9F (Probability)

Conditional on

N = n

, there are

(2 n)

pairs, and each is both friendly and birthday-matched with probability

1/ (2 \cdot 365)

. Since

E [N (N - 1)] = 36 5^{2}

for a Poisson variable,

E [matching friend pairs] = \frac{36 5 ^{2}}{2} \frac{1}{730} = \frac{365}{4} .

b

By Poisson thinning, allocating a Poisson

(365)

population independently and uniformly among

365

days gives independent Poisson

(1)

counts. Hence

P (Z_{1} = z_{1}, \dots, Z_{365} = z_{365}) = e^{- 365} i = 1 \prod 365 \frac{1}{z _{i} !}, z_{i} \in Z_{\geq 0} .

Given

Z_{i} = n

, the probability of no friendship among people sharing day

i

2^{- n (n - 1) /2}

. The day counts are independent, so

P (no matching friend pair) = [e^{- 1} n = 0 \sum \infty \frac{2 ^{- n (n - 1) /2}}{n !}]^{365} = (\frac{2 + C}{e})^{365} .

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