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Teams A and B are in a five-game playoff series; the team that wins three games is the team that wins the series. Assume that both teams are evenly matched (50/50 chance of winning).
A) Team A won the first game. What is the probability that team B will win the series? B) Continue to assume that Team A has already won one game, the the teams are not evenly matched. Assume that B is a better team. It's better in that its probability of beating team A in any one game is .55. What is the probability that team B will win the series?

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    Hint: the teams may continue playing a total of five games even after a winner has been determined. The winner of the series won either $3, 4$ or $5$ of the games in the series. Consider the use of a binomial distribution to continue. – 2017-02-14
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    Not for this question. It is the first to 3 victories wins the series. There will be no more games after a team is a winner. I calculated that the probability of team B winning 3 games is (1/2)^3. But is this answering the question? – 2017-02-14
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    Yes, I know that it is first to three victories wins the series, but they can play "*exhibition matches*" after the series has officially ended. The calculations are made much simpler by making this observation. The probability that team $B$ wins three games **in a row** is indeed $(1/2)^3$ but it is not asking about three games in a row, it is asking about three games before $A$ wins three games., which corresponds directly to **at least three games** out of five games (*where games could continue to be played even after an official winner of the series has already been determined*) – 2017-02-14
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    So what are you saying is the answer? I'm confused. – 2017-02-14
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    I'm saying you should think about my hint and come up with an answer yourself. I'm saying an incredibly convenient tool to use for the problem is the [binomial distribution](https://en.wikipedia.org/wiki/Binomial_distribution), and I'm saying that the way to turn your question into one that uses the binomial distribution is to allow the players to play exhibition matches after a winner has already been determined. What is the probability that $B$ wins exactly three matches out of the five given that $A$ won the first game? This is the same as $B$ winning three of the four final games... – 2017-02-14
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    @JMoravitz using the formula you gave me for the previous problem, would it be ∑k=3,4(4k)(1/2)^K*(1/2)^4-K??? – 2017-02-14
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    Once you fix all of your formatting issues, (*visit [this page](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) for information on how to type mathematical equations on this site*), yes $\sum\limits_{k=3}^4 \binom{4}{k}(1/2)^k(1/2)^{4-k}$ is the correct answer for part (a), and the answer for part (b) is achieved similarly by replacing each of the $1/2$'s with something more appropriate. (*for adding only two things, people often don't bother using summation notation and just write the two terms being added*) – 2017-02-14
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    Awesome. Thank you. I wasn't sure how to type the correct notation. Thank you for your help today @JMoravitz – 2017-02-14

1 Answers 1

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There are better ways of solving this problem, like drawing a probability tree, but a brute-force solution follows to determine $P\left(W\right)$, the probability that Team B wins the series.

For a small set of possible outcomes, it’s easy to list the outcomes that would result in Team B winning the series (ignoring the first game since it’s already happened), where $A$ is the event that team A wins the match and $B$ is the event that team B wins the match. We can ignore any scenario in which A wins twice since that would mean B loses the series, as A has a win already. Thus, the outcomes in which Team B wins the series form the set:

$$\{ BBB, ABBB, BABB, BBAB \}$$

Note that outcomes $BBBB$ and $BBBA$ are not listed; they can only occur after B has won and are thus irrelevant.

Each outcome in the set is a sequence of events, and each event has a probability of happening: $P\left(A\right)$ and $P\left(B\right)$. Since one event does not affect another’s probability, the probability of an outcome is the product of the probabilities of its constituent events, leaving us with the terms:

$$ P\left(B\right)\cdot P\left(B\right)\cdot P\left(B\right) \\P\left(A\right)\cdot P\left(B\right)\cdot P\left(B\right)\cdot P\left(B\right) \\P\left(B\right)\cdot P\left(A\right)\cdot P\left(B\right)\cdot P\left(B\right) \\P\left(B\right)\cdot P\left(B\right)\cdot P\left(A\right)\cdot P\left(B\right) $$

If any one of these outcomes occurs, Team B wins the series, so we can add these terms together:

$$ P\left(W\right)= \left[P\left(B\right)\cdot P\left(B\right)\cdot P\left(B\right)\right] +\left[P\left(A\right)\cdot P\left(B\right)\cdot P\left(B\right)\cdot P\left(B\right)\right] +\left[P\left(B\right)\cdot P\left(A\right)\cdot P\left(B\right)\cdot P\left(B\right)\right] +\left[P\left(B\right)\cdot P\left(B\right)\cdot P\left(A\right)\cdot P\left(B\right)\right] $$

This simplifies to $P\left(W\right)={P\left(B\right)}^{3}+3\cdot \left[{P\left(B\right)}^{3}\cdot P\left(A\right)\right]$. Plug your probabilities in: either $P\left(A\right)=P\left(B\right)=0.5$ or $P\left(A\right)=0.45$ and $P\left(B\right)=0.55$, depending on the part of the problem you’re solving. That should be 0.3125 for the part where the teams are equal and 0.39098125 for the part where Team B has a 55% chance of winning each match.