Six teams play a tournament in which every team plays every other team exactly once. No ties occur, and each team has a $\dfrac{1}{2}$ probability of winning any game it plays. Find the probability that no two teams win the same number of games.
This is what I have so far:
There are $\binom{6}{2} = 15$ pairs of teams, and $2^{15}$ possible outcomes. The min and max possible # of games won are from $0$ to $5$. If $h$ represents the # of games on by a certain team, than $0 \leq k \leq 5$. Because of this, there are $5!$ outcomes in which no two teams win the same number of games. Therefore, the probability is: $\dfrac{5!}{2^{15}}$. When simplified, we get $\dfrac{15}{4096}$. However, when I imputed this answer into the question, it was wrong. Where was my error, and how can I fix it?
NOTICE: The probability of a team winning in each game is 1/2 , NOT 1/12.
Best Answer
The total number of games won must add up to $15$, since there were $15$ games. Each team must have won between $0$ and $5$ games. It's fairly obvious that the only way for none of the teams to have the same number of wins as another is if the six teams won $0,1,2,3,4$, and $5$ games.
I think this is where you made the innocent mistake. There are $6$ numbers to be distributed among the $6$ teams, which can be done in $6!=720$ different ways (not $5!=120$). The answer you seek is $\frac{720}{2^{15}}=\frac{45}{2048}$.