# The Birthday Problem

At a gathering of 25 people, a gambler offers you an even money bet that there are two people in the room who have the same birthday. (Neither you nor the gambler know the birthdays of the other people ahead of time.) If there are no two people with the same birthday, you win \$100. If there are two or more shared birthdays, the gambler wins \$100 from you. Should you take the bet?

Later, you are at a gathering of 100 people. The gambler says that since it is now more likely that there are two people with the same birthday, he will bet \$1 million against your \$100. If there are no two people with the same birthday, you win \$1 million. If there are two or more shared birthdays, the gambler wins \$100 from you. Should you take the bet?

The birthday problem leads many astray, because our intuition suggests a different answer than the mathematics of probability provides. You may naturally think in terms of how likely it would be for someone to share your birthday. Actually, in both gatherings, the odds would be in your favor if you took an even money bet that no one else had your birthday.

However, when you consider all the possible pairs, the odds are multiplied. In a group of 25, the odds that any two people will have the same birthday are 56.8%. (This calculation ignores leap years and assumes all birthdays are equally likely.) In fact, for any group of 23 or larger, the odds favor at least one shared birthday. Therefore, you should not take the gambler’s first bet.

As the number of people increases, the odds continue to multiply and quickly approach 100% (although they do not actually reach 100% until there are 366 people present). For 100 people, the odds that there is at least one shared birthday are 99.99997%. The chance that there is not at least one shared birthday is one in 3,254,690. Therefore you should also not take the gambler’s second bet, even though he is offering to bet \$1 million against your \$100.

The Math

Let’s begin with the case that may pop into mind first, but trips us up in the end: the probability that someone shares your birthday.

It may be intuitively obvious that the chances of one random person having the same birthday as you are 1/365, and that is correct (ignoring leap years and assuming all birthdays are equally likely). However, future calculations will be easier if we begin by figuring out the chances of no match instead of the chances of a match. In the case of one other person, the chances that they will not share your birthday are 364/365, or 0.9973. So the probability of a birthday match is 1 – 0.9973 = 0.0027, or 0.27%.

To calculate the probability of independent events occurring together, you multiply the probabilities. So for the case of there being 24 other random people who do not share your birthday, we multiply 364/365 or 0.9973 by itself 24 times. The result is 0.937, so the probability of someone sharing your birthday in a group of 24 other people is 1 – 0.937 = 0.063, or about a 6.3% chance. For a group of 99 others, it’s a 23.8% chance, and even with 365 other people, there is only a 63.3% chance of someone sharing your birthday. These are the type of low probabilities that may come to mind when you think of sharing birthdays.

There is a crucial difference with the probability of any two people sharing a birthday: you must consider not only you and every other person, but everyone with every other person. Again beginning with the probability of no match, the case of two people is the same: 364/365. For the third person, however, the chances of no match are lower, because their birthday may match the first person or the second person. The odds of no match for the third person are 363/365. We will subtract one for each person present, so for 25 people, the odds of no match are (364/365) * (363/365) * (362/365) * … continuing to 341/365. This gives us 0.431. So the odds of a match are 1 – 0.431 = 0.569, or 56.9%. Once you reach 70 people, the odds of a match are 99.9%. For 100 people the odds of a match are 99.99997%. You can use Wolfram Alpha to calculate the odds for any size group.