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?


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...


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.

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