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The Chicago Cubs are playing a best-of-five-game series (the first team to win 3 games win the series and no other games are played) against the St. Louis Cardinals. Let X denotes the total number of games played in the series. Assume that the Cubs win 59% of their games versus their arch rival Cardinals and that the probability of winning game is independent of other games.

(a) calculate the mean and standard deviation for X.

In this problem I think I understand how to calculate the mean and standard deviation, but first you have to calculate the total combinations of plays. Is there a way using combinations or permutations to derive all possibilities rather than drawing a tree?

The tricky thing with this using combinatorics is that its a best of five. So I don't see how to use permutations directly because as soon as you have 3 wins the series is over--meaning you'll over count if you don't take this into consideration.

I'd appreciate any feedback. Thank you!


k=3

$$ \begin{array}{c|c|c} \text{Loss} & \text{Win} & \text{Combinations}\\ \hline 3 & 0 & 1\\ 2 & 1 & 3\\ 1 & 2 & 3\\ 0 & 3 & 1\\ \end{array} $$ Total Combinations = 8

k=4

$$ \begin{array}{c|c|c} \text{Loss} & \text{Win} & \text{Combinations}\\ \hline 1 & 3 & 4\\ 2 & 2 & 6\\ 3 & 1 & 4\\ 4 & 0 & 1\\ \end{array} $$

**Here have to subtract 1 from total combinations because cannot have WWWL.

Total combinations = 15-1 = 14

k=5

$$ \begin{array}{c|c|c} \text{Loss} & \text{Win} & \text{Combinations}\\ \hline 2 & 3 & 10\\ 3 & 2 & 10\\ 4 & 1 & 5\\ 5 & 0 & 1\\ \end{array} $$

**Here have to subtract from total combinations because cannot have WWWLL and LWWWL. Total combinations = 26-2 = 24

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  • $\begingroup$ I don't think it helps in this case, but in some similar situations it can be very helpful to pretend that the games go on to the end. $\endgroup$ – dfeuer Jun 6 '13 at 22:34
  • $\begingroup$ IMO, if you can solve a question more than one way, then do it. If you get the same answer both times, you can be more confident that your answer is correct. If you get a different answer with each approach, then you have the opportunity to double-check your assumptions and work. $\endgroup$ – Code-Guru Jun 6 '13 at 22:38
  • $\begingroup$ @Code-Guru What does IMO mean? $\endgroup$ – user1527227 Jun 6 '13 at 22:39
  • $\begingroup$ IMO = In my opinion. $\endgroup$ – Code-Guru Jun 6 '13 at 22:45
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Let us solve a more general problem. Imagine we have a series between two teams, A and B. Team A wins a game with probability $p$, and we make the independence assumption of the OP. Let us suppose that the first team to win $w$ games wins the series. Let $X$ be the number of games. Then $w\le X\le 2w-1$. Yours is the case $w=3$.

For any $k$ in this interval, we find the probability that $X=k$. In order for the series to have length exactly $k$, one of the teams has to have won $w-1$ of the first $k-1$ games, and has to win the last game.

The probability that A wins the series in $k$ games is therefore $$\binom{k-1}{w-1}p^{w-1}(1-p)^{w-k}p=\binom{k-1}{w-1}p^w(1-p)^{k-w},$$ and the probability that B wins the series in $k$ games is $$\binom{k-1}{w-1}(1-p)^wp^{k-w}.$$ The events A wins the series in $k$ games and B wins the series in $k$ games are disjoint, so to find $\Pr(X=k)$ we add the two expressions above.

Remark: A distribution closely related to the above calculations is the negative binomial. For details, Wikipedia has a discussion, as do many introductory probability books.

For the case $w=3$, we don't need all this machinery, and can simply draw a tree.

Let $p=0.59$. We have $\Pr(X=3)=p^3+(1-p)^3$. For $k=4$, we get $3p^3(1-p)+3(1-p)^3p$. For $k=5$ we get $6p^3(1-p)^2+6(1-p)^3p^2$. All these expressions can be simplified somewhat.

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  • $\begingroup$ +1 For a great explanation for applying the binomial distribution. $\endgroup$ – Code-Guru Jun 6 '13 at 22:40
  • $\begingroup$ "In order for the series to have length exactly 3, one of the teams has to have won 2 of the first 2 games and has to win the last game." In other words, one team won the first three games, so the series is over. $\endgroup$ – Code-Guru Jun 6 '13 at 22:41
  • $\begingroup$ I don't see why $k=6$ is mentioned, that can never happen. I would be careful about trying to guess general rules from the "win $3$" case. Best of $7$ gets more complicated, but the general analysis in my answer holds. Your addition did not address the probabilities. We are only interested in situations where the series is won in $3$, $4$, $5$. $\endgroup$ – André Nicolas Jun 6 '13 at 22:45
  • $\begingroup$ @user1527227: About your first comment, sure it is true. The series lasts $3$ games precisely if it is a "sweep." So yes, one team has to win the first two and win the third. Looks like a funny way of viewing things, but "series lasts exactly $4$" and "series lasts exactly $5$" falls under the same pattern. Perhaps I will add a remark at the end. $\endgroup$ – André Nicolas Jun 6 '13 at 22:51

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