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.


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?

  • 3
    $\begingroup$ 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. $\endgroup$
    – Doug M
    Feb 12 '19 at 18:15
  • $\begingroup$ 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. $\endgroup$
    – Nick Desai
    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 $\frac {{4\choose 2}}{365}$


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