Question

Sixteen players $S_1,S_2,....S_{16}$ play in a tournament. They are divided into eight pairs at random. From each pair a winner is decided on the basis of a game played between the two players of the pair. Assume that all the players are of equal strength. The probability that exactly one of the two players $S_1$ and $S_2$ is among the eight winners is $\frac{k}{15}$.Find the value of $k$?

Answer 1

If $S_1$ and $S_2$ are in the same pair, then exactly one wins:
If $S_1$ and $S_2$ are in two pairs separately, then exactly one of $S_1$ and $S_2$ will be among the eight winner.
If $S_1$ wins and $S_2$ losses, $S_1$ losses and $S_2$ wins
Now the probability of $S_1,S_2$ being in the same pair and one wins $=$ (probability of $S_1,S_2$ being the same pair) $\times$ (probability of anyone winning in the pairs)
And the probability of $S_1{,S}_2$ being the same pair $=\frac{n\left(E\right)}{n\left(S\right)}$
where $n\left(E\right)=$ the number of ways in which 16 person can be divided in 8 pairs.
$\therefore n\left(E\right)=\frac{\left(14\right)!}{{(2!)}^7.7!}$ and $n\left(S\right)=\frac{16!}{{(2!)}^8.8!}$
Therefore probability of $S_1$ and $S_2$ being in the same pair $=\frac{\left(14\right)!.{(2!)}^8.8!}{{(2!)}^7.7!\left(16\right)!}=\frac{1}{15}$
The probability of any one winning in the pairs of $S_1,S_2=P=1$
Therefore The pairs of $S_1,S_2$ being in two pairs separately and $S_1$ wins,$S_2$ losses + The probability of $S_1,S_2$ being in two pairs separately and $S_1$ losses, $S_2$ wins.
$=[1-\frac{\frac{\left(14\right)!}{{(2!)}^7.7!}}{\frac{16!}{{(2!)}^8.8!}}]\times \frac{1}{2}\times \frac{1}{2}+[1-\frac{\frac{\left(14\right)!}{{(2!)}^7.7!}}{\frac{16!}{{(2!)}^8.8!}}]\times \frac{1}{2}\times \frac{1}{2}$
$=\frac{1}{2}\times \frac{14\times \left(14\right)!}{15\times \left(14\right)!}=\frac{7}{15}$
Therefore required probability $=\frac{1}{15}+\frac{7}{15}=\frac{8}{15}$