Expected number of games in the Baseball World Series 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?
 A: Your expression is almost right with the choose formula on both terms to be ${(4+k-1)\choose k}$ the reason being, the last one of p or (1-p) is fixed and that the the k games of the loser on both terms should find their way within (total number of games - 1) slots with the last being fixed.  Other than that your logic is right and is no different than the solution that you referred to.
A: @BallBoy's answer explains why your calculation is incorrect. To fix this, note that you should only count combinations where the $k$ games won by the losing team do not include the last game (the losing team can't win the last game). So that gives
$$\sum_{k=0}^3(4+k)\binom{3+k}kp^k(1-p)^4+\sum_{k=0}^3(4+k)\binom{3+k}kp^4(1-p)^k\approx 5.78$$
A: 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.
