Playing tennis until one of the players wins 3 times (Binomial distribution)

Question

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:

$$ {5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2}+{5 \choose 4}\left(\frac{2}{3}\right)^{4}\left(\frac{1}{3}\right)+\left(\frac{2}{3}\right)^{5}=\frac{64}{81} $$

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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?

Answer 1

$ \displaystyle {5 \choose 3}\left(\frac{2}{3}\right)^{3}\left(\frac{1}{3}\right)^{2} ~ $ considers that there are five matches and Nadal wins exactly three of the five matches.

The book's solution considers that Nadal may win three or more of the five matches which is the correct solution. On your point about matches not continuing if Nadal won the first three matches or three out of four matches, here is an alternate way to think -

$(i)$ Nadal wins straight first three matches and we stop.

$(ii)$ He loses exactly one of the first three matches and wins fourth match. We then stop.

$(iii)$ He loses exactly two of the first four matches and wins fifth match.

So Nadal's chances of winning is:

$ \displaystyle \left(\frac{2}{3}\right)^{3} + {3 \choose 1} \cdot\frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3} + {4 \choose 2} \left(\frac{1}{3}\right)^{2} \left(\frac{2}{3}\right)^{3} = \frac{64}{81}$

Answer 2

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 $$\frac{72+72+48}{243}= \frac{192}{243} = \frac{64}{81}$$