Consider the following problem, from Understanding Probability by Henk Tijms.
In the World Series Baseball, the final two teams play a series
consisting of a possible seven games until such time that one of the
two teams has won four games. In one such final, two unevenly matched
teams are pitted against each other and the probability that the
weaker team will win any given game is equal to $p=0.45$. Assuming that
the results of the various games are independent from each other,
calculate the probability of the weaker team winning the final. What
are the expected value and the standard deviation of the number of
games the final will take?
The probability of the weaker team winning is:
$$
\sum_{k=0}^3 \binom{7}{k} (1-p)^kp^{7-k}\simeq 0.3917.
$$
For the expected value of the number of games, I consider the following table, where I list the number of possible wins for each of the two teams:
$$
\begin{array}{cc}
\text{strong team} & 4 & 4 & 4 & 4 & 0 & 1 & 2 & 3 \\
\text{weak team} & 0 & 1 & 2 & 3 & 4 & 4 & 4 & 4 \\
\end{array}
$$
which leads me to the expression:
$$
\sum_{k=0}^3 (4+k)\binom{4+k}{k}p^k(1-p)^4 + \sum_{k=0}^3 (4+k)\binom{4+k}{k}p^4(1-p)^k \simeq 8.622
$$
the result, however, is wrong. The only thing that I can think of is that I am counting in a wrong way the number of games with a given number of wins for the two teams.
I have found the solution of this problem here, but why is my approach wrong?
Best Answer
When you count all possibilities that the series finishes, e.g., 4-2 in favor of the stronger team, you count every time the stronger team wins 4 games and the weaker team wins 2, in any order, as a 6-game series. However, if the stronger team wins games 1,3,4,5, and the weaker team wins games 2,6, the series lasted only 5 games (and game 6 was never played). So not every arrangement of 4 strongenr-team-wins and 2 weaker-team-wins gives a legitimate 6-game series.