Calling statisticians: The Birthday Problem

[Mears]

Some of you may have heard of the original birthday problem which tries to find the probability that at least two people out of N people will share the same birthday. It is pretty easy to figure out. All you do is find the probability that nobody will have the same birthday and then do 1 - that probability. For example for 4 people:
1 - (365/365) *(364/365) * (363/365) * (362/365) = 1.64%

Now what I'm wondering is how to find the probability that at least THREE people out of N people will have the same birthday. I was thinking that all I would have to do is multiply each part of the top equation by the probability that no two people up to that point share a birthday like so:

1 - (365/365)*0*(364/365)*.00274*(363/365)*.0082*(362/365)*.0164

But I don't think that works. Any suggestions.

 

[TuffGirl]

I'll take a stab at this. I used to love statistics in college.

If you want to calculate the exact probability, you first count how many possible groups of 3 are in the room, which is combination(n,3) or [n*(n-1)*(n-2)]/6.

In any 3-person group the chances that one person's birthday will match the other 2 people's is (1/365)^2. So a probability of failure of 1 - (1/365)^2. When you raise the probability of failure to the power of the number of possible triples, then you get:

[1 - (1/365)^2] ^ [n*(n-1)*(n-2) / 6] as the probability of failure of 3 of the same birthday in a group of n.

So the probability of success is 1 - [1 - (1/365)^2] ^ [n*(n-1)*(n-2) / 6].