Re: Lotto chance calculations

Jul 15, 2003 5 Replies

on a similar vein, years ago I heard that it only took about 14 people in a room to have a 50/50 chance of one of them sharing a birthday with another.



The explanation was based on the chances of NOT sharing a birthday, but I must have been too drunk to follow the logic.



Do you know how the calcs work for that?



It's 20 to achieve slightly greater than an even chance, in fact.

Logic is very simple:

Pick 2 people - the chance of them having the same birthday is 1 in 365 (assuming all days are equally likely). In a group of 20 people, there are (20*19)/(1*2) or 190 different pairs. Multiply 1/365 by 190 and you get

0.52, or slightly over 50%. That includes all the probabilities that a birthday is shared between 2 or more.

No, you can't work it out like that. Try repeating your logic for 40 people in the room and you end up with a probability greater than 100% !

I think the reason your method doesn't work it you end up double counting when more than 2 people share the same birthday.

Whatever method you devise, the (well, one) acid test of whether it is (or can be) correct is to see what result it gives for 366 people. Assuming that, to avoid complications, people born on 29th February are disqualified, it is impossible for 366 people all to have different birthdays, and so the expression for the probability of there being at least one shared birthday in a collection of N people *must* give the answer 1 when N>365.

No, there should be a very small chance (remember the formula is for the chances of NO shared birthdays).

Surprisingly, 0! is actually 1, so the formula does work for n65 and gives the answer of 1.45*10^-157. Quite long odds I'd say.

With n66 or more the formula won't work, since you can't have factorials of negative numbers. But going back to the earlier version

365*364*...*(366-n)/365^n we can see we'll end up with 0 on the top line for n>365. Of course it doesn't account for leap years...

That's completely new and counter-intuitive to me - but Bill Gates seems to know it :)

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