Suppose we have a group of n people.There are 364 days in every 3 out of 4 years, and 365 days in 1 out of 4 years. The probability of any two randomly selected people having the same birthday is then
\[\frac{3}{4} \times \frac{1}{364} + \frac{1}{4} \times \frac{1}{365}\]
In a group of
\[n\]
people there are \[\frac{n(n-1)}{2}\]
pairings so the probability of any two people from this group having the same birthday is \[\frac{n(n-1)}{2}( \frac{3}{4} \times \frac{1}{364} + \frac{1}{4} \times \frac{1}{365})\]
We require that this be more than a half, so solve
\[\frac{n(n-1)}{2}( \frac{3}{4} \times \frac{1}{364} + \frac{1}{4} \times \frac{1}{365}) \gt \frac{1}{2}\]
This simplifies to
\[5836n^2-5836n-2125760=0\]
The solution to this equation are
\[n=19.59, \: -18.59\]
to 2 decimal places.Obviously we take the first of these and round it up, taking
\[n=20\]
.