Question

Ignore, for simplicity, the possibility of being born on February 29th. Assume that a person is just as likely to be born on any one date as on any other date, excluding February 29th, of course. Suppose 253 people are picked at random. (We use 253 since e” = 1/2 almost exactly. Use 1/2 in your computations.)

a.  What is the expected number of different birthdays of the group?

b. The expected number of days which are each the birthday of at least two of the people?

The answers for these are c 182.5 d 56

Answer 1

a)

[Let us assume there are D days in a year, in order to be more general.]

Actually, the expected number of unique birthdays is very easy to calculate. Let e(n) denote the expected number of unique birthdays when there are n people.

Choose one of the n people. Then either they have a different birthday than everybody else, or they don’t. They have a different birthday than the other n-1 with probability . Let us write for , and so the probability that they have a different birthday to everybody else is . In this case, the expected number of unique birthdays is 1+e(n-1). In the other case, which thus has probability  , the expected number is just e(n-1).

Hence,

As e(1) = 1, it follows that:

D = 365 , n = 253

hence

e(253) = 365*(1 - (364/365)^253)= 182.674

b)

X = number of days with at least 2 birthdays
X = sum Ii
E(X) = sum E(Ii)
= sum P(day i has at least 2 birthdays)
= sum (1- P(day i has 0 birthdays) - P(day i has 1 birthday)

365*(1 - (364/365)^253 - 253 (364/365)^252 * 1/365)
= 55.94768

=56