*Wow! You two have the same birthday!*

It isn't *that* surprising - in a group of 30 people, the chances are better than 50/50 that two people will have the same birthday. Yes, it is a bit hard to believe at first.

*Yeah, I'd put it closer to 200!*

Ok, so lets do some math. Oops - that's a four letter word. Trust me, it won't be that bad.

You've got Alice and Bob, thats a group of two - what are the chances that they have the same birthday?

*Thats easy! 1/365*

Yep. Rather low chance, but not too bad... so Charlie comes in, what are the chances that that two of them have the same birthday?

*Um...*

OK, lets take this bit by bit...

What are the chances that Charlie and Alice have the same birthday?

*I told you already... 1/365*

That you did. Now, what about Charlie and Bob?

*Again? 1/365*

So the chances that two of them from that group have the same birthday would be...

*
1/365 + 1/365 + 1/365? That would be 3/365
*

You would think so, but thats not quite the right answer. I apologize, I started things off a bit misleading. For any group of size `n`, the chances of two people having the same birthday would be:

`BP(n) = n-1 + BP(n-1)`

Extending this, we get the following table:

Group Bday-Prob
2 1/365
3 3/365
4 6/365
5 10/365
6 15/365

While this looks good, a problem occurs not too far down the road. At 30, according to this, the chances that two people have the same birthday is 325/365. Furthermore, it keeps going up and so just
a few more iterations, and its greater than one - which is clearly wrong.

Instead, lets take the probability that two people don't have the same birthday and use the 'and' nature of multiplication. Then, subtract the probability from 1 to show the probability that it *is* true.

So, back to Bob and Alice.. What are the chances they *don't* have the same birthday?

*That would be 364/365*

Now, add Charlie to this...

*How?*

Well, how many days left are there for Charlie to have a birthday different than Bob and Alice?

*365 - 2 is... 363!*

So, now, we multiple these together... and subtract it from 1

`1 - ((364/365) * (363/365))`

What about adding David?

*There are only 362 birthdays left so...*

`1 - ((364/365) * (363/365) * (362/365))`

Yep! So, for `n` people, the probability is:

`1 - ((364/365) * (363/365) * (362/365) * ... * ((365-n+1)/365))`

Want to see some magic? This is the same as...

`365!/((365-n)! * 365)`^{n}

*Epp! thats math!*

Yep, OK - the easier way to think of it...

`BP(n) = (365 - (n - 1))/365 * BP(n-1)`

So, going back to the original question... How many people does it take to have a group have 50% chance of two people sharing a birthday?

For 23 people, the chances that two of them have the same birthday is about 50.73% chance.

*
That few?! Wow!*

What about for three people to share the same birthday?

Well, that will take someone who has a bit more time and patience than I do.