[Combinatorics]
Problem: Blaise and Pierre will play 6 games of squash. Since they are equally skilled, each is equally likely to win any given game. (In squash, there are no ties.) The probability that each of them will win 3 of the 6 games is $\frac{5}{16}$. What is the probability that Blaise will win more games than Pierre?

Solution: There are two possibilities: either each player wins three games or one player wins more games than the other.

Since the probability that each player wins three games is $\frac{5}{16}$, then the probability that any one player wins more games than the other is $1-\frac{5}{16}=\frac{11}{16}$.

Since each of Blaise and Pierre is equally likely to win any given game, then each must be equally likely to win more games than the other.

Therefore, the probability that Blaise wins more games than Pierre is $\frac{1}{2} \times \frac{11}{16}=\frac{11}{32}$.
We consider the results of the 6 games as a sequence of 6 Bs or Ps, with each letter a B if Blaise wins the corresponding game or $\mathrm{P}$ if Pierre wins.

Since the two players are equally skilled, then the probability that each wins a given game is $\frac{1}{2}$. This means that the probability of each letter being a $B$ is $\frac{1}{2}$ and the probability of each letter being a $\mathrm{P}$ is also $\frac{1}{2}$.

Since each sequence consists of 6 letters, then the probability of a particular sequence occurring is $\left(\frac{1}{2}\right)^{6}=\frac{1}{64}$, because each of the letters is specified.

Since they play 6 games in total, then the probability that Blaise wins more games than Pierre is the sum of the probabilities that Blaise wins 4 games, that Blaise wins 5 games, and that Blaise wins 6 games.

If Blaise wins 6 games, then the sequence consists of 6 Bs. The probability of this is $\frac{1}{64}$, since there is only one way to arrange $6 \mathrm{Bs}$.

If Blaise wins 5 games, then the sequence consists of $5 \mathrm{Bs}$ and $1 \mathrm{P}$. The probability of this is $6 \times \frac{1}{64}=\frac{6}{64}$, since there are 6 possible positions in the list for the $1 \mathrm{P}$ (eg. PBBBBB,BPBBBB, BBPBBB, BBBPBB, BBBBPB, BBBBBP).

The probability that Blaise wins 4 games is $\left(\begin{array}{l}6 \\ 2\end{array}\right) \times \frac{1}{64}=\frac{15}{64}$, since there are $\left(\begin{array}{l}6 \\ 2\end{array}\right)=15$ ways for 4 Bs and 2 Ps to be arranged.

Therefore, the probability that Blaise wins more games than Pierre is $\frac{1}{64}+\frac{6}{64}+\frac{15}{64}=\frac{22}{64}=\frac{11}{32}$.

Final answer: $\frac{11}{32}$