# Why do these two methods of calculating the probability of winning a best-of-7 series give the same answer?

I was having a discussion with a friend about the probability, and we came up with very different methods to solve it that lead to the same answer. The problem is pretty simple: you have two teams A and B playing a best of 7 series where wins are independent and the probability that team B wins is .35. What's the probability of team B winning th series? Friend says, "this is just a binomal random variable $X$ with $n=7$ and $p=.35$ and we are looking for $P(X\geq 4)$, which according to my TI-83 here is .1998".

I was rather convinced that this must be wrong. If we think of the series as a sequence of A's and B's, then we are looking for the probability of obtaining a sequence of length $n=4,5,6,7$ with 4 B's and the last element is a B. I was sure that his method will include the probability of obtaining, say, {B,A,B,B,B,A,A}, which we are not interested in and in fact couldn't even ever occur. So I figure that what we really want for a sequence of length $n$ is the probability of obtaining a sequence of length $n-1$ with exactly 3 B's, and then tacking a B onto the end. So for a sequence of length $n$, the probability should be $\binom{n-1}{3}.35^4 .65 ^{n-4}$, and then the answer should be $\sum_{n=4}^{7}\binom{n-1}{3}.35^4 .65 ^{n-4}$. I was confident that I was right and he was wrong, but then I plugged that into Wolfram Alpha and got... .1998.

What's going on here? Is it a coincidence?

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Why do you say that {B,A,B,B,B,A,A} could not ever occur? This looks like a valid sequence of wins to me... –  Code-Guru Jul 18 '12 at 19:22
Once team B wins 4 games the series is over. –  crf Jul 18 '12 at 19:24
See David Spencer's answer below. –  Code-Guru Jul 18 '12 at 19:32
To make more money, the league has decided that the series will go $7$ games, but the usual rules for determining the winner (first to win $4$) apply. Then Team B wins the modified series iff it wins the real series. –  André Nicolas Jul 18 '12 at 20:54

This is not a coincidence, you are looking at the same problem in two different ways. The situation {B,A,B,B,B,A,A} is irrelevant in your friend's argument, because you are not counting the number of different possible sequences of events.

Rather, you are attempting to estimate the probability of at least $n$ successes in $k$ trials, which is exactly what the binomial distribution does.

In your second approach, you are essentially estimating the number of possible legal sequences terminating in a victory condition, and then computing the total fraction out of all possible sequences.

Both approaches are equivalent. One is a little easier to implement ;)

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He is correct. Although this does include series that would be over after fewer than $7$ games "in real life", they would still all result in team B winning, no matter what happens after team B gets $4$ wins.

In your example, the probability that one of

{B,A,B,B,B,A,A}

{B,A,B,B,B,A,B}

{B,A,B,B,B,B,A}

{B,A,B,B,B,B,B}

happening without terminating series after four wins is equivalent to the probability of {B,A,B,B,B} happening across the first five games because later events do not affect anything.

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