The Birthday Problem

The Birthday Problem

The Statement of the Problem

The birthday problem concerns the chance that two or more people in a room of randomly assembled folks have the same birthday (forgetting the year -- only the day).

The Assumptions

This is an example of a probability problem where events are independent:

• First of all, to make life easy, we are going to ignore leap days (so assume that no one is born on February 29th).
• Assume that each day of the year is equally likely to be the birthday of any particular person.

Therefore the chance (or probability) that any particular person has any particular birthday is equal, to 1/365.

The assumption of independence means that the probability that two people have the same birthday is the product of the probability that person 1 has any birthday -- that's easy, 365/365=1 -- times the probability that person 2 has the exact same birthday (=1/365).

The Chance that two given people have the same birthday

Therefore, the chance that a room containing two people has a matched pair is

$P(pair\ with\ same\ birthday)=P(any\ birthday)*P(same\ birthday)={\frac {365}{365}}*{\frac {1}{365}}={\frac {1}{365}}$.

That was easy! Conclusion: don't bet that two people out on a random date will have the same birthday!

The Chance that two people in a group have the same birthday

It gets a little tougher when you have more than two people, however, and it becomes easier to work with the complementary problem: what's the chance that all people in a room have different birthdays? Then the probability that two (or more) have common birthdays is just the complement --

$P(all\ different)=1-P(some\ same)$

Hence, we can calculate the probability that some have birthdays in common is

$P(some\ same)=1-P(all\ different)$

So it turns out to be easy to compute the probability that everyone has a different birthday. The first person you ask can have any birthday they like:

$\left.P(person\ 1\ birthday)=365/365\right.$

The next person may have any birthday but the one chosen by person 1: hence their probability is

$\left.P(person\ 2\ birthday)=364/365\right.$

They make their choice independently, and so the probability that the two have different birthdays is

$P(both\ different)={\frac {365}{365}}{\frac {364}{365}}={\frac {364}{365}}$

and so on, so that if we have have twenty people, say, in the room, the probability that they all have different birthdays is

$P(all\ different)={\frac {365}{365}}{\frac {364}{365}}\cdots {\frac {346}{365}}\approx 0.5886$. That means that the probability of two people sharing a birthday is

$P(two\ the\ same)=1-0.5886\approx 0.4114$.

(so it's a little less likely than not). By 23 in the room, it's a little more likely than not ($\approx 0.5073$). And if there are 37 in the room (as there are in a full section of Math for Liberal Arts at NKU), then the probability that two or more students will share a birthday is about 0.8487. Very good betting odds!

A practical application

The probability which appears above might seem like a rather arcane (i.e. useless!) application. But actually, it's the sort of calculation that one could make any week of the year.

Here's a weather forecast for the next several days (as of the time I wrote this), and I'm hoping that it's going to rain.

The grass is brown, my rain barrels are empty (so I have to actually pay for water to water my yard, garden, trees, flowers, etc. -- makes me mad!). So what's the chance that it's going to rain in the next four days?

The probability that it's going to rain by Saturday is actually one minus the probability that it's NOT going to rain by Saturday:

$P(rain)=1-P(no\ rain)$

And it's a lot easier to calculate the probability that it's not going to rain in the next several days. That's because we don't have to consider all the ways in which it might rain (e.g. rain just once; rain twice; rain three times). Let's assume that the weather forecasters at WeatherUnderground have got it right: there's a 20% chance of rain tomorrow, no chance Thursday, 20% chance on Friday, and 30% chance on Saturday. This is true if those chances are independent (and they're actually probably not -- but let's assume that they are, to simplify the analysis). When we say 20% chance, that means that the probability that it rains is .2 (2 out of 10), so the probability that it doesn't rain is .8 (= 1-.2). Because the events are independent, we can simply multiply the probabilities that it won't rain in the next four days: the probability of NO RAIN is

$\left..8*1*.8*.7=.448\right.$.

This means that the probability of rain is somewhat better than half (.552 = 1-.448). Hooray!