No problem. I totally understand this and Im only in 7th grade! woo hoo. But yea. Im gonna reccomend to my friends some of them are doing the birthday paradox project too 
Thanks again Kalid!
[…] fact, all you need is a pool of 75 athletes for a 99.9 percent chance that two of them will end up running around the Olympic Village in their birthday […]
OK, I’m awful at math. I’m writing a piece right now about Olympians sharing the same birthday. Every day I log on to the official Olympic website and check the birthdays. There are roughly 10,960 athletes competing. From what my untrained eye can tell, it averages to about 30 birthdays per day. Can anyone help me out and explain the math a little better for me? Writing is my thing. Math makes my brain hurt. Thanks!
[…] and it took only 36 or so to find them. How is this possible ? Well, it’s called the Birthday paradox, and it is based on the inability of our brains to think in exponentials, and to consider scenarios […]
Solution below is much simpler. Just find probabilty of each one having diff B’day and then subtract from 1 to get 0.507297234 answer. Any thoughts?
364/365363/365362/365361/365360/365359/365358/365357/365356/365355/365354/365353/365352/365351/365350/365349/365348/365347/365346/365345/365344/365*343/365=0.492702766
1 -0.492702766 = 0.507297234
[…] 57 så är det 99 % sannolikhet. En trevlig och enkel förklaring till hur det går till hittas på better explained. Vill du läsa mer om problemet och andra logiska kullerbyttor rekommenderas ett besök på David […]
@Abdul: Great question. Offhand, I don’t think people being born in the same year should change things.
My son used the birthday paradox for his science fair project. I do have one question. Is it possible to calculate the odds of 3 people in a group of 25 having the same birthday (month/day)? Could I do this using Appendix C, by simply changing the /2 to /3 in the R3 line {pairs = (people * (people -1)) / 2}?
Thanks so much the explanation. I apologize–I’m confused on one point which indicates: "…we could list the pairs and count all the ways they could match. But that’s hard: there could be 1, 2, 3 or even 23 matches!"
I don’t understand why wouldn’t the limit on matches be 22? The “target” can’t match with himself, (or can he?) Sorry, I’m sure I’m missing something obvious.
"…we could list the pairs and count all the ways they could match. But that’s hard: there could be 1, 2, 3 or even 23 matches!
Hi Mark, great catch. Yes, that should be 22 matches, appreciate the correction.
Hi Shark – the equation is a probabilistic argument. In fact, you hit “100%” (i.e., the limit of the javascript programming language) at around 90 people. At that point, the difference between 99.9999999… and 100.0 is too small to represent on the computer!
So it is theoretically possible to have 365 random people with 365 different birthdays. Practically, at around 100 randomly chosen people, you are virtually guaranteed to have a match (i.e., the probability of not having one is tiny, too small to be shown on a computer :)).
Khalid-thanks for the clarification and the website–it’s a remarkable service and resource.
So if I ask 23 people to pick a number between 1 and 365, there’s a 50% chance that at least two of them would choose the same number. This is the scariest thing I’ve ever heard of.
Hey.
I was wanting to take leap day into account and so I figured I should use 365.25
And by the way my sample size is 50 people
Would this work…
50*49=2450
2450/2=1225
so 1225 combinations
364.5/365.5=99.7264022%
so 99.7264022 is the chance of a combination not matching
.997264022^1225=3.48%
so 3.48 is the chance that all 1225 combinations don’t match
1-3.48%=96.5131327%
so 96.5131327 is the cance of at least one of them matching
That is how I worked it out and I’m not sure if it is correct so help me out please.
Hey.
I just realized that in my math I used 364.5 and 365.5 instead of 364.25 and 365.25 and that messed it up, but if I change that, did I have the right idea?
Haha. Oops.
My 3rd grade daughter is testing the birthday paradox for a science fair project. The math itself is a little difficult, but testing the paradox is easier to understand. We have tested 14 samples and 12 samples produced a birthday match. Wouldn’t you expect it to be closer to half the samples? Is there an optimal sample size? I saw someone mention they did 40 samples.
[…] Wil Reimer, author of the Historical Connections books, did the Birthday Paradox problem [http://betterexplained.com/articles/understanding-the-birthday-paradox/] in our class of 30+ students and darn it, there was a pair of students with the same birthdate. […]
If you’re interested in “triples”… a room with 88 or more people has an over-50% chance of at least one birthday being held by at least three people. For a room with 733 or more people, it’s guarenteed! (Per the pigeonhole principle. Hypothetically, a room of 732 people could consist of 366 pairs of people each sharing a birthday unique to that pair. The next person to walk into the room must have a birthday belonging to one of the pairs.)
I’ve been trying and failing to solve the problem for uneven distributions, such as if we assume all birthdays except Leap Day are equally likely and Leap Day is 25% as likely as the rest. There doesn’t seem to exist a simple formula for it on the Net…
I just completed the following survey: I asked people in the office to pick a number between 1 and 365 inclusive. I moved around office so nobody could hear any other person’s response.
It took 20 tries before one person matched an earlier response! Interestingly, the 18th person and the 20th person I asked picked “6.” So, if I had chosen the respondents in a different order, I could have gotten the match earlier.
Somebody explain i still dont get it
[…] a room with 23 people, there’s a 50/50 chance that two people share the same […]