Curious observation

[--Pete]

Hi,

There are, counting leap day, 365.25 days a year. So, the probability that a person's birthday falls on any given day is 1 in 365.25 (1/365.25). The probability that a person's birthday doesn't fall on a given day is thus 1 - 1/365.25 = 364.25/365.25 as I said in the previous post.

Now, consider two people. If they are not twins, the probability of both not having their birthday on a given day is simply
(the probability of the first not having a birthday on that day) times
(the probability of the second not having a birthday on that day).

I looked for a good explanation of this, but all I found where either too technical (if you can read the notation, you probably already know the results) or too superficial. The best I could find was here.

Add a third person. The probability that all three don't have a birthday on a given day is
(the probability of the third not having a birthday on that day) times
(the probability that the first two don't have a birthday on that day) (which we've already worked out.)

Now, this gives us (364.25/365)^1, (364.25/365)^2, (364.25/365)^3 for one, two, and three persons respectively. We can continue to the case of n people by induction. It's a simple argument. We've shown the case for 1 already. Assume that it is true for (n-1), then for n it is
(the probability of the nth person not having a birthday on that day) times
(the probability that the first (n-1) people don't have a birthday on that day)

So, we get (364.25/365)^(n-1) * (364.25/365)^1 = (364.25/365)^n as required.

The odds are easy to calculate given the number of people. It's the number of people that would be, I think, difficult to calculate.

--Pete