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#1




The odds of 3 people in a room having the same birth date
I know it takes 23 people in a room to have a 50% chance of 2 people having the same birth date.
I was just at a party with a total of 12 people. THREE of the people had the same birth date (but different years). What are the numerical odds of that occurring? J. p.s., I'm putting this in GQ since I assume it has a factual answer. 
#2




The math for two people sharing a birthday is pretty simple, but the math for three is a bit more of a chore, but Monte Carlo simulations are easy.
The odds of 3 (or more) people sharing a birthday among 12 people, assuming that birthdays are randomly distributed (not true) and that leap years don't exist (also not true), are about 0.16%. The number of people at which the odds of having three share a birthday is 50% or greater under the same assumptions is 88. My code if anyone wants to check it: PHP Code:
Last edited by iamthewalrus(:3=; 02162020 at 12:09 AM. 
#3




Total asspull, but wouldn't it be (12/365)*(12/365) = .1% ?

#4




No need to simulate anything, it's as easy as for 2 people, just compute 12! times the coefficient of x^{12} in (1 + x/365 + (x/365)^{2}/2)^{365}



#5




Er, I mean, that's the probability there aren't 3 people with the same birthday. So the probability there are is 1 in 617 or whatever it works out to.

#6




Or .15%

#7




There's a simple way to get a goodenough approximation when there are n=12 people.
Start by counting the number of distinct trios in the room. This is C(12,3) = 12*11*10/(3*2*1) = 220 Multiply that by the chance the other two members of a trio have the same birthday as the first member, which is (1/365)*(1/365) The approximate answer is therefore 220 * (1/365)*(1/365) = .00165 or 0.165%. Note that this is only an approximation. There is a 0.001% chance that the room contains a quadruple birthday, and the simple calculation shown above counts those quadruplets 4 times instead of just once. We can subtract out these overcounts to get a better approximation: 0.165%  3*0.001% = 0.162% (aka 616to1 odds against) (There's also a minuscule chance of TWO trios; again these are overcounted in the formula above.) If n is much larger than 12, there will be much more overcounting — you'll have to worry about quintuplets, three trios, two trios and a quadruplet and so on. Horrible; you'd want to use the method of DPRK or of iamthewalrus(:3= instead. But for n=12, the 0.162% estimate (or even the simpleryet 0.165% estimate) may be good enough. 
#8




Here's a calculator for generalizations of the birthday problem:
https://www.dcode.fr/birthdayproblem Suppose you have c different categories and x different objects which each belong to exactly one of those categories. Suppose you expect that the chance that any one of those objects will be in any one of those categories is the same as the chance that it will be in any other one of the categories. Suppose you want to know what the chances are that at least n of the objects will be in one of the x categories. Then put c in for number of days in a year, x in for number of people in the room, and n in for number of people sharing the same birthday. Fill in the "That at least n people" dot in the probability section. Then click "Calculate". 
#9




I knew somebody that had three children with the same birthday (no multiple births). But, if you do the math, they were "Christmas presents".
There would be slightly different odds for the birthday being on a specific day of the year. 


#10




Yep. 0.162% and the probability calculator agrees given by Wendell Wagner agrees with you, as does iamthewalrus(:3='s Montel Carlo sim and septimus's approximation.
Last edited by pulykamell; 02162020 at 07:58 AM. 
#11




Thanks, everyone!
I guessed that adding that third person would greatly raise the odds to over 1 in a million. So it's surprising that the odds are only about 1 in 617. I'll let everyone else at the party know! Thanks again, J. 
#12




For your size group (12) four people sharing the birthday would be about 1 in 100 000, and five people would be 1 in about 22.8 million.

#13




I would image if we were talking about not only the same birthday, but also the same birth year, the odds would go off the charts.
If you were at a party of 12 people, most likely, you would have individuals of similar ages, so the odds would be better than simply 12 random people off the street where you might have individuals from new born babies to the elderly. I was waiting to get on a plane that held 12 people. We all had our passports out and I noticed that the guy in front of me had the same birthday and birth year as me. (He was Australian, I'm from the US.) However, the airport was in Lukla, Nepal and we were flying to Kathmandu. This limits the ages of the individuals involved because its unlikely that you would have young or elderly people in an extremely high altitude location as Lukla (9383'). Anyway of coming up with odds of such an occurrence?
__________________
Its only funny until someone gets hurt, then its fuckin' hilarious! 
#14




I was once at a popular math lecture and the speaker mentioned the birthday problem and decided that to actually test it with the audience of maybe 100. I interrupted to point out that two people in the audience were twins. However it turned out that someone else in the audience shared their birthday. I once directed a research group of about 10 and two of the members were born on the same day, one in Chile, one in Argentina. It was Halloween, 1938.
Sorry, I can't help mentioning that Feb. 12, 1809 was a very auspicious day for the world. Abe Lincoln and Charles Darwin were born on that day. 


#15




I had a high school teacher with a somewhatphotographic memory, so that he could call roll on the first day (the roll sheet somehow listed everyone's birthday) and be able to note all the duplicate birthdays. There were three people having one birthday. And I was one of them. (One of the others I had already known about.)
AND they were all in the same year! How about that! Well, that's not surprising. It was a high school class, where everybody was in the same year, thus all about the same age. 
#16




I think that is known as a birthday. A birth date includes the year.

#17




Quote:
Now, if we're talking what's the possibility that one of the 11 other passengers share your birthdate, well, that should be 1(10949/10950)^11, which is about 1 in 1000. Each person has a 10949/10950 likelihood of not sharing your birthdate (given a 30year span of birthdates). And if we want to add all these probabilities together, we multiply them. Now, if you're asking what are the odds that any two people in a group of 12 with 10950 possible birthdates share a birthdate, it's 1 in 166. That's a different question than trying to match up your specific birthdate with someone else in the group. (Which is the distinction that is the essence of the birthday problem.) Somebody feel free to check my work. Last edited by pulykamell; 02172020 at 11:48 AM. 
#18




The opposite of the "birthday problem" is: how many people do you need to gather in order to have a chance (for instance, 50%) to cover every single birthday? For 365 days you already need over 2000 people on average.
Last edited by DPRK; 02172020 at 12:04 PM. 
#19




Quote:
1(364/365)^253 = 0.5005 


#20




Sounds about right: You'd expect that number to be of the same order of magnitude as 365. What's surprising about that?

#21




Quote:
Thinking about it, I can understand why that is not the case — it just was a little surprising to me. Last edited by pulykamell; 02172020 at 03:07 PM. 
#22




Not half, you'd expect ln(2)

#23




All of these are just theoretical calculations; they are off from actuality in several ways.
First, they are using 1/365.25 in the calculations, as if every single day of the year had the exact same chance of being a birthday. Which is far from actual reality. In North America, Christmas Day is a very infrequent day for births. By far, most births are in late summer  fall (JulyOctober), reflecting conception happening in midwinter thru early spring. Even weekly, it varies: births are more common midweek (TuesdayThursday) and rarer on weekends, especially Sundays. [This may be due to doctors/hospitals schedules, since 20% of births are induced and 33% are Csections.] And it's a very significant variation; there are literally twice as many babies born around Labor Day as on Christmas Day. Also, these calculations assume that the 'people in the room' are an independent crosssection of the population. Obviously, they aren't  they are all gathered together in one room, presumably for some common purpose. Like the educational classroom examples: all the people will be about the same age. Similarly if you pick 12 people at a private dinner, or 12,000 people at a public concert  all likely to be of a similar age. Also residing in a similar area, of similar educational level, economic background, probably even similar political views. And for a small group like a dinner party, probably some of them are related, either biological relatives or related by marriage(s). So far from an independent sample. So while this 'birthday problem' is a fun item, and a good illustration of how for most people statistics refute their 'common sense', these statistical calculations don't strictly match reality. 
#24




The "2 people sharing a birthday" odds was our first lesson in my finite mathematics class in college. As it happened, despite having enough people to put the odds over 0.5, there were no shared birthdays, so it was a little awkward for the professor, but he still showed us the math to underscore that "intuitive" answers are often wrong.
Last edited by KneadToKnow; 02172020 at 04:15 PM. 


#25




Quote:

#26




Yes, if you have a strong mathemetical/statistical background. If you are a lay person, I think you'd intuit around 183, especially as most people have no idea what a natural log is. That's precisely also why the answer to the birthday problem is so surprising to most. If you know your math and look at it, you can see why it makes sense. But I doubt the vast vast majority of the populace would guess anything around 23 if they have never heard this problem before.

#27




It really shouldn't be any more awkward than throwing a heads when calling tails. I'm sure the professor expects this not to work about half the time (assuming a class size around 23.) Meanwhile, in a class of 41, we're up to a 90% chance.
Last edited by pulykamell; 02172020 at 04:36 PM. 
#28




Different but related birthday probability question:
Given two people chosen at random, what is the probability that they have the same birthday? (At first thought, it seems like approx. 1/365. Is that right?) The situation I'm thinking of is this: Given a married couple, who probably chose each other without regard to their birthdays, what is the chance of them having the same birthday? 
#29




Yes, so around 0.27%

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