Calculate the expectation of games won by a team I've come across this problem on a forum, seems interesting.
Assume a soccer team has played $N$ soccer games and has scored 0 goals and also has conceded 0 goals. Therefore we know the team won't be able to win any of the games. Now if the soccer team has scored $N$ goals and has also concede $N$ goals, what is the expectation of the games won by this team? 
 A: Let me first rephrase the question:
Let A and B be two soccerteams that have played exactly $N$ games against each other. It is known that after all these games, each team scored exactly $N$ goals. What is the expected number of games won by team A?
I will assume the following: if we consider, say, team A and one goal scored by that team, the probability that goal has been scored is the same for every game played and does not depend on the other goals scored. I think this is the most natural way to think about the problem, but it of course depends on the precise mathematical interpretation of something belonging to the real world.
Now for the solution: If $N_A$ is the expected number of games won by team $A$ and similarly for $B$ then by symmetry, $N_A= N_B$. Also, if $N_D$ is the expected number of draws, then $N_A + N_B + N_D = N$, so $N_A = \frac12 (N-N_D)$.
By symmetry again, $N_D = N \times P$, where $P$ is the probability the first match will be a draw. Then let $P[i]$ denote the probability that the number of goals scored by team $A$ in the first game equals $i$. (In particular $P[0] + \dots + P[N] = 1$.) By symmetry again, the same probabilities apply to team B and in particular, $P = \sum_{i=0}^N P[i]^2$, i.e. a draw occurs when both teams happen to score the same amount of goals in the first match. So we try to determine $P[i]$.
$P[i]$ is given by the number of "good combinations" / the total number of combinations. Here a "combination" is a sequence of $N$ numbers ranging from $1$ to $N$, such that the $k$'th number records in which match the $k$'th goal was scored. (E.g. for $N=3$ the sequence 1-2-1 would mean that of the 3 goals scored by team A, the first was scored in the first game, the second in the second game, the thirth in the first game.) There are of course $n^n$ possible combinations. The ones that contain exactly $i$ times $1$ are "good". To find these combinations, we may determine $\binom{N}{i}$ positions first and then fill in the other positions in $(N-1)^{N-i}$ ways with the remaining numbers from $2$ to $N$. This gives us
$$ P[i] = \frac{ \binom{N}{i} (N-1)^{N-i} }{N^N}. $$
Therefore the answer is given by
$$ N_A = \frac{N}2 \left[1 - \sum_{i=0}^N \left( \frac{\binom{N}{i}(N-1)^{N-i}}{N^N}\right)^2  \right].$$
Presumably this can further be simplified, and quietly I'm hoping there is a clever combinatorial argument that counts this number in another way, meanwhile simplifying this equation, but I'll leave it just here.
