# Probability of number of sets in tennis game

Player A and player B are playing tennis. Determine the probability of the game ending with 3, 4 and 5 sets played respectively, given that the probability of player A winning a set is 0.6 and that the game ends when one of the players has won 3 sets.

Since the probability of player B winning is $0.4$, the probability of the game ending with 3, 4 and 5 sets played should be

$P(3) = 0.6^3 + 0.4^3$

$P(4) = 0.6^3\times0.4 + 0.4^3\times0.6$

$P(5) = 0.6^3\times0.4^2 + 0.4^3\times0.6^2$

However, the values given in my book are

$P(3) = 0.6^3 + 0.4^3$

$P(4) = 3\times0.6^3\times0.4 + 3\times0.4^3\times0.6$

$P(5) = 6\times0.6^3\times0.4^2 + 6\times0.4^3\times0.6^2$

• What do you mean by "Since the probability of player A winning is $0.6*(1-0.4) = 0.6^2$"? The probability of $A$ winning a set is simply $0{,}6$. May 15, 2014 at 9:26
• Of course, I meant to say "the probability of player A winning and player B not winning" May 15, 2014 at 9:51
• But if A wins, B is not winning... The events are not independant, they are the contrary of one another. You are just writing $P(A \wedge A)=P(A)^2$, which is of course wrong. May 15, 2014 at 10:07
• You're right of course! Thank you May 15, 2014 at 13:00
• In tennis, unlike baseball, basketball, and football (whether American-style in which contact between the ball and a foot generally means failure or settling for second-best outcome or rest-of-the-world style, i.e. futbol), whomsoever wins the last point/last game/last set wins the match. You forgot to take this into account in calculating $P(4)$ and $P(5)$. May 15, 2014 at 13:34

It's because of re-ordering. For the case of the game ending in 4 sets, there are 3 slots where the set won by the losing player can be placed. BAAA, ABAA, AABA

There are six ways for the losing player to place the two wins in the 5 sets situation.

I think you're forgetting to "count the possibilities" here.

What you gave as P(4) for example, is the probability of one specific way of ending the game after 4 sets. But in fact, there are more of them - let's denote A the sets where A wins, and B the sets where B wins. Then the game ends after 4 sets in the following cases:

AABA ABAA BAAA $\Rightarrow$ each has p = $0.6^3*0.4$, and there are 3 of them, so multiply by 3

BBAB BABB ABBB $\Rightarrow$ each has p = $0.6*0.4^3$, and again there are three of them

This gives you the factor of 3 you need (and indeed, for P(5) you will find 6 different outcomes for both players winning).

Hope this helps!

Anwer:

P(A) = .6

P(B) = .4

Player A winning in the 3rd Set = $0.6^3$

Player B winning in the 3rd Set = $0.4^3$

Player A winning in the 4th Set = BAAA, ABAA, AABA, in other words, B could have won in any of the three first games = ${3\choose 1} (.6)^3 (.4)$

Similarly B winning in the 4th Set = ABBB,BABB,BBAB, in other words, A could have won in any of the three first games = ${3\choose 1} (.4)^3 (.6)$

Player A winning in the 5th Set = B could have won in any of two games in the first four games = ${4\choose 2} (.6)^3 (.4)^2$

Similarly B winning in the 5th Set = A could have won in any of two games in the first four games = ${4\choose 2} (.4)^3 (.6)^2$

Add each pairs and you get P(3), P(4) and P(5)

Thanks Satish