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Is the following correct? My friend tells me i need to (add up the binomial probabilities of 2 wins in 2 sets, 2 wins in 3 sets and 2 wins in 4 sets THEN multiply the sum obtained by p)

P(3) = 5C3 x 0.5^3 x 0.5^2

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Are matchups independent? – Macro Jan 13 '13 at 6:03

The probability of winning a 5 set match is the sum of the probabilities of the events $$\begin{align*} \text{win in}~ 3:& WWW\\ \text{win in}~4:& LWWW,WLWW,WWLW\\ \text{win in}~5:& LLWWW, LWLWW,LWWLW, WLLWW, WLWLW, WWLLW \end{align*}$$ If the probability of winning a set is $p$ and the probability of losing a set is $q = 1-p$, the probability of winning a 5 set match works out to be $$p^3 + 3p^3q + 6p^3q^2 = p^3(1+3q+6q^2)$$ which of course is $\frac{1}{2}$ when $p=\frac{1}{2}$ as Thierry Silbermann has already pointed out to you. See also an article titled "The Drunken Tennis Player" in Ian Stewart's Game, Set and Match.

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Well to win a 5 sets match, a tennis player needs to win at least 3 sets.

During a match you can have $2^5$ possibles outcomes, and on these 32 possible outcomes, the player will win on 16 configurations (10 if he wins 3 sets, 5 if he wins 4 sets and 1 if he wins 5 sets)

Each outcome have the same probability, so the result is:

$P$(3 winning sets) + $P$(4 winning sets) + $P$(5 winning sets) $= 10/32 + 5/32 + 1/32 = 16/32$ so $1/2$ to win

You can look at this to train:

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