In a tournament with 10 teams, each of them plays against each other exactly once. Additionally, there are no ties, and each team has a $50 \%$ chance of winning any match. What is the probability that, after tallying the scores from $\frac{10 \cdot 9}{2}=45$ games, no two players have the same number of wins?
Solution

Since each of the 45 matches has two possible outcomes, there are $2^{45}$ distinct ways to distribute the wins in the tournament. Each team can win from 0 to 9 matches. Since there are 10 teams, for all the numbers of wins to be distinct, each integer in the set $\{0,1,2, \ldots, 9\}$ must be associated as the number of wins for a single player. This can be done in 10! ways. It remains to determine if this distribution of win numbers is achievable in this tournament. If $j_{i}$ is the player who is supposed to win only $i$ matches, with $i \in\{0,1,2, \ldots, 9\}$, then make $j_{i}$ win against all players $j_{0}, j_{1}, \ldots, j_{i-1}$ and lose to all players $j_{i+1}, j_{i+2}, \ldots, j_{10}$. This will ensure the planned distribution. Therefore, the probability is $10!/ 2^{45}$.