<s>[INST] Playing tennis until one of the players wins 3 times (Binomial distribution) Question
Nadal and Federer playing tennis against each other.
Nadal's probability to win one match is $\frac{2}{3}$, independently from the previous results.
The two are playing until one of them wins 3 matches.
What is the probability for Nadal's victory?
My Take
Nadal's probability to win is Nadal's probability to win exactly 3 matches which is:
$$
{5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}=\frac{80}{243}
$$
The Book's Answer
According to the book I should take into consideration Nadal's chances of winning 3\4\5 times:
$$
{5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}+{5 \choose 4}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)^{5}=\frac{64}{81}
$$

I can't tell why that's the right answer.
As far as I understand the game should stop after Nadal gets 3 points, so why adding the probability of him winning 4\5 matches?
 [/INST] The final score could be 3-2 for Nadal in which case we must have seen 5 experiments with 3 Nadal victories and 2 Federer ones. So $\binom{4}{2}=6$ ways to "place" two of those victories among the first $4$ games and then the final game must be Nadal's. So the probability of that happening is
$$6 \times (\frac{2}{3})^2  \times (\frac13)^2 \times \frac23 = \frac{48}{243}$$
To get 3-1 we need one Federer win among the first 3 and then a final Nadal one and so we get
$$3 \times (\frac23)^2 \frac13 \times \frac23  = \frac{24}{81} = \frac{72}{243}$$
For 3-0 we just have
$$(\frac23)^3 = \frac{8}{27}=\frac{72}{243}$$
So in total $$\frac{72+72+48}{243}= \frac{192}{243} = \frac{64}{81}$$
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