Birthday Problem using Complement Probability [duplicate]

Question

I'm having trouble attempting this question. In a group of $n$ people, what is the probability that at least $3$ have the same birthday?

I started off by saying Pr(At least $3$) $= 1 -$ (Pr(no $3$) + Pr($2$ share a birthday)).

However, how would i calculate that exactly $2$ people have the same birthday, or is this wrong? Any suggestions on how to continue?

Answer 1

I think you could proceed in this way: $P(at \ least \ 3) = P(at \ least \ 2) - P(exactly\ 2)$

since $P(at \ least \ 2) = 1 - P(no \ one)$

thus $P(at \ least \ 3) = 1 - P(no \ one) - P(exactly\ 2)$

where

$$P(no \ one)=\frac{365!}{(365-n)!365^n}$$

and

$$P(exactly\ 2)=\binom{n}{2} \frac{365!}{(365-n+1)!365^n}$$

for the derivation of the last equation the ingredients are as follow:

1. Select a pair, the probability that they share the same birthday date is: $\frac{1}{365}$
2. The probability that the remaning "n-2" people have different birthday date is: $\frac{364\cdot363\cdot...\cdot(364-n+3)}{365^{n-2}}$
3. As to select a pair there are $\binom{n}{2}$ choices for the pair, the totalprobability is given by the following product:

$$\binom{n}{2} \frac{1}{365} \frac{364\cdot363\cdot...\cdot(364-n+3)}{365^{n-2}}=\binom{n}{2} \frac{364!}{(364-n+2)!365^{n-1}}=\binom{n}{2} \frac{365\cdot364!}{(364+1-n+2-1)!365\cdot365^{n-1}}=\binom{n}{2} \frac{365!}{(365-n+1)!365^n} \ \square$$ references

Probability question (Birthday problem)