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Probability of winning a best of $5$ game series given that the first game is lost.

Assuming that the two teams (A and B) are equally likely to win a game and the games are independent events, so I considered it like this:

If the first game team A loses $L????$, they must win $3$ out of the remaining $4$ games, so I got $\binom43$ ways that this could happen. I then divided it by the total number of ways to win $3$ out of the $5$ games $\binom53$, I then get $\binom43/\binom53$ as my final answer. Is this method appropriate?

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    $\begingroup$ What about winning 4 games? In the title you specified "Probability of winning a best of 5 game series" $\endgroup$
    – CSch of x
    Commented Oct 23, 2020 at 17:31
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    $\begingroup$ I don't understand the division, nor am I clear who "they" refers to. There are $4$ games left to be played. Assuming "they" means "$A$" then the probability that $A$ wins at least three of them is $\binom 43 \times \frac 1{2^4}+\binom 44 \times \frac 1{2^4}$. Note: convince yourself that it's fine to imagine that all $4$ remaining games are played, even if the series might be decided prior to game $\#5$. $\endgroup$
    – lulu
    Commented Oct 23, 2020 at 17:31
  • $\begingroup$ Hi, thanks for the help, I understand now $\endgroup$
    – Carefully unknown
    Commented Oct 23, 2020 at 17:49

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Your method isn’t quite right. There are indeed $\binom{4}{3}$ ways for the remaining 4 games to play out. But the probabilities for each possibility are determined as a binomial distribution:

$\Pr(LWWW) =\frac{1}{16}$

$\Pr(WLWW) =\frac{1}{16}$

$\Pr(WWLW) =\frac{1}{16}$

$\Pr(WWW) =\frac{1}{8}$ (Here the 5th game is not played.)

Then sum the probabilities to get a final answer of $\frac{5}{16}$.

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