4.5.1 problem 69

There are $\left(\frac{1000}{2}\right)$ pairs of sampled individuals - each pair has a $\frac{1}{10^6}$ chance of being the same person. Therefore, we can estimate the "rate of occurrence" of a pair being the same person as $\left(\frac{1000}{2}\right)\ast \frac{1}{10^6}=\frac{1000\ast 999}{2\ast 10^6}\approx \frac{10^6}{2\ast 10^6}=1/2$. Therefore, the number of pairs in the sample that are the same person can be approximated by Pois(1/2).

Then the probability that there is at least one pair in the sample that are the same person is $1-e^{-0.5}=0.393$. This can be verified as a close approximation in R - the probability that every individual in the sample is unique is the last value resulting from the command cumprod(1-(0:999)/1000000), which is .6067. 1 minus this value gives .3933, the actual probability some two sampled individuals are the same person, which is very close to our Poisson approximation.