A "five nations" tournament problem I got this problem from the book Weighing in odds by D. Willians (page 97)
Each of five people plays each of the others in a fair game: the probability of winning is 1/2. Find the probability that each of them wins exactly two games. 
I got by numerical calculations involving a 10 x 5 matrix of range 4 that the probability should be $24\times 2^{-10}$ but I guess that there must be an easier answer 
 A: There are $10$ games, so $2^{10}$ possible results.   
Start with A.  There are six ways to choose the players he beats.  Now rename the players so he beats B and C.   
There are two results for the B vs C game.  Rename them so B beats C.   
Now C has two losses, so has to beat D and E.   
We have two possible results for D vs E, rename them so D beats E.   
E now has two losses, so has to beat B.  B has two losses, so must beat D.  That finishes the schedule.   
There were $6\cdot 2 \cdot 2=24$ choices, so the chance is $24 \cdot 2^{-10}$
A: There is $10$ game so $2^{10}$ possible configurations. To count the favourable odds, to begin we can consider only the player $a_1$ he has ${4\choose2}=6$. Now we studied the chance of $a_2$, a player who lost the match with $a_1$, he have two chance:
1) he win with $a_3$, the other player who lost the match with $a_1$, and this has $2$ possibility, so the configuration is complete because $a_3$ have to win with both the other player. In total $12$ possibility.
2) he lost with $a_3$, and in this case $a_3$ has to win one of two possible match ant the configuration is complete. So also in this case $12$ possibility.
Totally we have $24$ favourable odds so the probability is $\frac{24}{2^{10}}$.
A: If the five players are the vertices of a pentagon, the pentagon has five sides and five diagonals. So, there are ten games. Each game has two outcomes. So, there are $2^{10}$ total outcomes for all players and games.
The tree diagram shows all $3!=6$ possible outcomes among $ABC$:
$\hspace{4cm}$
The first outcome $A>B>C$ implies $A$ beats both $B$ and $C$, consequently loses to $D$ and $E$. Similarly, $C$ loses to both $A$ and $B$, consequently beats $D$ and $E$. Now we have $B$ losing to $A$ and beating $C$. In the next trial we will consider the outcomes for $BDE$. Since $B$ needs to win once, it can be $D$ or $E$. Finally, it remains to determine the relationship between $D$ and $E$.
Due to symmetry, there are $6\cdot 2\cdot 2=24$ outcomes. Hence, the required probability is: $\frac{24}{2^{10}}.$
