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

\[n=20\]

.