SOLUTION TO PROBLEM 1

Let us denote this probability by p(n). If q(n) is the probability of the event that no two people have their birthday on the same day, then p(n) = 1 - q(n).

Let' s find q(n). We know that the probability can be computed as the ratio of the number of favorable cases to the number of all possible cases.

How many favorable cases are there? The first person has 365 choices. The second person has only 365 - 1 choices since he/she should not be born on the same day with the first person. The third person has 365-2 choices since this person's birthday should be different than the birthdays of the first two people. And so on .... The n-th person has 365 - n +1 choices by the same argument. Therefore the number of favorable cases is

$$365 \times 364 \times 363 \times \dots \times (365-n+1).$$

The number of total possible cases is

$$365 \times 365 \times \dots \times 365 = 365 ^n$$

by the argument that any person could be born in any day of the year.

Thus,

$$q(n) = {{365\times 364 \times \dots \times (365-n+1)}\over {365 ^n}}$$

and

$$p(n) = 1 - {{365\times 364 \times \dots \times (365-n+1)}\over {365 ^n}}.$$

By a straightforward computation one gets (Surprise!!!)

$$p(25) = 0.57, \quad p(40) = 0.89, \quad p(55) = 0.99.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

    Back to the Problem of the Week Page
 
 

Last modified:   Fri Aug 24 14:03:39 CDT 2001