# What is the probability of winning a best of 5 game, if I already won the first game?

For example, if I am playing a game with a friend and we both are equally likely to win each game, then what is the probability of me winning the best of 5 match, given that I already won the 1st game? I did some work on this question, but I don't think this method seems so sound.

$$P$$(I win in 3 games) = $$(\frac{1}{2})^2$$

$$P$$(I win in 4 games) = $$(\frac{1}{2})^3 \cdot \frac{2!}{1! \cdot 1!}$$

$$P$$(I win in 5 games) = $$(\frac{1}{2})^4 \cdot \frac{3!}{1! \cdot 2!}$$

$$P$$(me winning) = sum of the above probabilities = 0.6875

You did that exactly right:

to win in 3 games, you must win the next 2, so that's a 1 in 4 chance

To win in 4 games, you must win game 4, and of the two games in the middle four you must win one and lose one, and there are two ways to do that, so that's 2 out of 8

To win in 5 games, you must win game 5, and 1 out of the 3 in between, and there are 3 ways to do that, so that's 3 out of 16

And, since these are all independent events, you can add them all up. Well done!

To win your friend in the best $$5$$ games, given that you have already won $$1$$ game, you must win at least $$2$$ out of remaining $$4$$ games.

Let $$X$$ be the number of your wins in these $$4$$ games. This is binomial experiment, where $$X\sim Bin(4,0.5)$$. So: $$P(X\ge 2)=P(X=2)+P(X=3)+P(X=4)=\\ {4\choose 2}\cdot \frac1{2^4}+{4\choose 3}\cdot \frac1{2^4}+{4\choose 4}\cdot \frac1{2^4}=\\ \frac{6+4+1}{16}=\frac{11}{16}\approx 0.6875.$$