I don't think this is exactly the same as the proposed problem. What it's giving you is the expected value. This includes in its calculation the lower probability cases and the higher probability cases. For instance, there's a slim change of this happening when the number of employees is 365. It's been a while since I took probability, but I think the expected value is going to be larger than the minimum value needed to achieve a .5 probability if the probability graph is skewed, which in this case it is.
I have no answer for you. I mean... the solution seems to be 2121 or 2122 (no, it isn't, see below) determined by brute force, but I don't know why.
But, it's worth noting that if this interests you, Project Euler almost certainly will as well and is worth checking it out if you don't already know about it:
EDIT: So if you're wondering why I got a different brute force result from other users, it's because I had a stupid bug. I was looking at the ratio of meeting the criteria to the ratio of not meeting it, not the ratio of meeting it to total.
Brute force gives me about 2287/2288. Trying each number 1,000,000 times (distribute X balls in 365 bins, count number times all bins are filled) so it seems reliable, but maybe I have bug somewhere:
import java.util.Random;
public class Birthday {
private static Random rand = new Random(System.currentTimeMillis());
public static void main(String... args) {
final int n = 365;
boolean[] bins = new boolean[n];
final float maxNumberOfAttempts = 1000000;
for (int numberOfBalls = 2285; numberOfBalls <= 2400; ++numberOfBalls) {
int attempts = 0;
int numberOfTimesAllBinsFilled = 0;
while (attempts < maxNumberOfAttempts) {
if (distribute_X_Balls_in_N_bins_randomly_and_return_the_number_of_filled_bins(
numberOfBalls, bins) == n) {
numberOfTimesAllBinsFilled++;
}
++attempts;
}
float prob = numberOfTimesAllBinsFilled / maxNumberOfAttempts;
System.out.println((prob > .5 ? "PASSED:" : "FAILED:") + numberOfBalls + ":" + prob);
}
}
public static int distribute_X_Balls_in_N_bins_randomly_and_return_the_number_of_filled_bins(
int x, boolean[] bins) {
for (int inx = 0; inx < bins.length; ++inx) {
bins[inx] = false;
}
int whichBin = 0;
int total = 0;
while (--x >= 0) {
whichBin = rand.nextInt(bins.length);
if (!bins[whichBin]) {
bins[whichBin] = true;
++total;
if (total == bins.length) {
return total;
}
}
}
return total;
}
}
This isn't the birthday problem. OP is asking how many people to hit every possible birthday at least once. I thought Wikipedia might cover this on their birthday problem page, but they don't.
http://en.wikipedia.org/wiki/Coupon_collector%27s_problem
The formula in the article (1/2 + n*gamma + n log n) gives 2365 for the expected number, but this is different than "at least 50% chance".