[mikeisgus]

I have 200 Facebook friends and 5 of them share the same birthday. None are related to one another. I thought this was a bit strange.

[s_cianci]

Why do you think it's so strange? There are only 365 dates in a year (366 if you want to count leap years). With the world population at around 6 billion people, by doing some quick arithmetic, that amounts to over 16 million people per date. While this is not a complete solution to the entire question of the probability of "x out of y" people sharing the same birth date, it gives you at least a basic idea of approximately how many people would share a common birth date.

[Perito]

Figure out the probability that they won't share the same birthday and subtract it from 1.

For the person, the probability is 365/365 (we ignore leap year here). For the next person, the probability is 364/365. For the third person, the probability of is 363/365. Continue to the 200th person. Multiply all of the probabilities together and subtract from 1. You'll find that it's not so improbable. If you had 366 friends, it's sure that at least two of them would share the same birthday.

[galactus]

It may seem rather counterintuitive, but if we put 23 people in the same room, the probability that at least 2 of them have the same birthday is about

.

EDIT:

This is tougher than I previously stated.

This can indeed be represented by a Poisson where we choose 5 from the 200 that share birthdays.



If n=200, then



Most of these birthday problems deal with 'at least' 2 or 3 sharing a birthday or something like that.

[WWIIVet]

This is the same as tossing 200 balls into 365 boxes. Divide the balls into two groups of 5 and 195. Put the 5 balls into the first box. This can be done in C(200,5) ways. The other 195 balls can be distributed in 364^195 ways, The total number of possible ways is 365^200. Thus the desired probability is C(200,5)*364^195 divided by 365^200. Using logarithms will help the computations

[galactus]

This is the same as tossing 200 balls into 365 boxes. Divide the balls into two groups of 5 and 195. Put the 5 balls into the first box. This can be done in C(200,5) ways. The other 195 balls can be distributed in 364^195 ways, The total number of possible ways is 365^200. Thus the desired probability is C(200,5)*364^195 divided by 365^200. Using logarithms will help the computations

That is similar to what I had posted previously, but edited/changed it. I got to looking into this and found perhaps there was more to it. But then again... perhaps not.

The method I outlined in my previous post is from a stats and probability book.

I used Poisson and binomial to arrive at the same solution as WWIIVET.



Binomial(same as WWIIVET's solution).



I agree WWIIVET. I should have stayed with my original post.

[mikeisgus]

Thank you all for your input. I do not understand enough about math to follow your equations, but I do appreciate that there are people like you who do. LOL