A birthday problem
You are at a party with 22 other people.
1. Estimate what the chances are of at least two people at the party having the same birthday.
2. Now calculate it. (For simplicity, feel free to assume no one was born on Feb. 29)
Yes, I was surprised.
Re: part 2. You wrote that like you expect most people to know calculus. They don’t. Dimwit.
Reply to The MascotAAAAAAAAAAAAAAAAAAAAAAAAA!!!!!!
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.
Reply to annieI 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%
Reply to likesforestsCalculation = 5.86%
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.
Reply to wangAre 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.
Reply to chesstigerAfter 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:
Reply to likesforests1 – ( (365-(22/2) )/365 )^22 = 49%
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.
Reply to Donnie