Probability of 3 people in a room of 30 having the same birthday

The birthday problem with 2 people is quite easy because finding the probability of the complementary event "all birthdays distinct" is straightforward. For 3 people, the complementary event includes "all birthdays distinct", "one pair and the rest distinct", "two pairs and the rest distinct", etc. To find the exact value is pretty complicated.

The Poisson approximation is pretty good, though. Imagine checking every triple and calling it a "success" if all three have the same birthdays. The total number of successes is approximately Poisson with mean value ${30 \choose 3}/365^2$. Here $30\choose 3$ is the number of triples, and $1/365^2$ is the chance that any particular triple is a success. The probability of getting at least one success is obtained from the Poisson distribution: $$ P(\mbox{ at least one triple birthday with 30 people})\approx 1-\exp(-{30 \choose 3}/365^2)=.0300. $$

You can modify this formula for other values, changing either 30 or 3. For instance, $$ P(\mbox{ at least one triple birthday with 100 people})\approx 1-\exp(-{100 \choose 3}/365^2)=.7029,$$ $$ P(\mbox{ at least one double birthday with 25 people })\approx 1-\exp(-{25 \choose 2}/365)=.5604.$$

Poisson approximation is very useful in probability, not only for birthday problems!

An exact formula can be found in Anirban DasGupta, The matching, birthday and the strong birthday problem: a contemporary review, Journal of Statistical Planning and Inference 130 (2005), 377-389. This paper claims that if $W$ is the number of triplets of people having the same birthday, $m$ is the number of days in the year, and $n$ is the number of people, then

$$ P(W \ge 1) = 1 - \sum_{i=0}^{\lfloor n/2 \rfloor} {m! n! \over i! (n-2i)! (m-n+i)! 2^i m^n} $$

No derivation or source is given; I think the idea is that the term corresponding to $i$ is the probability that there are $i$ birthdays shared by 2 people each and $n-2i$ birthdays with one person each.

In particular, if $m = 365, n = 30$ this formula gives $0.0285$, not far from Byron's approximation.

As being pointed out by Micheal Lugo the formulation given by Anirban DasGupta is a exact answer for this problem, however a formal proof is needed. I have found and verified a solution by Doctor Rick from Math Forum, below is the link

http://mathforum.org/library/drmath/view/56650.html

His approach is to partition the sample space as following:

   1.   none share a birthday
   2.   one pair shares a birthday
   3.   two pairs share different birthdays
   4.   three pairs share different birthdays
   :
 1+N/2. N/2 pairs share different birthdays
 2+N/2. three or more share a birthday

Then he points out a clever way to count for each partition by picking different birthday for each pair of person. I have tried and arrived with the same formulation as Anirban DasGupta's. For more detail please take a look at the link above!