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Birthdays in the Same Month

by e(ho0n3
Tags: birthdays, solved
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e(ho0n3
#1
Nov21-07, 11:38 AM
P: 1,367
1. The problem statement, all variables and given/known data
How many people have to be in a room in order that the probability that at least two of them celebrate their birthday in the same month is at least 1/2? Assume that all possible monthly outcomes are equally likely.


2. Relevant equations
Axioms and basic theorems of probability.


3. The attempt at a solution
Let E_n be the event that at least two of n people in a room celebrate their birthday in the same month. I'm basically asked to determine the value of n for which P(E_n) >= 1/2.

It must be easier to calculate the complement of E_n, ~E_n, and from that calculate P(E_n). Note that ~E_n is the event that nobody celebrates their birthday in the same month.

Now, ~E_n is the union of the events that nobody celebrates their birthday in the ith month of the year, which I will call F_i, i = 1 to 12. Since the F_i's are mutually exclusive, P(~E_n) = P(F_1) + ... + P(F_12).

If month j has d days, then P(F_j) = (365 - d)^n / 365^n. For months 1, 3, 5, 7, 8, 10, 12 d = 31, for months 4, 6, 9, 11 d = 30, and for month 2 d = 28 (assuming no leap years).

So P(~E_n) = 7 * 334^n / 365^n + 4 * 335^n / 365^n + 337^n / 365^n.

P(E_n) = 1 - P(~E_n) >= 1/2. I don't know how to determine n analytically, but I did obtain it numerically (using a simple basic program) and got n = 37.

The answer according to the book is 5. I must have done something wrong, but what?
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danni7070
#2
Nov21-07, 11:48 AM
P: 92
Aren't you supposed to assume that monthly outcomes are equally likely? So that you don't have to take february as 28 days and so on.
e(ho0n3
#3
Nov21-07, 12:43 PM
P: 1,367
Quote Quote by danni7070 View Post
Aren't you supposed to assume that monthly outcomes are equally likely? So that you don't have to take february as 28 days and so on.
Is that what that means? I always overcomplicate. Sigh.

I think this way is much easier:

If there are more than 12 people, then P(E_n) = 1 (by the pigeonhole principle). If there's only one person, then P(E_n) = 0. Thus 1 < n < 13. Assuming the latter, then P(~E_n) = 12 * ... * (12 - n) / 12^n.

Of course, this doesn't make it easier to calculate P(E_n) analytically. Numerically I get that P(E_4) = 0.427083 and P(E_5) = 0.618056 so n = 5. Sweet!


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