Double birthday problem

Question

I know the single coincidence has been thoroughly discussed here and elsewhere, but I happen to have a group of 20 close friends where we have not one but TWO birthday coincidences. Being the data artist in the group, I've decided to explore this and share. It is extremely unique, I believe, but how rare do probs say it is?

Answer 1

If you want to get exactly two matches (on different dates), the probability is:

$$\frac{{{365}\choose{2}}\cdot {{363}\choose{16}}\cdot{{20}\choose{2}}\cdot{{18}\choose{2}}\cdot16 ! }{365^{20}}\approx .0713$$

Here is the reasoning for the numerator: Choose two dates of year for the matches. Choose $16$ remaining dates for nonmatching birthdays. Choose two of $20$ slots for the earlier match date. Choose two of remaining $18$ slots for the later match date. Make an arrangement of the $16$ unmatched dates in the remaining $16$ slots.

I ran a Python simulation one million times, and the results seemed in agreement.

Answer 2

Let's discard the 29th of February for simplification.

Number of cases: $365^{20}$

Number of cases without two birthsday at the same date: $365!/345!$

Number of cases with two birthsday at the same date and no other coincidence: $\binom{20}{2}*365!/346!$

Number of cases with two birthsday at one date, two birthdays and another date, and no other coincidence: $\binom{20}{2}\binom{18}{2}*365!/347!/2$

Answer 3

Assuming that all $365$ birthdays are equally probable (ignoring leap years for example):

The probability that all birthdays are unique is $$1\times \frac {364}{365}\times \cdots \times \frac {365-19}{365}\approx 0.589$$

The probability that there is exactly one match is $$\binom {20}2\times \frac 1{365}\times \frac {364}{365}\times \cdots \times \frac {365-18}{365}\approx 0.323$$

Thus the probability that there are two or more matches is $$1-0.589-0.323\approx 0.088$$