When there are 22 people in a room, what is the probability that two of them have the same birth days?

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- Feb 12th 2010, 05:40 PMPunchReal life probability...
When there are 22 people in a room, what is the probability that two of them have the same birth days?

- Feb 12th 2010, 07:42 PMRandom Variable
This is a classic problem referred to as the birthday problem

The probability that at least two people share a birthday is 1 minus the probability that no one shares a birthday.

Choose the first person. He can have any birthday. Now choose a second person. The probability that the second person doesn't' share the same birthday as the first person is 364/365 (excluding a birthday on Feb 29). Now choose a third person. The probability that the third person doesn't share the same birthday as the first or second person is 363/365. And so on.

P(none of the 22 people share a birthday) = $\displaystyle 1 \times \frac{364}{365} \times \frac{363}{365} \times...\times \frac{344}{365} = \frac{365 \times 364 \times 363 \times ... \times 344}{365^{22}} =\frac{365!}{343! \times 365^{22}}$

so P(at least two of the 22 people share a birthday) = $\displaystyle 1 - \frac{365!}{343! \times 365^{22}} \approx 0.476$ - Feb 13th 2010, 01:04 AMmr fantastic
- Feb 13th 2010, 05:55 AMPunch
- Feb 13th 2010, 11:00 AMRandom Variable
Choose 2 people. That can be done in $\displaystyle \binom{22}{2} = \frac{22 \times 21}{2} $ = 231 ways.

The probability that the two people share a birthday is 1/365. Chose a third person. The probabilty that the third person doesn't have the same birthday as the first two is 364/365. Choose a fourth person. The probability that the fourth person doesn't have the same birthday as the first two or the third is 363/365. And so on.

P(exactly 2 people share a birthday) = $\displaystyle 231 \times \frac{1}{365} \times \frac{364}{365} \times \frac{363}{365} \times ... \times \frac{345}{365} = 231 \ \frac{365!}{344! \times 365^{22}} \approx 0.352$