birthday odds

[low_delta]

So let's say you have a group of ten people, and you want to determine the odds of two of them having the same birthday. I'm not certain, but I think it goes like this...

One person has 357 of 365 chances not to match. So that's 355/365. Each person has this chance for each of the other nine people. So anyone's odds of matching any other person are 1-(356/365)^9.

Maybe.

But I have no clue how to calculate this for three people matching.

Comments

[vwip.livejournal.com]

I think the chance of there not being two people in a group of n people who share a birthday is:

365!/((365-n)!*365^n)
(*100 if you want a percentage)

For 10, this would be:
365*364*363*362*361*360*359*358*357*356 / 365*365*365*365*365*365*365*365*365*365

1 - this would give you the chance of there being at least one shared birthday

[low-delta.livejournal.com]

Okay. I came up with a couple different answers. This was one of them. That's about a 12% chance.

Any idea how to attack the other problem?

[vwip.livejournal.com]

The chance of any three people in a group sharing any one birthday?

Can't be sure, but I guess it'd be the chance of any one of the other 8 sharing a birthday with the two who already share a birthday, multiplied by the 12% chance of those two sharing a birthday in the first place.

[marswalker.livejournal.com]

Combinatorics... so much fun!

http://en.wikipedia.org/wiki/Binomial_coefficient

[mummm.livejournal.com]

Yup... we are certainly odd... *L* (or at least I am...)

I have another friend on Facebook (a former student), a teacher friend (same AGE!), and met another of my students when we were out doing errands yesterday who also has the same birth date.

[low-delta.livejournal.com]

I gotta learn me some maths.

http://en.wikipedia.org/wiki/Birthday_problem

[cynnerth.livejournal.com]

My brain is crying.