# Discrete Mathematics Geometric Distribution

Many sports leagues decide their champions using a best-of-seven format (e.g., the MLB, NBA, NFL, etc.) There is a Team A and a Team B. The two teams play each other for at most seven games. The first team that wins 4 games is declared the winner. Assume that the games are in dependent of each other.

Suppose Team A is the better team and has a $$60\%$$ chance of winning each game.

a. What is the probability that Team A will win the championship in $$7$$ games?

b. What is the probability that Team A will win the championship?

I think that for part A it is $$(4/10)^6 \cdot (6/10)$$ because that would mean they win on the 7th game, but I'm not sure if that's what it is asking for. And for Part B wouldn't it just be $$60\%$$?

Any help would be nice, thank you!

• Welcome to MathSE. This tutorial explains how to typeset mathematics on this site. For team $A$ to win the tournament in exactly seven games, team $A$ must win exactly three of the first six matches and then win the seventh game. Your factor of $(\frac{4}{10})^6$ is the probability that team $B$ wins all six of the first six games. Of course, after team $B$ won the first four games, the tournament would have been over, with team $B$ winning. Dec 2, 2018 at 23:51
• Hey, thank you I need look at the tutorial im sorry, so would it be something like (4/10)^3 * (6/10)^4, that would mean that team B wins 3 games and then team A wins 4 games out of 7. Dec 3, 2018 at 1:02
• The quantity $(\frac{4}{10})^3\left(\frac{6}{10}\right)^4$ represents the probability that team $B$ wins three games and then team $A$ wins four games in that order. Dec 3, 2018 at 1:21

For team $$A$$ to win the tournament in exactly seven games, team $$A$$ must win exactly three of the first six games and then win the seventh game.

Since team $$A$$ has probability $$\frac{6}{10}$$ of winning each game, the probability that team $$A$$ wins exactly three of the first six games can be determined using the binomial distribution. The probability that exactly $$k$$ successes occur in $$n$$ trials, each of which has probability $$p$$ of success, is given by the formula $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $$\binom{n}{k}$$ is the number of ways exactly $$k$$ successes could occur in $$n$$ trials, $$p^k$$ is the probability of $$k$$ successes, and $$(1 - p)^{n - k}$$ is the probability of $$n - k$$ failures. Thus, the probability that team $$A$$ wins exactly three of the first six games is

$$\binom{6}{3}\left(\frac{6}{10}\right)^3\left(\frac{4}{10}\right)^3$$

Multiplying this result by the probability that team $$A$$ wins the seventh game gives the probability that team $$A$$ wins the tournament in exactly seven games.

$$\binom{6}{3}\left(\frac{6}{10}\right)^3\left(\frac{4}{10}\right)^3 \cdot \frac{6}{10}$$

For the second part, observe that if team $$A$$ wins the tournament, it must win in exactly four, five, six, or seven games.

• Okay I really understand now. My professor never referred to this as Binomial Distribution, after reading and taking a look at the equation it makes sense. The probability that they win in game 7 is .1659, and for part 2 it is .7102 that they win the championship I believe. What I did was use the same equation but change the combination C(n + 1 , k) where k was always 3. Dec 3, 2018 at 2:15