Comments on: A birthday problem

By: Donnie

Donnie — Thu, 30 Oct 2008 15:05:49 +0000

The answer is here: http://mathforum.org/dr.math/faq/faq.birthdayprob.html. I thought it would be around 5%, it’s actually ~50.7%

@Mascot: Probability, not calculus. Dimwit.

@Annie: What was the scary part? Being in the same room with 22 people?

@Wang: Fortunately, I won’t have to show the math. Wikipedia will do it for me.

@Chesstiger: Yes, I should have recognized few readers would want to take the time solving the problem.

@likesforests: Hmmm, by estimate I more meant a rough feeling without calculation. But I didn’t make that clear. (Mine was 5%. Boy was that off.)

Your rough calculation is roughly correct.

By: likesforests

likesforests — Thu, 30 Oct 2008 07:42:56 +0000

After a few drinks, considering the trickier problem:

Estimate: 60%

(Work in my Head)

Change Guy#1 has unique birthday: 100%
Change Guy#22 has unique birthday: (365-22)/365 = 17/18
Change Guy#x (random 1-22) has unique birthday = 35/36
Chance everyone has unique birthday = (35/36)^22
Gross approximation = (36-22)/36 = 14/36 = 42%
So chance of at least one non-unique birthday = 58%

Rough Calculation:
1 – ( (365-(22/2) )/365 )^22 = 49%

By: chesstiger

chesstiger — Thu, 30 Oct 2008 06:23:32 +0000

Are you crazy? I am not solving your math test. Do it yourself!

Me and calculus, two hands not on the same belly. I hate, despice, gag, calculus. With other words. Tell me to write something and i might come up with something. Tell me to count x and y minus z and you will be waiting your whole life for the outcome.

By: wang

wang — Thu, 30 Oct 2008 05:50:08 +0000

50%

I remember this (or at least I think I do) from stats. But I haven’t the slightest on how to calculate this anymore.

If I remember correctly it jumps up to something like 99% with 50 people in the room.

Looking forward to you showing us the math on this one.

By: likesforests

likesforests — Thu, 30 Oct 2008 03:50:41 +0000

I misread your question and considered a simpler one–what’s the chance that at least one of the twenty two others at the party has the same exact birthday as you?

Estimation = 6%
Calculation = 5.86%

By: annie

annie — Thu, 30 Oct 2008 00:55:21 +0000

AAAAAAAAAAAAAAAAAAAAAAAAA!!!!!!

I had half a bottle of zinfandel before I qrote this and I still don’t know. you’ve scarred me for life. and that’s hard to do. good job.

By: The Mascot

The Mascot — Wed, 29 Oct 2008 21:22:36 +0000

Re: part 2. You wrote that like you expect most people to know calculus. They don’t. Dimwit.