# Real life probability...

• Feb 12th 2010, 05:40 PM
Punch
Real 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 PM
Random 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 AM
mr fantastic
Quote:

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

Did you mean exactly two or at least two (the answers will be different for each case ....)
• Feb 13th 2010, 05:55 AM
Punch
Quote:

Originally Posted by mr fantastic
Did you mean exactly two or at least two (the answers will be different for each case ....)

Sorry, I meant at least two. However, is it possible to solve for two?
• Feb 13th 2010, 11:00 AM
Random Variable
Quote:

Originally Posted by Punch
Sorry, I meant at least two. However, is it possible to solve for two?

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) = $\displaystyle 231 \times \frac{1}{365} \times \frac{364}{365} \times \frac{363}{365} \times ... \times \frac{345}{365} = 231 \ \frac{365!}{344! \times 365^{22}} \approx 0.352$