Probability that at least 3 people share the same birthday?

Question

There are 15 people in a room. Calculate the probability that at least 3 of them have a common birthday.

The answer should be ~ 0.0033.
The problem I have with the binomial distribution is that it overcounts the cases of multiple birthdays. Nevertheless for this problem, the error is not very large.

This is a tough one to choose a best answer to, since each answer is based on a different approach and contributes in a unique way. I'm going to let the voters decide.

Answer 1

edit
at most there should be 13 distinct birthdates.
14 and 15 dates won't give you 3 commoners.

P(at least 3 common)
= 1 - P(none common) - P(pair(s))
= 1 - 365P15/365^15 - [0.5^1*(15!/1!1!13!)*365P14/365^15 + 0.5^2*(15!/2!2!11!)*365P13/365^15 + 0.5^3*(15!/3!3!9!)*365P12/365^15 + ...(nearly 0 probability of 4-7 pairs)]
= 1 - 0.7471 - [0.2235 + 0.01238 + 0.00021]
= 0.01681
= 1.68%

Answer 2

If we consider their birth date to only be the day and month, not year then we'll have something to work with. If we consider the year as part of the birthday then I would say the probability will be nearly zero unless we control the situation a little be controlling the general age group of the room.

Let's consider a four year set of dates. there is a 4 / 1461 probability of having any birthday other than feb 29th and a 1/1461 probability of having this birthday.

Let X be the number of people with a given birth day.
use the binomial mass function to find:

P(X ≥ 3 | day is not feb 29) and P(X ≥ 3 | day is feb 29)

the unconditional probability is:

365 * P(X ≥ 3 | day is not feb 29) + P(X ≥ 3 | day is feb 29)
= 0.003325434

this assumes that a person as equal probability of being born on any given calender day.

to support this I wrote a simulation in R where a list of 1461 possible days in four years is listed to cover all possible birthdays. a random sample with replacement of size 15 is taken from this list and if the mode of the list is greater than or equal to three then I index a count of the time three or more people have the same birthday. after 50000 simulations i find the estimated probability to be 0.0034. the code for the simulation is below.

/// === R code === \\\

answer <- 365 * (1 - sum(dbinom(0:2, 15, 4/1461 ))) + (1 - sum(dbinom(0:2, 15, 1/1461)))

birthdays <- seq(1, 365 ,1)
birthdays <- c(birthdays, birthdays, birthdays, birthdays, 366)

sims <- 50000
sets <- 0

for(i in 1:sims)
{
class <- sample(birthdays,15,TRUE)
if( -sort(-table(class))[[1]] >= 3) {sets <- sets + 1}
}

cat("calculated probability is:",answer,"\n")
cat("estimated probability is:",sets / sims,"\n")

Answer 3

This is an application of the binomial distribution where p=1/365 and q=364/365 and n=15. There are the usual 16 terms, and you want to compute the last three (q^15 + 15pq^14+ 105p^2q^13) and subtract the sum of these from 1. That will be the answer.

Answer 4

Number of ways for three to have a common birthday = (15C3) (365) (365)^12 - obviously getting some double counting from cases where more than three people share a birthday, or two sets of three people share birthdays.

So baseline probability is (15C3) (365)^13 / 365^15 = 0.342% (3s.f.); this is too high, but the correction terms should be pretty small; let's estimate some.

If four people share a birthday: (15C4) (4C3) (365)^12 / 365^15 = 0.011% (2 s.f.), which will again be slightly high since it includes cases where five or more people share a birthday. (The correction for this is about 0.0004% so we can ignore it).

If two sets of three people share a birthday: (15C3) (12C3) (365) (364) (365)^9 / 365^15 = 0.0006%, so we can neglect this term.

So the probability is 0.33% to 2 s.f.

Answer 5

I'm thinking we should take a complementary counting approach. P(at least 3 people share the same bday) = 1 - P(all have different birthdays) - P(at least one pair of people share the same birthday). For simplification purposes, let's not take leap years into consideration.

Of the two probabilities we need to compute P(all have different birthdays) is simpler to determine. That probability is 365P15 / 365^(15).

For P(at least 1 pair of people share the same birthday) it seems like we should try using the principle of inclusion/exclusion (PIE).

Addition: Actually, a casework approach like the one used by ksoileau may be a better way to finish.

Answer 6

Not good with Math, but will give you a real life story.
There were 125 in my high school grad class.
5 of us (girls) had/have the same b-day. We all went to different elementary schools, so we didn't know each other until 7th grade.
2 of us have been best friends since 7th grade. None of the others were/are friends, with each other or with us.
My friend and I each have 2 boys and 1 girl.
Our 1st husbands have the same first name.
Our second husbands have the same last name.
Our 2nd husbands each retired from the military and we recently discovered that they are distant cousins.
We both live in houses with a lake view.
Calculate these probabilies!!

Answer 7

P(at least 3 have the same birthday)
=1-P(all 15 have different birthdays)-P(some pairs of people have the same birthday, but no 3 do)
=1-P(all 15 have different birthdays)
-P(exactly one pair of people have the same birthday, but no 3 do)
-P(exactly two pairs of people have the same birthday, but no 3 do)
-P(exactly three pairs of people have the same birthday, but no 3 do)
-P(exactly four pairs of people have the same birthday, but no 3 do)
-P(exactly five pairs of people have the same birthday, but no 3 do)
-P(exactly six pairs of people have the same birthday, but no 3 do)
-P(exactly seven pairs of people have the same birthday, but no 3 do)
=1-(365C15)*15!*365^(-15)
-365*(15C2)*2!*365^(-15)
-(365C2)*(15C4)*(4C2)*2!
*365^(-15)
-(365C3)*(15C6)*(6C2)*(4C2)
*3!*365^(-15)
-(365C4)*(15C8)*(8C2)*(6C2)
*(4C2)*4!*365^(-15)
-(365C5)*(15C10)*(10C2)*(8C2)
*(6C2)*(4C2)*5!*365^(-15)
-(365C6)*(15C12)*(12C2)*(10C2)
*(8C2)*(6C2)*(4C2)*6!
*365^(-15)
-(365C7)*(15C14)*(14C12)
*(12C2)*(10C2)*(8C2)*(6C2)
*(4C2)*7!*365^(-15)

Answer 8

(1-14/365)
(1-.0383561)
or probability of .9616
or round to (1) set of (3)

Answer 9

well excuse me!...answers chose the section according to the content...sometimes it's wrongly interpretated...

Answer 10

This is impossible as of now, because there is no formula to possibly find the answer. And as we all know, to calculate, your going to need a formula.