This is a good bar bet
I bet that I can take a room of 60 people (at random, for e.g. combining two classrooms of 30 students at your school), I can find 2 people in the room with the same birthdate by month and day.
Any takers that I cannot?
2007-01-24 03:16:45 · 14 answers · asked by srrl_ferroequinologist 3 in Science & Mathematics ➔ Mathematics
In response to Drocks 27 experiment, my birthdate is Aug 11, lets see how many people responding, before we get a match.
2007-01-24 03:29:23 · update #1
I'll do this ignoring Feb 29th birthdays.
If you have 366 people in the room, two people with the same birthday will be a certainty. If you have 1 person in the room, two people with the same birthday will be an impossibility. For all the other possibilities it is easier to figure out the probability of no one sharing a birthday and simply taking the inverse.
n is the number of people in the room.
For n=2
364/365
The first person in the room has a specific birthday, so to not match the first person in the room, the second person must have a birthday on one of the other 364 days in the year. The probability of this is 364/365.
For n=3
(364/365)(363/365)
The first two people in the room do not share a birthday, so for the third person to not share a birthday, his birthday must fall on one of the other 363 days of the year. The probability of this is 363/365. We multiply this by the probability of the first two people not sharing a birthday to get the probability of three people not sharing a birthday.
This can be generalized with the following equation.
365!/((365-n)! * 365^n)
With 60 people in the room, this comes out to less than a one percent chance. That means, with 60 people in the room there is more than a 99 percent chance that some pair of people will share a birthday.
2007-01-24 03:38:39 · answer #1 · answered by OMGWTFBBQ!!1 3 · 0⤊ 0⤋
P=1/365*365
2007-01-24 03:21:43 · answer #2 · answered by JAMES 4 · 0⤊ 0⤋
This is a bet with pretty good chances for you.
The possibility for not finding two persons with the same birthday is quite low.
There are only 365*364*363*...*(366-n) out of 365^n possible combinations for all different birth dates in a group of n persons.
The possibility to win the bet is
p=1-(365*364*...*306)/365^60
=99.2%
I would only accept the bet, if your name is Murphy. ;-)
2007-01-24 03:58:36 · answer #3 · answered by schmiso 7 · 0⤊ 0⤋
You may or may not it's probability - but your chances are good since just one single person woudl have ~ 59/365=.16 chance of finding a match and you extend that out since any 2 of teh 60 can match.
Fun problem I'll have to do the math on that later.
2007-01-24 03:22:29 · answer #4 · answered by G's Random Thoughts 5 · 0⤊ 0⤋
You don't need 60 people. You can get 2 people with the same birthday, with just under 30 people.
2007-01-24 03:21:14 · answer #5 · answered by Anonymous · 0⤊ 0⤋
The wiki aricle is a readable explaination of how this is derived. A formula for an approximate answer is:
P = 1 - e^( -n^2 / 730) = 1 - e^( - 3600/730) = 1 - e^( - 4.93) = 99.27%
2007-01-24 03:21:44 · answer #6 · answered by Scythian1950 7 · 2⤊ 0⤋
I bet you can, I mean chances are theres going to be one, you should maybe change your bet, to within two days of the birthday.
Such as the 9th and the 7th of July would be acceptable
2007-01-24 03:21:07 · answer #7 · answered by Fiona M 3 · 0⤊ 0⤋
I once read somewhere, that the mathematical likelihood is 100% among only 12 or 13 people.
2007-01-24 03:33:16 · answer #8 · answered by Anonymous · 0⤊ 0⤋
Thats a sucker bet. The same people that would take you up on that bet buy lottery tickets and play slot machines.
2007-01-24 03:59:54 · answer #9 · answered by davidosterberg1 6 · 0⤊ 0⤋
It's an old parlor trick and I think you only need about 23 people to have the odds in your favor.
2007-01-24 03:21:19 · answer #10 · answered by Gene 7 · 0⤊ 0⤋