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I am trying to brush up my knowledge on probability in the library, and I found this question which got me stumped:

Two tennis professionals, A and B, are scheduled to play a match; the winner is the first player to win three sets in a total that cannot exceed five sets. The event that A wins any one set is independent of the event that A wins any other, and the probability that A wins any one set is equal to .6. Let x equal the total number of sets in the match; that is, x = 3, 4, or 5. Find p(x).

Here's my take:

x=3

For the match to end in 3 rounds, I calculated the probability of A winning all three rounds and B winning all three rounds. $$ p(x=3) = \binom{3}{3}(0.6^3)(0.4^{0}) + \binom{3}{0}(0.6^0)(0.4^{3}) = 0.2800 $$

x=4

For the match to end in 4 rounds, I calculated the probability of A winning all three rounds and B winning all three set, but both losing one set. $$ p(x=4) = \binom{4}{3}(0.6^3)(0.4^{1}) + \binom{4}{1}(0.6^1)(0.4^{3}) = 0.4992 $$

x=5

Did the same thing here: $$ p(x=5) = \binom{5}{3}(0.6^3)(0.4^{2}) + \binom{5}{2}(0.6^2)(0.4^{3}) = 0.5760 $$

I think p(x=3) is correct, but for p(x=4) it seems like it includes p(x=3), the same for p(x=5) which includes p(x=3) and p(x=4). The sum of p(x) = 1.3552, so clearly it's wrong.

I'd really appreciate it if you could help.

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    $\begingroup$ By cases $x=4,5$ you overlooked the extra condition that the last set must be won by the winner of the match. $\endgroup$ – drhab Apr 5 '17 at 9:56
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If $X = 4$, there are two cases, namely, $A$ wins or $B$ wins. For the first case, $A$ must win in the $4$th set and win exactly two sets in the first three sets. Therefore, $$ \Pr(X = 4, A \text{ wins}) = 0.6 \cdot \binom{3}{2}(0.6)^2(0.4)^1 $$ Similarly, $$ \Pr(X = 4, B \text{ wins}) = 0.4 \cdot\binom{3}{2}(0.4)^2(0.6) $$ The case of $X = 5$ is left as an exercise.

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  • $\begingroup$ Thanks! I understand much better now. $\endgroup$ – bot1131357 Apr 5 '17 at 11:31

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