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Mathematics 82 Online
OpenStudy (amistre64):

How many people have to be in a class so that there is a 50% chance that at least 2 people have the same birthday?

OpenStudy (anonymous):

is this a riddle?

OpenStudy (saifoo.khan):

it's a riddle.. satellite is stuck in the middle.. o_O

OpenStudy (anonymous):

xD

OpenStudy (goformit100):

haha @saifoo.khan

OpenStudy (amistre64):

more of a conundrum :) a coworker is taking an online stats course and asked me this question there is no "correct" answer that i know of, but is meant to exhibit intuition versus reality i think

OpenStudy (anonymous):

i will be quiet because i know the answer will post a problem from dylan that took much of the morning

OpenStudy (anonymous):

Reality is a bit painful, I do say.

OpenStudy (anonymous):

oh so it is an honest question, not a riddle answer is 23

OpenStudy (lgbasallote):

well i know if there are 366 people then 100% 2 people share same birthday :)

OpenStudy (goformit100):

its easy

OpenStudy (amistre64):

23 could be plausible, any logic to back it hto?

OpenStudy (amistre64):

lgba, lol; yes 266 to be 100%

OpenStudy (amistre64):

366 that is

OpenStudy (anonymous):

Logic is overrated.

OpenStudy (amistre64):

but the "at least part" hmmm

OpenStudy (goformit100):

correct @amistre64

OpenStudy (lgbasallote):

then maybe 183 people then 50% chance? hehe

OpenStudy (anonymous):

I got this! \[\huge1-e^{-x^2(730.5)}\]

OpenStudy (anonymous):

it is a calculation, for which you definitely need a calculator

OpenStudy (anonymous):

Yay for paying attention in class :D

OpenStudy (amistre64):

id say intuitively, you need at least 2 people :)

OpenStudy (anonymous):

FYI: 365.25 days in a year

OpenStudy (lgbasallote):

but isnt that 1/365 x 1/365 chance? :O

OpenStudy (amistre64):

thats what i thought too

OpenStudy (anonymous):

Leap year yo :-P

OpenStudy (anonymous):

pass to the compliment, i.e. consider the probability that no two people have the same birthday am using 365 with no stupid leap year to make calculation easier

OpenStudy (amistre64):

ah yes, the compliment

OpenStudy (anonymous):

Aww nobody wants to dispute my expression eh? :-P

OpenStudy (amistre64):

its a cute expression lol

OpenStudy (anonymous):

if there are say \(n\) people then the probability no two people have the same birthday is \[\frac{365\times 364\times 363\times ...\times (365-n+1)}{365^n}\]

OpenStudy (anonymous):

@amistre64, I've never understood this. Why is your picture an integral?

OpenStudy (anonymous):

(I should also be asking why @satellite73 is using a bike.)

OpenStudy (amistre64):

its not, thats my profile angle ;)

OpenStudy (anonymous):

now it is a matter of computing for various choices of \(n\) and recalling that this is the compliment.

OpenStudy (amistre64):

having a + sign just seemed to remedial

OpenStudy (zepp):

42.

OpenStudy (anonymous):

this is the bike that i took from amistre because he did not let me ride the bike they gave him for this 1000 medal (back in the day)

OpenStudy (anonymous):

You two have such a good relationship. I am jealous.

OpenStudy (amistre64):

it tooks years of mistrust and antagonism to build this relationship as strong as it is lol

OpenStudy (anonymous):

he is the older brother i never had

OpenStudy (zepp):

o_o

OpenStudy (zepp):

Loki and Thor :D

OpenStudy (amistre64):

cain and able is what i had pictured, but pagan dieties work too :)

OpenStudy (anonymous):

@agentx5 is that for real?

OpenStudy (anonymous):

Medalling @amistre64 for his contribution to @satellite73's brotherly needs.

OpenStudy (anonymous):

Is what for real? The expression I wrote?

OpenStudy (anonymous):

i was thinking more of abbot and costello http://www.youtube.com/watch?v=KVn0aksCzNE&feature=related

OpenStudy (anonymous):

Every time I hear "for real" on this site, I immediately think of \(\mathbb{R}\)...

OpenStudy (anonymous):

If you make it into a standard y=f(x) function I believe it gives you the answer for any input, provided you use a ceiling or floor function to make it into whole #'s :-)

OpenStudy (anonymous):

yeah the integral so probability you get no two birthdays the same with \(x\) people is \(e^{-750x^2}\)?

OpenStudy (anonymous):

poisson?

OpenStudy (anonymous):

no that can't be it

OpenStudy (anonymous):

Of course somebody has already done this and uploaded it :-D I was trying to give you all a graph in Wolf but it wasn't understanding what I was asking it to do I guess, kept graphing the wrong thing

OpenStudy (anonymous):

Hehe, now this is more interesting: http://en.wikipedia.org/wiki/Birthday_attack

OpenStudy (radar):

23 according to wikipedia

OpenStudy (radar):

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

OpenStudy (radar):

I knew that it was a surprisingly few, but didn't realize it could be that small of a number.

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