Groups of three

Topic: Random variables and distributions · more questions on this topic
‹ Expected wait for double six A memoryless component ›

One million people are divided, as far as possible, into disjoint groups of three. Birthdays are independent and uniform on 365 days.

1. How many groups are expected to have all three birthdays equal?

2. Approximate the probability that at least one such group occurs.

Worked answer (click to show)

One million is not divisible by 3, so there are 333333 groups and one person left over. The probability that three independent birthdays coincide is 1/365^2.

1. The number N of matching groups is \operatorname{Bin}(333333, 1/365^2), so \mathbb{E} N=333333/365^2\approx 2.50.

2. \mathbb{P}(N>0)=1-(1-1/365^2)^{333333}\approx 0.92. The same figure comes from the Poisson approximation with mean 2.5: 1-e^{-2.5}\approx 0.918.