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#1
02-15-2020, 11:07 PM
 Member Join Date: Oct 2004 Posts: 1,714

## 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
02-16-2020, 12:08 AM
 Guest Join Date: Jul 2000 Location: Santa Barbara, CA Posts: 12,411
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:
``` #!/usr/bin/pythonimport randomN = 100000n = 88m = 3successes = 0for i in range(N):    days = dict()    for j in range(n):        day = random.randint(1,365)        if not days.has_key(day):            days[day] = 0        days[day] += 1        if (days[day] == m):            successes += 1            breakprint successes * 100.0 / N  ```

Last edited by iamthewalrus(:3=; 02-16-2020 at 12:09 AM.
#3
02-16-2020, 12:15 AM
 Member Join Date: Dec 2000 Location: Aptos CA Posts: 9,173
Total asspull, but wouldn't it be (12/365)*(12/365) = .1% ?
#4
02-16-2020, 12:40 AM
 Guest Join Date: May 2016 Posts: 4,751
No need to simulate anything, it's as easy as for 2 people, just compute 12! times the coefficient of x12 in (1 + x/365 + (x/365)2/2)365
#5
02-16-2020, 12:51 AM
 Guest Join Date: May 2016 Posts: 4,751
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
02-16-2020, 12:52 AM
 Member Join Date: Dec 2000 Location: Aptos CA Posts: 9,173
Quote:
 Originally Posted by DPRK 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.
Or .15%
#7
02-16-2020, 03:32 AM
 Guest Join Date: Dec 2009 Location: the Land of Smiles Posts: 21,101
There's a simple way to get a good-enough 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 over-counts to get a better approximation:
0.165% - 3*0.001% = 0.162% (aka 616-to-1 odds against)
(There's also a minuscule chance of TWO trios; again these are over-counted in the formula above.)

If n is much larger than 12, there will be much more over-counting — 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 simpler-yet 0.165% estimate) may be good enough.
#8
02-16-2020, 06:34 AM
 Charter Member Join Date: Jul 1999 Location: Greenbelt, Maryland Posts: 14,549
Here's a calculator for generalizations of the birthday problem:

https://www.dcode.fr/birthday-problem

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
02-16-2020, 07:22 AM
 Guest Join Date: Jun 2019 Posts: 337
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
02-16-2020, 07:56 AM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by DPRK 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.
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; 02-16-2020 at 07:58 AM.
#11
02-16-2020, 08:25 AM
 Member Join Date: Oct 2004 Posts: 1,714
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
02-16-2020, 09:03 AM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by jharvey963 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.
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
02-16-2020, 12:25 PM
 Guest Join Date: Jul 2005 Location: Namche Bazaar Posts: 2,703
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?
__________________
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#14
02-16-2020, 02:20 PM
 Member Join Date: Mar 2002 Location: Trantor Posts: 13,666
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
02-17-2020, 01:35 AM
 Guest Join Date: Sep 2011 Location: Sunny California Posts: 15,764
I had a high school teacher with a somewhat-photographic 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
02-17-2020, 11:07 AM
 Charter Member Join Date: Mar 1999 Location: NY (Manhattan) NY USA Posts: 20,980
Quote:
 Originally Posted by jharvey963 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).
I think that is known as a birthday. A birth date includes the year.
#17
02-17-2020, 11:47 AM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by Mangosteen 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?
It depends on how you look at it. I'm going to start simple with just birthdays. If you're asking what are the odds that one specific person (the person in front of me) shares the same birthday as me, it's about 1/365, as you would expect (or 1/365.25 if you want to be a little more precise.) If we're going birthdate, then we'd have to estimate how many possible birthdates we have, and the odds are 1 in that number. So, let's assume a 30 year range, so we have about 10950 possibilities (let's forget leap years, it's not that important to get a general sense), ergo 1 in 10950.

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 30-year 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; 02-17-2020 at 11:48 AM.
#18
02-17-2020, 12:03 PM
 Guest Join Date: May 2016 Posts: 4,751
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; 02-17-2020 at 12:04 PM.
#19
02-17-2020, 12:51 PM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by DPRK 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.
I was actually also surprised to find, assuming I did my math correctly, that you need about 253 people in a room for there to be just over a 50% chance somebody shares a particular birthdate (like, say, your own.)

1-(364/365)^253 = 0.5005
#20
02-17-2020, 01:02 PM
 Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 87,302
Sounds about right: You'd expect that number to be of the same order of magnitude as 365. What's surprising about that?
#21
02-17-2020, 03:05 PM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by Chronos Sounds about right: You'd expect that number to be of the same order of magnitude as 365. What's surprising about that?
Sure, same order of magnitude, I just felt like the number should be significantly less. Like I would think a natural intuition would be that you’d need half of 365 for there to be a 50-50 shot of someone sharing your birthday, but it’s not 183, but rather 253, significantly more.

Thinking about it, I can understand why that is not the case — it just was a little surprising to me.

Last edited by pulykamell; 02-17-2020 at 03:07 PM.
#22
02-17-2020, 03:13 PM
 Guest Join Date: May 2016 Posts: 4,751
Not half, you'd expect ln(2)
#23
02-17-2020, 04:02 PM
 Charter Member Join Date: Mar 2003 Location: Minneapolis, MN Posts: 15,271
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 (July-October), reflecting conception happening in mid-winter thru early spring. Even weekly, it varies: births are more common mid-week (Tuesday-Thursday) and rarer on weekends, especially Sundays. [This may be due to doctors/hospitals schedules, since 20% of births are induced and 33% are C-sections.] 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 cross-section 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
02-17-2020, 04:15 PM
 Voodoo Adult (Slight Return) Charter Member Join Date: Jul 2000 Location: Charlotte, NC, USA Posts: 27,063
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; 02-17-2020 at 04:15 PM.
#25
02-17-2020, 04:30 PM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by Tim@T-Bonham.net 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 (July-October), reflecting conception happening in mid-winter thru early spring. Even weekly, it varies: births are more common mid-week (Tuesday-Thursday) and rarer on weekends, especially Sundays. [This may be due to doctors/hospitals schedules, since 20% of births are induced and 33% are C-sections.] 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 cross-section 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.
Yes, but these calculations are fine for a first order approximation. I would be surprised if the actual "birthday" problem odds differ significantly from the theoretical answer of 23. And, if it does, one should think that the answer is actually 21 or 22 (as some birthdays are slightly more likely to show up), so even more common and surprising than the 23 answer.
#26
02-17-2020, 04:32 PM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by DPRK Not half, you'd expect ln(2)
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
02-17-2020, 04:35 PM
 Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 49,335
Quote:
 Originally Posted by KneadToKnow 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,
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; 02-17-2020 at 04:36 PM.
#28
02-18-2020, 10:57 PM
 Guest Join Date: Sep 2011 Location: Sunny California Posts: 15,764
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
02-18-2020, 11:53 PM
 Guest Join Date: May 2016 Posts: 4,751
Quote:
 Originally Posted by Senegoid 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?)
Yes, so around 0.27%

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