QUESTION: Suppose two equally strong tennis players play against each other until one player wins three games in a row. The results of each game are independent, and each player will win with probability $\frac{1}2$. What is the expected value of the number of games they will play?

MY APPROACH: I have tried to set up some kind of recurrence relation here but could not succeed.. Observe that there can be at most a winning streak of $2$. A winning streak of $3$ means that the game ends.. If we assume that the number of $1$ game winning streak is $x$ and the number of $2$ game winning streak is $y$ then $x+y+1$ obviously yields the desired answer..

Note: A $1$ game winning streak simply means that they win alternately.. Since each one of them has a $50\%$ chance of winning therefore we can do this..

Now, somehow, we have to find the value of $x$ and $y$.. But here I am stuck.. With so less information, neither can I set up a recurrence relation, nor do I see any way to compute two variables..

Any help will be much appreciated.. :)

  • $\begingroup$ @lulu Okay.... But I have doubt.. $E(A,A,2)$ means that $A$ has won the last two games and it is $A's$ turn now.. So the probability of winning the first game is half and the second game is also half.. So, since $A$ has to compulsorily win them, isn't it $\frac{1}2 . \frac{1}2 = \frac{1}4$.. From where do you get $(1+E(B,B,1))$ ? $\endgroup$ Aug 7, 2020 at 7:07
  • $\begingroup$ My comment was needlessly complex, as we don't actually care who holds the current winning streak. (Also in the first version of the comment I was considering different rules then the ones you wanted, so I deleted the comment). I posted something simpler below. $\endgroup$
    – lulu
    Aug 7, 2020 at 7:10
  • $\begingroup$ @lulu Yes.. I am trying to get what your answer wants to say.. $\endgroup$ Aug 7, 2020 at 7:13
  • $\begingroup$ here is a duplicate question (phrased differently). $\endgroup$
    – lulu
    Aug 7, 2020 at 7:17

2 Answers 2


Recurrence works fine.

All we care about is the length of the current win streak, we don't even care who has been winning. Accordingly, let $E_i$ denote the expected number of games it will take if one player currently has a winning streak of length $i$. The answer we seek is $E_0$.

We get: $$E_2=\frac 12\times 1+\frac 12\times (1+E_1)=1+\frac 12\times E_1$$

Similarly $$E_1=1+\frac 12\times (E_1+E_2)$$ and $$E_0=1+ E_1$$

This is easily solved and yields $$\boxed {E_0=7}$$

  • $\begingroup$ I have just one doubt.. discussed in the comment above... $\endgroup$ Aug 7, 2020 at 7:10
  • 1
    $\begingroup$ My earlier comment pertained to a different rule set than you wanted (so I deleted that comment). What doubt do you have for the argument I presented here? If the current win streak is $1$, say, then with equal probability either the current winner wins again, moving us to $E_2$ or the current winner loses, keeping us in $E_1$ and in either case we have played one game. $\endgroup$
    – lulu
    Aug 7, 2020 at 7:13
  • $\begingroup$ Oh yes!! Thank you so much.. $\endgroup$ Aug 7, 2020 at 7:16
  • $\begingroup$ The addition of $1$ in every recurrence denotes that they are playing one game right?.... But what does that $\frac{1}2 \times 1$ in $E_2$ mean? $\endgroup$ Aug 7, 2020 at 7:24
  • 1
    $\begingroup$ For the first equation I wrote I considered two paths, according to who won. If the current streak holder wins it takes $1$ game to finish. If the current streak holder loses it takes $1+E_1$ games to finish, and both paths have probability $\frac 12$ of occurring. $\endgroup$
    – lulu
    Aug 7, 2020 at 7:25

Suppose the winner of the last game is on a $1$-game streak. How many more games until someone is on a $2$-game streak? This is just a geometric random variable with parameter $1/2$, so has expectation $2$.

Now, once someone is on a $2$-game streak, either they get a $3$-game streak next game, or you go back to someone being on a $1$-game streak. So from a $1$-game streak, after an expected $3$ games either someone completes a $3$-game streak or you are back where you started. The number of times this happens before you get a $3$-game streak is also exponential with parameter $1/2$, so has expectation $2$. Crucially, the number of stages of this form you have to go through is independent of the length of each stage. So the total time for all stages has expectation $2\times 3=6$. This is the expected time from the position where someone is on a streak of $1$, i.e. the expected number of games needed after the first game, so the total expectation is $7$.

  • $\begingroup$ What do you mean by 'geometric random variable with parameter' ? $\endgroup$ Aug 7, 2020 at 7:25
  • $\begingroup$ @StrangerForever Have a look at geometric distribution $\endgroup$
    – drhab
    Aug 7, 2020 at 7:40

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