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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!

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  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Dec 2 '18 at 23:51
  • $\begingroup$ 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. $\endgroup$ – STPM222 Dec 3 '18 at 1:02
  • $\begingroup$ 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. $\endgroup$ – N. F. Taussig Dec 3 '18 at 1:21
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No. For the first you want the probability that team A wins their fourth victory on the seventh game. Thus that they win thrice among the first six, then win on the seventh game.

For the second part, you want the probability that team A wins their fourth victory on the fourth, fifth, sixth, or seventh game.

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  • $\begingroup$ Hey, so for part A would that be (4/10)^3 * (6/10)^4 ? $\endgroup$ – STPM222 Dec 3 '18 at 1:19
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    $\begingroup$ Count the ways three victories may be arranged among six games. $\endgroup$ – Graham Kemp Dec 3 '18 at 1:25
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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.

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  • $\begingroup$ 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. $\endgroup$ – STPM222 Dec 3 '18 at 2:15
  • $\begingroup$ Your answers are correct. $\endgroup$ – N. F. Taussig Dec 3 '18 at 8:41

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