Suvrat is perfectly correct, but in the interest of clarity, and yes pedantry, I'd like to add my \$0.02.

It may not be clear from above how to combine probabilities. You should associate AND with multiplication and OR with addition. Thus the probability of the person 1 having a unique birthday AND person 2 having a unique birthday AND etc. is

``` 365   364   363         365-(n-1)     365!     1
--- X --- X --- X ... X ---------  = ----- X -----
365   365   365            365         n!     365^n
```

Indeed, using this formula, here are the probabilities that in a group of n people there will be at least two people sharing the same birthday.

```    n     probability
-----  -------------
10       11.6948%
20       41.1438%
23       50.7297%
30       70.6316%
40       89.1232%
50       97.0374%
```

Well, what we neglected was leap day. This is a good approximation, but let's see if we can get the absolutely correct answer including leap day. Person X has one of 366 possible birthdays, but not with equal weight. To make it easier, let us group 4 consecutive years together. The chance of his/her birthday being January 21 is 4/(3*365 + 366) vs. the chance it's February 29 which is 1/(3*365 + 366) since there are 3*365 + 366 (=1461) days in 4 years, with 4 Jan 21th's and 1 Feb 29th. Note 3*365+366 = 4*365+1.

OK, remember we are going to compute the probability that every one of the n people have unique birthdays, than subtract that probability from 100%. So in our computation we have two cases to consider: (A) one where no one is born on Feb 29 and (B) one where 1 person is born on Feb 29. Let me drop the denominator, which will be n factors of 1461.

Case (A) is the same as our approximation before, except over 4 years instead of 1:

Num(PA) = (4*365 + 1)*(4(365-1)+1) *(4(365-2)+1)*...*(4(365-(n-1))+1).
Believe it or not, this product has a name, and is called the Pochhammer function, a ratio of Gamma functions.

Case (B) is more complicated. Let's say that the k'th person is the one with the Feb 29th birthday. Then

Num(PB)(k) = (4*365 + 1)*(4(365-1)+1)*...* (4(365-(k-2))+1)*(4(365-(k-2))+0)*(4(365-(k-1))+0)*...* (4(365-(n-2))+0).
This is so convoluted it's just a unnamed ratio of Gamma functions, as far as I know.

OK, I'm getting tired. Still, we aren't done since we have to sum over all possible k's. I'm not going to write that down, since there isn't a pretty closed form that would be enlightening. Schematically, it looks like

Prob = (1460/1461)*Num(PA)/1461 + (1/1461)SumkNum(PB)(k)/(1461*n)
Instead I wrote a python program to compute both the inexact and exact results. /msg me and I'll e-mail it to you. Someone checking it would be useful. In fact I'll check it more when I get time. Here's a table of the EXACT results (to 6 digits)
```    n     probability
-----  -------------
10       11.6867%
20       41.1213%
23       50.7045%
30       70.6056%
40       89.1057%
50       97.0297%
```

As a physicist I have to question my sanity and priorities in life for spending 1.5 hours of my life to compute a 10-3% effect in such an academic problem, but I did actually enjoy it.

More pedantry: what makes this a paradox??? Nothing! It's just farking cool!

Even more pedantry: derobert reminds me that leap day is suspended on years divisible by 100 unless the year is also divisible by 400, so my calculations are only exact for a room full of people born no earlier than 1 January 1901. Since 2000 had a leap day, you can use my code reliably until 2100, when your grandkid can add the appropriate correction factor.