How many people must be present before two of them are likely to have the same birthday? Since there are 365 days in a year, you might think you need 100 people or more. Actually 23 people gives you an even chance of finding a duplicate birthday. A few more, and the odds are definitely in your favor. This surprising statistic is called the birthday paradox.
The math is easy if you turn the problem around. What are the odds that every birthday is unique? The first one is free. The second person has a new birthday with probability 364/365. (I'm ignoring leap year.) The third person has a new birthday with probability 363/365, then 362/365, and so on. By the time we get to 343/365, the ratio is 49.27%. So we have a 50 50 shot at finding a duplicate birthday.
Assume there are n choices, and we are looking for duplicates. The odds of having no duplicates after k selections is the product of 1-i/n, as i runs from 0 to k-1. If k is small relative to n, the product is approximately 1 minus the sum of i/n. This in turn is 1 minus the sum of the first k integers divided by n. Set k to the square root of n, and the sum of the first k integers is half of n. Therefore, you have an even chance of finding a duplicate amongst k selections, where k is the square root of n.