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

I have been looking at the birthday problem ( and I am trying to figure out what the probability of 3 people sharing a birthday in a room of 30 people is. (Instead of 2).

I thought I understood the problem but I guess not since I have no idea how to do it with 3.


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!

Source : Link , Question Author : irl_irl , Answer Author : Community

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