Birthday problem, the hard way(not using 1-unfavourable outcomes).

Question

How would you go about calculating the chance of two people having the same birthday in a room of 3 people and a year consisting of 365 days?

Answer 1

There are $3$ main cases:

• person 1 = person 2,
• person 2 = person 3, and
• person 1 = person 3.

Each case has a $365^2$ number of possibilities and they are independent so you add them not multiply. However we end up adding the special case of person 1 = person 2 = person 3 twice more than intended so you must subtract it out.

$$365\cdot 365\cdot 3=399675$$

$$399675-2\cdot 365=398945$$

$$398945/(365^3)=0.008204$$

Answer 2

Hint: $$P(A)=1-\frac{_{365}P_n}{365^n}$$ in this case $n=3$, see Birthday Problem.