# Two players until one player wins three games in a row. Each player will win with probability $\frac{1}2$. How many games will they play?

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.. :)

• @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 $(1+E(B,B,1))$ ? Aug 7 '20 at 7:07
• 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.
– lulu
Aug 7 '20 at 7:10
• @lulu Yes.. I am trying to get what your answer wants to say.. Aug 7 '20 at 7:13
• here is a duplicate question (phrased differently).
– lulu
Aug 7 '20 at 7:17

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 $$\boxed {E_0=7}$$

• I have just one doubt.. discussed in the comment above... Aug 7 '20 at 7:10
• 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.
– lulu
Aug 7 '20 at 7:13
• Oh yes!! Thank you so much.. Aug 7 '20 at 7:16
• 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? Aug 7 '20 at 7:24
• 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.
– lulu
Aug 7 '20 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 $$7$$.

• What do you mean by 'geometric random variable with parameter' ? Aug 7 '20 at 7:25
• @StrangerForever Have a look at geometric distribution Aug 7 '20 at 7:40