### Re: Probability of people with the same birthday

Area: Other
Posted By: Michael Onken, WashU
Date: Wed Jul 16 17:15:57 1997
Area of science: Other
ID: 867408144.Ot
Message:

This question is one of the most common brain-teasers in statistics. The secret to the question is that it does not ask for a specific day of the year. So the more people you have, the more likely the next person added to the group will match any one of the other people. The statistical trick to the mathematical solution is to calculate the opposite question: If you have a room of 50 people, what is the probability of no one having a birthday on the same day? This is much easier to figure:

If the 50 people enter the room one at a time, then each person who enters has a probability of having a unique birthday equal to the number of possible unique days divided by the total number of available days, i.e. 365 (days in a year = total possible birthdays) minus the number of birthdays already taken (the number of people already in the room) divided by 365 (total possible birthdays).

```or:    365 - (n-1)
365
```

Where n is the number of the person entering the room, so n-1 is the number of people already in the room. Thus, the second person to enter would have a 364/365 probability of having a different birthday (n=2, so n-1=1, and 365-1=364), and the 26th person to enter would have a 340/365 probability of having a different birthday (n=26, so n-1=25, and 365-25= 340).

To get the total probability of all 50 people having different birthdays, we simply multiply the probabilities of each together:

```or:     365 x 364 x 363 x ... x 316
365 x 365 x 365 x ... x 365
```

Notice that the final probability is 316/365 for n=50, n-1=49, and 365-49= 316. Also notice that the first person to enter has a 365/365 or absolute probability of having a unique birthday: since there is no one else around, the birthday is unique. If you have either a good calculator with statistical functions or a cheap calculator with a lot of patience (guess which I had), you will get a solution of 0.02962642 for the above equation. This means that the probability of all 50 people having different birthdays is about 3%, or said another way:

• If you have a room of 50 people, there is a 97% probability of two people having a birthday on the same day.

That is a very high probability! In fact, with only 22 people, the probability is already better than 50%!

The statisticians in the audience will recognize the numerator of the equation as the permutation of 365 and 50, or 365!/(365-50)!, and the denominator as 365 ^ 50. So for any number n of people in the room, the probability of any two having the same birthday is:

```     365Pn
1 - ------
365n
```

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