How many people must there be in a room before the probability that someone has the same birthday as you do is at least $\frac{1}{2}$? How many people must there be before the probability that at least two people have a birthday on July 4 is greater than $\frac{1}{2}$?

The probability of a single individual having the same birthday as me is $\frac{1}{365}$ so the complementary event has probability $1 - \frac{1}{365} = \frac{364}{365}$. The probability that none of $k$ individuals have the same birthday as me is $\frac{364^k}{365^k}$. This probability exceeds $\frac{1}{2}$ when $k \geq \log_{\frac{364}{365}}(\frac{1}{2}) \approx 253$. So there must must be $253 + 1$ people (counting myself) in a room before the probability that someone has the same birthday as me.

It is simpler to calculate the probability that 0 or 1 people have a birthday on July 4 and then take the complement. So the probability that $k$ out of $n$ people have a birthday on July 4 is ${n \choose k} \frac{364^{n-k}}{365}$. So when $k = 0$ we have $\frac{364}{365}^n$ and when $k = 1$ we have $\frac{n364^{n-1}}{365^n}$. Solving this normally would be pretty obnoxious, so I opted to write a quick python script to determine that the first value of $n$ for when the sum of these two values exceeds $\frac{1}{2}$ is $n = 613$.

x = 1
result = 1
while result > .5:
    result = (((364/365)**x) * (1 + x/364))
    print(str(x) + ": " + str(result))
    x += 1