# Birthday Problem without using complement

I have a solution to the birthday problem without using complements that is arriving at the wrong answer. I'd like to understand what I am doing wrong. I am not looking for alternate solutions to the problem.

Problem

Assuming there are only 365 days (ignore leap year), and each day is equally likely to be a birthday, what is the probability that at least 2 people have the same birthday in a room of N people?

Sample Space: $$365^N$$

Event Space

• $${N\choose 2 }$$ pairings for people with the same birthday
• for each pair, $$365$$ possible birthdays
• for remaining $$N-2$$ people, $$365^{(N-2)}$$ permutations which we can basically ignore (but still must be counted since they are part of the event space)

So I would expect the answer to be:

$$\frac{{N\choose 2 } * 365 * 365^{(N-2)}}{365^N} = \frac{{N\choose 2 }}{365}$$

With $$N=23$$, I get 69% chance of $$2$$ people having same birthday, but correct answer is ~50%. So where am I over-counting?

• If you have two pairs of people with the same birthday, or 3 people with the same birthday they will be over-counted. You will need to use inclusion - exclusion to remove the over-counts. Feb 12 '19 at 18:15
• Assume we have 3 people P1,P2,P3, and P1,P2 have the same birthday on Jan 1st. Then both of these data points should be included the event space: (1) P3 has birthday on Jan 1st (2) P3 does not have birthday on Jan 1st It doesn't seem like over-counting. Basically, once we narrow in on pair+birthday (ie: P1,P2 on Jan 1st), we still have 365 different birthdays for P3 and irrespective of what that day is, our event has at least 2 people with the same birthday. Feb 12 '19 at 18:31

If we have 4 people

We could have 4 different birthdays. $$\frac {365!}{(365-4)!} \frac {1}{365^4}$$

1 pair of birthday twins. $$\frac {365!}{(365-3)!} {4\choose 2}\frac {1}{365^4}$$

2 pair of birthday twins. $$\frac {365!}{(365-2)!} {4\choose 2,2}\frac {1}{365^4}$$

1 set of birthday triplets . $$\frac {365!}{(365-2)!} {4\choose 3}\frac {1}{365^4}$$

All 4 on the same day: $$\frac {365!}{(365)!} {4\choose 4}\frac {1}{365^4}$$

Some pair of same birthdays...

$$\frac {6}{365} - \frac {11}{365^2} + \frac {6}{365^3}$$

Which is less than $$\frac {{4\choose 2}}{365}$$