Question

If people are present in a room, what is the probability that no two of them celebrate their birthday on the same day of the year? how large need be so that this probability is less than ?

1. diemkieu
The number needed so that this probability is less than 1/2 is 23.

### How to illustrate the information?

It should be noted that 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.
To find the probability that both the second person and the third person will have different birthdays:
(365/365) * (364/365) * (363/365) = 132
132/133 = 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)
= 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).
365! / ((365-n)! * 365^n).
We know that the probability of finding at least two people with the same birthday is 1 minus the probability that everybody has a different birthday. It turns out that the smallest class where the chance of finding two people with the same birthday is more than 50% is… a class of 23 people. The probability is then about 50.73%.