# How do I finish this math problem?

So, I managed to start the problem and got somewhere. I think what I've done so far is at least right.

The problem goes as following:

How big does a group have to be for the probability that at least 2 in the group share the same birthday surpasses 50 percent?

So to start, I assumed that there are n people in the group. And the chance of at least 2 sharing a birthday is 1 minus the chance that no one shares a birthday.

Calculating the probability that no one shares a birthday, the first person would be allowed to have 365 out of 365 possible birthdays, and the second person is allowed to have 365-1=364 birthdays out of 365 since 1 birthday is already taken. So the probability will be

(365/365) * (364/365) * (363/365) * .... * (365-n+1/365) for n people, or simplified: (365! / ((365^n)(365-n)!))

and the probability of there existing overlapping birthdays is 1 - (365! / ((365^n)(365-n)!)) and putting it greater that 0.5 gives us

1- (365! / ((365^n)(365-n)!)) > 0.5 or that (365! / ((365^n)(365-n)!)) < 0.5

But I am stuck here and have no idea whatsoever as how to solve for the inequality. Been trying for an hour to get anywhere. Does anyone have any ideas?

The answer to the problem is supposed to be 23, so the equation should give an answer that is close to 23.

### 5 Answers

- PuzzlingLv 76 months agoFavorite Answer
The easiest way to solve that is to iteratively try larger values of n, rather than trying to "solve" that complicated inequality.

You know you are looking for an integer, so try some guesses. You can even narrow it down using a binary search.

You'll find that 22 is one one side of the boundary and 23 is on the other. So the answer is 23 people.

I treated it as an equality and gave the problem to WolframAlpha and it came back with:

n ≈ 22.7676903156182

But you can't have a fraction of a person so round up to 23 people.

- Log in to reply to the answers

- DixonLv 76 months ago
You have done the hard work but you can't solve that equation algebraically, it just needs trial and error.

Note that since you are looking for an integer solution it is not one of these iterative solutions where you just get endless decimal places. You will find one integer gives below 0.5 and the next gives above 0.5

- Log in to reply to the answers

- Mike GLv 76 months ago
The Probability of having at least 2 sharing a birthday in a group of n people

Your method is correct and if you continued you would end up with (for exactly P = 0.5)

P = 1 - nPr(365,n)/365^n= 0.5

when n = 23

P = 1-nPr(365,23)/365^23

P = 0.507 = 50.7%

- Log in to reply to the answers

- What do you think of the answers? You can sign in to give your opinion on the answer.
- Barkley HoundLv 76 months ago
Rather that repeat it here look at this page.

- Log in to reply to the answers