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

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?

• What about winning 4 games? In the title you specified "Probability of winning a best of 5 game series" Commented Oct 23, 2020 at 17:31
• 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 $\#5$.
– lulu
Commented Oct 23, 2020 at 17:31
• Hi, thanks for the help, I understand now Commented Oct 23, 2020 at 17:49

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 $$\frac{5}{16}$$.