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You and an opponent are playing tennis - first to get 2 wins in a row wins. The probability of you getting a win is .6. The probability of him getting a win is .4. What's the probability of you winning the game?

I think this can be modeled as a Markov chain with 5 states (2 Losses, 1 Loss, 0 net, 1 Win, 2 Wins). Therefore, I think I could write out some equations to solve this. I'm trying out a different approach; does this make sense:

P(you win right off the bat) = (.6)(.6) = .36 P(he wins right off the bat) = (.4)(.4) = .16

P(you win) = .36/(.36+.16)

Edit: additionally, how can I relate this problem to the equations of gamblers ruin, if at all?

Other edit: Online, someone said the answer is .91, using the following argument. Why am I wrong? I even checked my answer using equations with the Markov chain.

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  • $\begingroup$ there are multiple ways to do this. One of the ways is to find probability of win after $2$ flips, after $3$ flips, after $k$ flips and so on... you get the same formula as in the method I have described below. $\endgroup$
    – Math Lover
    Aug 15, 2020 at 18:47
  • $\begingroup$ btw just saw your edit of gambler's ruin example where John starts with \$2. That probability will be higher as your problem of tennis match is less forgiving. Your wins stay with you in case of this example but in the tennis match problem that you have stated your wins reset the moment you lose. Example: say you win and get to \$3, you can lose two consecutive and get to \$1 but you are still in the game. In the tennis match problem that is not the case. I have detailed out the solution for tennis match in my answer. I hope this clarifies. $\endgroup$
    – Math Lover
    Aug 16, 2020 at 4:53

3 Answers 3

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Say $J$ and $M$ are playing a tennis match ($J$ and $M$). Whoever wins two consecutive games wins the match. We have to find the probability of two consecutive $J (JJ)$ before two consecutive $M (MM)$.

Probability of J winning a single game $= p = \dfrac{3}{5}$
Probability of M winning a game $= q = 1-p = \dfrac{2}{5}$

There are four possibilities after two matches - $JJ, JM, MJ, MM$

Probability of $JJ = p^2$ and you win.

Probability of $JM = pq$ and say your probability of win from there is $P(W|JM)$

Probability of $MJ = pq$ and say your probability of win from there is $P(W|MJ)$

Probability of $MM = q^2$ (you lose)

$i)$ If it is $JJ$, you win with probability $p^2$.

$ii)$ From $JM$, you can get to $MJ$ with probability $p$ and from there you can win with probability $P(W|MJ)$.
$P(W|JM) = p \times P(W|MJ)$

$iii)$ From $MJ$, you can win with probability $p$ or get to $JM$ with probability $q$ and then win with probability $P(W|JM)$ from there.

$P(W|MJ) = p + q \times P(W|JM)$

Solving $ii)$ and $iii)$,

$P(W|MJ) = \dfrac {p}{1-pq}$

$P(W|JM) = \dfrac {p^2}{1-pq}$

Probability of you winning

$ \begin {align*} P(W) &= P(JJ) + pq \times P(W|JM) + pq \times P(W|MJ) \\ &= p^2 + pq \times \dfrac {p^2}{1-pq} + pq \times \dfrac {p}{1-pq} \\ &= \dfrac {p^2(2-p)}{1-p(1-p)} \space \text {, replacing } q \text { with } (1-p)\\ \end {align*} $

As $p = \dfrac{3}{5}$, $P(W) = \dfrac{63}{95}$

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Your approach is wrong. What you calculated is $P(you\ win | the\ games\ lasted\ 2\ points)$ which is clearly different than $P(you\ win)$.

As for your edit: once you find the correct $P(you\ win)$ you can relate the problems since the tennis match deduced to a Bernoulli trial with $p = P(you\ win)$.

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  • $\begingroup$ Wait, but you can basically always assume that the game only lasts 2 games. The moment it lasts more than 2 games, that means you're coming back to the origin (where both people need to win 2 more games), therefore, you can just ignore the previous wins/losses and start fresh. $\endgroup$ Aug 15, 2020 at 6:48
  • $\begingroup$ Oh yes. I read the tennis part too fast. You are assuming that it is 40-40 in tennis terms, right? $\endgroup$
    – cgss
    Aug 15, 2020 at 7:02
  • $\begingroup$ James: No, it’s two wins in a row. After LW my chances are much better than after WL. $\endgroup$
    – gnasher729
    Aug 15, 2020 at 19:24
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Every sequence of games where you win overall ends in a $WW$, and either starts with $WL...$ or $LW....$, so the probability you eventually win is $$P(WW) + P(LWW) + P(WLWW) + P(LWLWW) + P(WLWLWW) + ... = \frac{P(WW) + P(LWW)}{1 - P(W)P(L)} = \frac{0.36 + 0.144}{1 - 0.24} = 0.663215789... = \frac{63}{95}.$$

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  • $\begingroup$ This is not correct. Probability of winning from $WL$ state is not the same as probability of winning from $LW$ state. $\endgroup$
    – Math Lover
    Aug 15, 2020 at 18:53
  • $\begingroup$ @MathLover You're right, I was thinking about it as if whoever got a lead of 2 wins overall won. I'll fix it. $\endgroup$ Aug 15, 2020 at 18:54

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