Question

Each of n people are randomly and independently assigned a number from the set {1, 2, 3, . . . , 365} according to the uniform distribution. We will call this number their birthday. (a) What is the probability that no two people share a birthday? (b) Use a computer or calculator to evaluate your answer as a decimal for n = 22 and n = 23.

Answer 1

a)

• P(event happens) + P(event doesn't happen) = 1
  P(two people share birthday) + P(no two people share birthday) = 1
  P(two people share birthday) = 1 - P(no two people share birthday).

So, what is the probability that no two people will share a birthday?

Again, the first person can have any birthday. The second person's birthday has to be different. There are 364 different days to choose from, so the chance that two people have different birthdays is 364/365. That leaves 363 birthdays out of 365 open for the third person.

To find the probability that both the second person and the third person will have different birthdays, we have to multiply:

(365/365) * (364/365) * (363/365) = 132 132/133 225,
which is about 99.18%.

If we want to know the probability that four people will all have different birthdays, we multiply again:

(364/365) * (363/365) * (362/365) = 47 831 784/ 48 627 125,
or about 98.36%.

We can keep on going the same way as long as we want. A formula for the probability that n people have different birthdays is

• ((365-1)/365) * ((365-2)/365) * ((365-3)/365) * . . . * ((365-n+1)/365).

If you know permutation notation, you can write this formula as

• (365_P_n)/(365^n).

That's the same as

• 365! / ((365-n)! * 365^n).

b) if n = 23

365! / ((365-23)! * 365^23)

P = 0.4927

if n = 22

365! / ((365-22)! * 365^22)

P = 0.5243