You randomly choose four people. What is the probability that two or more of them have the same birthday? It's easy to calculate the probability that none of them have the same birthday (($365*364*363*362)/365^{4})$, and then subtract that number from $1$ to get the answer.
But is there a way to directly calculate the probability, instead of calculating the opposite event and subtracting from one? The choosing of the people is presumably an independent event so I thought calculating the probability directly wouldn't be difficult, but whatever I've tried has not given the right answer...
P(2 or more having the same birthday) = P(2 having the same birthday) + P(3 having same birthday) + P(all 4 having same birthday)
$={4 \choose 2}(\frac{1*364*363}{365^3})+{4 \choose 2}(\frac{1*1*364}{365^3})(\frac{1}{2})+{4 \choose 3}(\frac{1*1*364}{365^3})+{4 \choose 4}(\frac{1*1*1}{365^3}) = .01635591246$