Discussion

Answer: A

The mean is 12 pairs of throws and the variance is 100, so the answer is 12 and 100.

Track the distance between the frisbees around the pentagon: 1 (adjacent) or 2, with 0 ending the game. Each frisbee moves one step left or right, so the gap changes by -2, 0 or +2 with probabilities 1/4, 1/2, 1/4. From distance 1 the gap stays at 1 with probability 3/4 and becomes 2 with probability 1/4; from distance 2 it ends with probability 1/4, becomes 1 with probability 1/4 and stays at 2 with probability 1/2.

Let \(E_{1}\), \(E_{2}\) be the expected numbers of throws from each distance:

\[ \begin{aligned} E_{1} &= 1 + \tfrac{3}{4}E_{1} + \tfrac{1}{4}E_{2} \\ E_{2} &= 1 + \tfrac{1}{4}E_{1} + \tfrac{1}{2}E_{2} \end{aligned} \]

\[ \begin{aligned} E_{1} &= 4 + E_{2},\quad E_{2} = 2 + \tfrac{1}{2}E_{1} \\ E_{1} &= 12,\quad E_{2} = 8 \end{aligned} \]

For the variance, the fundamental matrix of the absorbing chain is \(N = \begin{pmatrix} 8 & 4 \\ 4 & 4 \end{pmatrix}\), and the variance of the absorption time from distance 1 is the first entry of \((2N - I)t - t^{2}\) with \(t = (12, 8)\):

\[ 15(12) + 8(8) - 12^{2} = 244 - 144 = 100 \]

The other pairs do not satisfy these equations; 6 and 50 or 24 and 200 come from halving or doubling both values.