# Real Life Birthday Question

At work, we had a room of 50 people, 5 of them had the same birthday? What are the odds of that?

We can just keep all assumptions standard (365 days, no leap year, etc.) It's already complicated and don't want to over complicate it.

• The odds for exactly $5$ same birthdays or at least $5$ same birthdays? – barak manos Jan 9 '17 at 17:32
• One day five people? Or two days, one with two people and one with three? – Henry Jan 9 '17 at 17:37
• The poisson approximation is quite helpful here. Using the mean $\binom {50}5\times \frac 1{365^4}$ it gives us $.000119367$, so very surprising indeed. Of course it's just an approximate form. (Note: I am assuming you meant at least one day occurs at least $5$ times). – lulu Jan 9 '17 at 17:44
• I don't know how to do it exactly, To get an order approximation, I'd use poisson to get the probability < 5 on any day (with parameter 50/365) then raise that to the power 365 to give the prob of never having 5 on one day, then take that from 1 – Cato Jan 9 '17 at 17:47
• this paper gives an exact formula (which I have neither checked nor implemented). For problems like these, the approximate forms are usually a lot more useful. – lulu Jan 9 '17 at 17:48

Total no. of cases or the sample space size will be $365^{50}$ as all the $50$ people can have their birthdays in $365$ different cases on $365$ different days. Explanation : First person can have a birthday on any of the 365 days in an year. Same is the case with rest of the 50 people. Hence total cases $= 365*365*...(50\space times)$.

Now, favorable outcomes are the cases in which any of the exactly $5$ people have their birthday on the same day. No. of such cases will be :$$^{50}C_5\cdot365\cdot364^{45}$$ because we can choose any $5$ of the $50$ people to have the same birthdays multiplied by $365$ because there can be $365$ cases for $5$ people to have same birthdays, some elements of that set of 5 people birthdays on the same day will be {$5\space Jan, 5\space Jan, 5\space Jan, 5\space Jan,5 \space Jan$}, {$25\space Jan, 25\space June, 25\space June, 25\space June,25 \space June$} etc. multiplied by $364^{45}$ as the rest of the 45 people can have their birthday on any day except the day other 5 people chose.

Now, the required probability will be $$\frac{^{50}C_5\cdot365\cdot364^{45}}{365^{50}}$$ $$\frac{^{50}C_5\cdot 364^{45}}{365^{49}}$$

Now, for calculating the odds number of unfavorable cases is $$365^{50}-(^{50}C_5\cdot365\cdot364^{45})$$ So the odds of exactly 5 people out of 50 having their birthday on the same day is$$(^{50}C_5\cdot365\cdot364^{45}):(365^{50}-(^{50}C_5\cdot365\cdot364^{45}))$$

Assumption : All the days in an year are equally likely to be a person's Birthday. (Credit : Brian M. Scott)

• This works out to a probability of a little over $10^{-4}$. However, you ought to mention that you’re assuming two things that aren’t actually true: that no one is born on 29 February, and that the other 365 possible birthdays are all equally likely. (I’m not objecting to the assumptions: they’re certainly a reasonable first approximation. I just think that they should be mentioned.) – Brian M. Scott Jan 9 '17 at 19:25
• @BrianM.Scott 1)The OP provided the constraint that we don't have to consider leap year.2)All birthdays are equally likely obviously. – Gyanshu Jan 9 '17 at 19:33
• It is an empirical fact that all birthdays are not equally likely; in particular, there is some seasonal variation. I’ve seen various attempts to explain this, none entirely convincing. – Brian M. Scott Jan 9 '17 at 19:35
• @BrianM.Scott Ok i'll add it if you want but i still consider it obvious. – Gyanshu Jan 9 '17 at 19:39
• @BrianM.Scott, the real problem here, I think, is that $^{50}C_5\cdot365\cdot364^{45}$ overcounts the number of ways in which there can be a birthday that is common to exactly $5$ people. For example, a case in which Barb, Bob, Bill, Betty, and Bozo share one birthday and Carla, Chuck, Charlene, Connie, and Creighton share another (and everyone else is born on other days) gets counted at least twice in this approach. Some inclusion-exclusion is called for. – Barry Cipra Jan 9 '17 at 19:59

For getting the same birthday you must remember that Out of 365 days in a year (irrespective of the number of people you have in the experiment) the probability 'P' will always be P= 1/365. That is because no matter how many people are there they can share only one exact date of birth in a year(because everyone has only one birthday).

$50^{365}$ is total number of outcomes.

$45^{364}*365*50!/(45!*5!)$ is number of outcomes when there are 5 people having same birthday.

Please check my solution, I havent solved this type of problems for ages, thanks!

How about a slightly easier problem that illustrates how to solve this exactly: Probability of three people out of $50$ with the same birthday.

I take this to mean that there's at least one group of three people that share a birthday. (No leap year, all birthdays equally likely.)

The negation of this is that each person in the group of $50$ shares their birthday with at most one other person.

If everyone has a different birthday, this is easy: $N_1 = _{365}P_{50}$. But then it gets a lot more tedious.

There are $25$ separate cases for at most one other person sharing a birthday, consisting of $1$ to $25$ pairs of people:

$$N_2 = \sum_{k=1}^{25} {_{(365-k)}P_{(50-2k)}} \left[\prod_{j=0}^{k-1}{50-2j \choose 2} \cdot _{(365-j)}P_{(2)}\right].$$

So the probability is

$$N_{\geq 3} = 1 - \frac{N_1 + N_2}{365^{50}}.$$

Now, to solve your stated problem, you'd need to calculate $N_3$ and $N_4$. For $N_3$, you'd consider all $192$ cases that have at most three people sharing a birthday like, say, $6$ groups of three, $10$ pair, and $12$ singletons.

If your head isn't already hurting, then think about $N_4$, and it will.

I think that because of the endless conundrums of leap years and likely bias of birthdates to certain months (which changes over the year, at this time for example, anyone born in February of a year has been alive longer than the people born in the following months of the same year, and has had longer to die - yes morbid isn't it).

For these reasons, go for a poisson approximation - poisson gives the probability of a number of independent events occurring - in this case the events are not actually independent, but they almost behave as if independent for estimation.

for numbers of birthdays on any day

$\lambda=50/365.25$

$P(0) = \exp -\lambda = 0.872063934$

$P(1) = (\lambda^1 / 1!) \exp -\lambda = 0.119379046$

$P(2) = (\lambda^2 / 2!) \exp -\lambda = 0.00817105$

$P(3) = (\lambda^2 / 2!) \exp -\lambda = 0.000372852$

$P(4) = (\lambda^2 / 2!) \exp -\lambda = 0.000012760162$

for any day

$P(>= 5) = 1 - P(1) - P(2) - P(3) - P(4) = 0.0000003574835$

We can then use poisson again with this new value to find the probability of zero days with 5 birthdays $\nu = 0.0000003574835$

$P = 1 - P(0) = 1 - (\nu^0 / 0!) e^{-365.25\nu}$

$=0.000130562$

My approximation = .00013

over 1 in 7000 chance it can happen