# Another Birthday Probability Problem

1. Nov 21, 2007

### e(ho0n3

[SOLVED] Another Birthday Probability Problem

1. The problem statement, all variables and given/known data
Given 20 people, what is the probability that among the 12 months in the year there are 4 months containing exactly 2 birthdays and 4 containing exactly three birthdays.

2. Relevant equations
Axioms and basic theorems of probability.

3. The attempt at a solution
The sample space of this problem is a set of 20-tuples (n_1, ..., n_20) where n_i is an integer from 1 to 12 representing the month person i was born in. The sample space has u = 12^20 elements.

Divide the people into 8 distinct groups, 4 of 2 and 4 of 3. This can be done in

$$v = \frac{20!}{2!^4 3!^4}$$

ways. Now each of these 8 groups needs to be assigned to a distinct month. Pick any ordering of the groups. The first group has 12 months to choose from, the second has 11 months to choose from, etc. There are thus w = 12 * 11 * ... * 5 = 12!/4! assignments.

The probability sought is vw/u. Unfortunately, when I calculate this I get a different answer than the book's 1.0604 x 10^-3. Where could I have gone wrong?

2. Nov 21, 2007

### Avodyne

Your counting of the ways to assign the groups to months is not correct. Four months get assigned 4 people each (call these months "red"), four get assigned 2 people each ("blue"), and four get assigned 0 people each ("green"). How many distinct ways are there to "color" the 12 months?

When I use this count for w, I get your book's answer.

3. Nov 21, 2007

### e(ho0n3

OK. I know what the problem with my counting is: In v, I'm counting permutations of possible groups. For example, if A, B, C, D are four 3-person groups, v is counting all possible 4-permutations of those groups as being distinct. v should actually be

$$\frac{20!}{2!^4 3!^4 4!^2}$$

and that will produce the correct answer.

If I leave v unchanged however and consider the orderings of the groups, then I would have to ignore certain group-month assignments. For example, if groups A, B, C, D are assigned to months 1, 2, 3, 4, then I would have to ignore the assignment of D, C, B, A to months 4, 3, 2, 1 which I'm incorrectly counting in w. To ignore these, w needs to equal C(12,4) * C(8,4). That will produce the correct answer.

Thanks for the pointer Avodyne. I will mark this thread as solved now.

4. Dec 9, 2011

### tomz

Re: [SOLVED] Another Birthday Probability Problem

Sorry, I am new to these....But I have a quick question..
If two people from the same group (same month) is picked and switched.
That would be considered as another situation in u isnt? because u have order
But that would not count on v? because v is combination?
Then how can this be true?