4.5.4 problem 73

(a)

Let $X$ be the number of people that play the same opponent in both rounds. Let $I_j$ be the indicator variable that person $j$ plays against the same opponent twice. $P(I_j=1)=\frac{1}{99}$. Then, $\text{E}\left(X\right)={\sum }_{j=1}^{100}P(I_j=1)=100/99$.

(b)

There is a strong dependence between trials. For instance, if we know that the first $50$ players played the same opponent twice, then all of the players played the same opponents twice. Moreover, knowing each of the $I_i$ gives us perfect information about one other $I$ - they are strongly pairwise dependent.

(c)

Consider the $50$ pairs that played each other in round one. Let $I_j$ be the indicator variable for pair $j$ playing each other again in the second round. $P(I_j=1)=\frac{1}{99}$. Then, the expected number of pairs that play the same opponent twice is $\text{E}\left(Z\right)=\frac{50}{99}\approx \frac{1}{2}$.

We can approximate the number of pairs that play against one another in both rounds with $Z\sim \text{Poiss}\left(\frac{1}{2}\right)$. Note that $X=2Z$. $P(X=0)\approx P(Z=0)=e^{-\frac{1}{2}}\approx 0.6$.

$P(X=2)\approx P(Z=1)=\frac{{\left(\frac{1}{2}\right)}^1{e}^{-\frac{1}{2}}}{1!}\approx 0.32$.

Note that the approximation in part C is more accurate - the independence of the same pairs playing against each other is much stronger than the independence of individuals who play the same opponent. Knowing that the players in Game 1 of round 2 played against each other in round 1 gives us very little information about whether players in any other games also played against each other. Whereas, knowing that Player 1 in round 2 plays against the same player (say, player 71) guarantees that we know that that player 71 also plays against the same player.