Also having birthday today?  math puzzle
Out of how many people should a group at least consist in order to have that the probability that two persons out of that group are having birthday on the same day, is larger than 1/2? 
Explanation
The to many people surprising answer is that you only need at least 23 people in order to assure that the probability that two people are having birthday on the same day is larger than 1/2.
The probability that at least two people are having birthday on the same day is 1 minus the probability that everybody is having birthday one a different day. The probability that everybody out of a group with x people is having birthday on a different day is equal to P(x) = [365*364*363*....*(366x)]/(365^x). x = 23 is the smallest number for which 1  P(X) > 1/2. This proves that the group should contain at least 23 people in order to have a probability that at least two people are having birthday on the same day is larger than 1/2.
The probability that at least two people are having birthday on the same day is 1 minus the probability that everybody is having birthday one a different day. The probability that everybody out of a group with x people is having birthday on a different day is equal to P(x) = [365*364*363*....*(366x)]/(365^x). x = 23 is the smallest number for which 1  P(X) > 1/2. This proves that the group should contain at least 23 people in order to have a probability that at least two people are having birthday on the same day is larger than 1/2.
