Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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 ;)

share|cite|improve this answer

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





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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.