0
$\begingroup$

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

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

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

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

$\endgroup$
6
  • 1
    $\begingroup$ 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 $0{,}6$. $\endgroup$ May 15, 2014 at 9:26
  • $\begingroup$ Of course, I meant to say "the probability of player A winning and player B not winning" $\endgroup$ May 15, 2014 at 9:51
  • 1
    $\begingroup$ 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. $\endgroup$ May 15, 2014 at 10:07
  • $\begingroup$ You're right of course! Thank you $\endgroup$ May 15, 2014 at 13:00
  • $\begingroup$ 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 $P(5)$. $\endgroup$ May 15, 2014 at 13:34

3 Answers 3

2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.