# Probability

• Jul 17th 2013, 07:28 AM
alexander9408
Probability
If a year is taken to have 365 days, including 29 February, and there are n (1<n<=365) people, show that the probability that at least two people have the same birthday is 1-(365!)/( (365^n)(365-n)! ).
• Jul 17th 2013, 07:59 AM
MINOANMAN
Re: Probability
• Jul 17th 2013, 08:16 AM
ebaines
Re: Probability
Rather than read how others solve this I suggest thinking it through. The probability of at least two people sharing a birthday is 1 minus the probability of there being no shared birthdays at all. Imagine just one person in the room, who has a particular birthday. Now a second person walks in - the probability of him not having the same birthday as the first person is 364/365, so the probability of them having the same birthday is 1-(364/365). Now imagine a third person entering the room - in order for him to not share a birthday with either of the first two people his birthday must fall on one of the 363 remaining open dates, and the probability of that is 363/365. Hence the probability of three people all having different birthdays is (364/365)(363/365), and the probability of at least one shared date is 1 minus that, or 1-(364 x 363/365^2). Continue on for 4 people in a room, etc. For the general case of n people in the room the probabiliy of no one sharing a common birthday is (364 x 363 x 362 x ... x (365-n+1))/(365^n). One minus this is the probability of at least one duplicate birthday:

$P(at\ least\ one\ shared\ date) = 1 - \frac {364 \times 363 \times 362 \times ... \times (365-n+1)}{365^n}$

You need to clean this expression up a bit to get it in the form they want, but hopefuly this helps you understand how it's arrived at.
• Jul 18th 2013, 08:16 AM
alexander9408
Re: Probability
Quote:

Originally Posted by ebaines
Rather than read how others solve this I suggest thinking it through. The probability of at least two people sharing a birthday is 1 minus the probability of there being no shared birthdays at all. Imagine just one person in the room, who has a particular birthday. Now a second person walks in - the probability of him not having the same birthday as the first person is 364/365, so the probability of them having the same birthday is 1-(364/365). Now imagine a third person entering the room - in order for him to not share a birthday with either of the first two people his birthday must fall on one of the 363 remaining open dates, and the probability of that is 363/365. Hence the probability of three people all having different birthdays is (364/365)(363/365), and the probability of at least one shared date is 1 minus that, or 1-(364 x 363/365^2). Continue on for 4 people in a room, etc. For the general case of n people in the room the probabiliy of no one sharing a common birthday is (364 x 363 x 362 x ... x (365-n+1))/(365^n). One minus this is the probability of at least one duplicate birthday:

$P(at\ least\ one\ shared\ date) = 1 - \frac {364 \times 363 \times 362 \times ... \times (365-n+1)}{365^n}$

You need to clean this expression up a bit to get it in the form they want, but hopefuly this helps you understand how it's arrived at.

Thank you, very well explained. But I still can't get the final answer.......
• Jul 18th 2013, 08:41 AM
ebaines
Re: Probability
I apologize, but I now realize that I made a mistake with the exponent in the denominator here:

Quote:

Originally Posted by ebaines
... For the general case of n people in the room the probabiliy of no one sharing a common birthday is (364 x 363 x 362 x ... x (365-n+1))/(365^n).

This should have been (364 x 363 x 362 x ... x (365-n+1))/(365^(n-1))

which changes the formula to this:

$P(at\ least\ one\ shared\ date) = 1 - \frac {364 \times 363 \times 362 \times ... \times (365-n+1)}{365^{n-1}}$

You should be able to get it now.
• Jul 18th 2013, 05:40 PM
Prove It
Re: Probability
The original statement is false, if you include February 29, there are 366 days.

You can start with ANY false statement and conclude that something is true.