTHH) + P(THTT) + P(HTTT)+P(THHH) $$

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

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