Real Life Birthday Question

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.

Answer 1

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).

Answer 2

$50^{365}$ 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!

Answer 3

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)

Answer 4

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.

Answer 5

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