Posted tagged ‘birthday’

Birthdays

July 6, 2010

Ah yes, birthdays. That time of the year where we mark time’s relentless march onwards. Where the young are dragged kicking and screaming into responsibility and maturity and the old are led stumbling and murmuring from their ability to kick and scream. Still, they’re not all bad. Statistics show that those who have the most tend to live the longest.

There are 365 possible birth dates (366 if we’re being picky) and it was equally likely for you to be born on any one of those days (well, not quite, but it certainly isn’t this blog’s place to go into that). This means that the chance of having your birthday is $\frac{1}{365}$ 0r about 0.27%. Makes you feel pretty special, doesn’t it?

So, if I asked you for the minimum number of people I’d need in a room to have a better than 50% chance that two of them shared the same birthday, you wouldn’t even blink, would you? I mean, it all seems very simple. It must be half the number of days in the year. This would give $\frac{183}{385}$ which, as required, is over 50%. Feeling pretty confident about this, aren’t we?

Pictured: You

The answer, in fact, is only 23. That’s right, having just 23 people in your room gives a better than 50% chance that two of them share the same birthday. 23! That’s a small bus of people, or the entire population of “soccer” fans in the USA. And yet, surprisingly, that’s all it takes.

Not convinced? It’s understandable. You’ve been in groups of 23 or more many times and only very rarely have you discovered that someone shares your birthday. But that wasn’t the question, was it? If the question was how many people do you need to have a better than 50% chance they share a birthday with you, the answer is of course 183. With 23 people there’s only a 6.3% chance that one of them shares a birthday with you.

However, I didn’t specify which two people had to share a birthday, it’s not all about you, you know. What we are doing is comparing each person to every other person, giving us 253 distinct pairs of people. While this doesn’t help much with the calculation, it does make the answer seem slightly less surprising.

The comparison process

Want some proof? Lets first think about the probability that none of the 23 people share the same birthday, and think about each person in turn.

The probability person 1 has a different birthday than people we’ve previously analysed is 1 (as he haven’t analysed anyone else).

The probability  person 2 has a different birthday than person 1 is $\frac{364}{365}$ (this is the same as saying there is only a $\frac{1}{365}$ chance of them sharing a birthday).

The  probability person 3 has a different birthday from person 1 and 2 is $\frac{363}{365}$.

And so on and so on, you get the idea. This continues until we reach the 23rd person.

This means that the probability of no one in a room of 23 people sharing a birthday is

$1*\frac{364}{365}*\frac{363}{365}*\ldots*\frac{343}{365}=0.49270276$

The probability that two of group do share a birthday is therefore

$1-0.49270276=0.507297$ or 50.7297%

See, I told you so. Interestingly, you only need 57 people to give a 99% chance that two of them share a birthday. The percentage gets unbelievably close to 100% the closer you get to 366 (99.9999999999999999999999999998% at 200 people) at which point the pigeonhole principle dictates it reaches 100% (if 365 people all have different birthdays, what is the chance that a new member has a different birthday to all of them, excluding leap years?)

Like this, but with people

So there you have it. Want to amuse yourself and impress your friends at parties? Simply badger everyone for their birthday, neatly arrange all of your results and wave goodbye to your social life. Who wants to be popular anyway?