Difficult Probability question for any math wizzes out there!

[Minotar]

Ok, here is a very difficult math problem for any prob/stat wizzes out there...

What is the probability of the following event to occur:

Suppose you have a random group of 30 people. What is the probability that 2 each of those people will have a birthday in each month January through June, and 3 each of those people will have a birthday in July through December? Assume that the probability of a person being born in any particular month is equally likely.

Now, what if any 6 months had 2 people each born in them and the other 6 months had 3 people each born in them.... How is this probability different than the above?

Any help is greatly appreciated!!!!!!!!!

 

[djhuber82]

I'll take a shot at it, although it's been a little while since I did any combinatorics so go easy on me...

Let N =the number of ways to split people up into 12 groups that satisfy the constraints
p = the probability of a particular combination = (1/12)^30

then the probability of meeting the constraints is just N*p, right?

Let C(n,k) = "n choose k" = number of ways to choose k things from n things = n!/[k!(n-k)!] (this might normally be written C(k,n), I can't remember)

For the 1st part, we've got 12 groups (1=jan, 2=feb, etc) and we know the number in each group so:
N1 = C(30,2)*C(28,2)*C(26,2)*C(24,2)*C(22,2)*C(20,2)*C(18,3)*C(15,3)*C(12,3)*C(9,3)*C(6,3)*C(3,3)
and the probability is N1*p.

For the 2nd part, we just repeat the above for each distinct ordering of six 2's and six 3's. There are C(12,6) such orderings, so
N2 = N1*C(12,6)
and the probability is N2*p

Does this sound right?

 

[Minotar]

Originally posted by: djhuber82
I'll take a shot at it, although it's been a little while since I did any combinatorics so go easy on me...

Let N =the number of ways to split people up into 12 groups that satisfy the constraints
p = the probability of a particular combination = (1/12)^30

then the probability of meeting the constraints is just N*p, right?

Let C(n,k) = "n choose k" = number of ways to choose k things from n things = n!/[k!(n-k)!] (this might normally be written C(k,n), I can't remember)

For the 1st part, we've got 12 groups (1=jan, 2=feb, etc) and we know the number in each group so:
N1 = C(30,2)*C(28,2)*C(26,2)*C(24,2)*C(22,2)*C(20,2)*C(18,3)*C(15,3)*C(12,3)*C(9,3)*C(6,3)*C(3,3)
and the probability is N1*p.

For the 2nd part, we just repeat the above for each distinct ordering of six 2's and six 3's. There are C(12,6) such orderings, so
N2 = N1*C(12,6)
and the probability is N2*p

Does this sound right?

When I tried it this way, I got very low numbers....like 10^-39 order magnitude, which is a very low probability. For some reason, this doesn't seem right to me? Any thoughts from anyone else? Thanks in advance!

 

[djhuber82]

While this does give a low number, it's not THAT low. But there are some very large numbers involved and if you try to do it on a calculator you might overflow a register and get garbage results. If you write out N1 above most of the terms drop out and you get:

N1 = 30!/[(2^6)*(6^6)]
N1*p = 30!/[(2^6)*(6^6)*(12^30)] = 3.74E-7

C(12,6) = 924, so N2*p = 3.45E-4

While these still seem a bit low (0.03% chance of the 2nd situation, I would have guessed something more like 1%), they're not 10^-39...

 

[Minotar]

Originally posted by: djhuber82
While this does give a low number, it's not THAT low. But there are some very large numbers involved and if you try to do it on a calculator you might overflow a register and get garbage results. If you write out N1 above most of the terms drop out and you get:

N1 = 30!/[(2^6)*(6^6)]
N1*p = 30!/[(2^6)*(6^6)*(12^30)] = 3.74E-7

C(12,6) = 924, so N2*p = 3.45E-4

While these still seem a bit low (0.03% chance of the 2nd situation, I would have guessed something more like 1%), they're not 10^-39...

How about doing it this way...

Part a= [6*(1/12)^2]+[6*(1/12)^3]=13/288
Part b= [6*(C(12,1)/C(12,12))^2]+[6*(C(12,1)/C(12,12))^3]=312/1331

To me these probabilities seem more reasonable, but the way you listed sounds reasonable too Very difficult problem for sure. Really, the probability should not be all too low, I don't think. If the prob for someone to be born in any given month is equal, then the most common scenario is for 2.5 people to be born in each month out of the 30 people. 2 for 6 months and 3 for the other 6 months should not be far off from this. In any case, thanks for offering all of the help. This is a hard problem!!!!!! Any other thoughts?

 

[DrPizza]

Here's my take on it, but I haven't had time to verify it

Let's start with the probability of exactly 2 people out of the 30 have a birthday in January

(30*29 / 2)*(1/12)^2*(11/12)^28
http://faculty.vassar.edu/lowry/binomialX.html
This page will give you help with a binomial probability
(and in my calculations, the 30*29 / 2 is the combination of 2 out of 30)

That's about equal to .2642695
That link will verify it for you (probability of exactly 2 out of 30) which is the probability of 2 of them having a birthday in January.

oh crud. the probability in each month isn't 1/12.
Oh well, I'll start over, but instead of 1/12, I'll have to use 31/365 for January.
Hmmm... or do I need to use 365.25
In which case, February would be 28.25/365.25

___________________________________________________

Okay, better explanation anyway:

You have 365 tiles, labeled with the day of the year (1 through 365)
Pull out a tile. What are the odds that it falls in January?
31/365.
Return that tile to the pile. Now pull out another numbered tile.
What are the odds that this tile is a day that is in Feb? 28/365

So, what were the odds of pulling out a January and then a Feb?
31/365 * 28/365.
BUT, you could have also pulled out a Jan and Feb by pulling them in this order:
Feb then Jan.
So, if the order mattered, it would be 31*28 / 365^2
But, if the order you select the months doesn't matter, it would be 2 * 31*28/ 365^2

Now, suppose we do it for J, F, and M (one each)
The probability in that order is
31/365 * 28/365 * 31/365
However, if the order doesn't matter, there are
3*2*1 or 3! ways to rearrange 3 months. So, the probability would be 6 times greater.

Now, for your probability, if you had to choose, in order J, J, F, F, M, M, A, A, M, M, Ju, Ju, Jy, Jy, Jy, A, A, A, S, S, S, O, O, O, N, N, N, D, D, D
(ugh), the probability would be:
31/365 * 31/365 * 28/365 * 28/365 *31/365 *31/365 * 30/365 * 30/365....
(oh, heck with writing it out that way!)
the denominator is obviously going to be 365^30
Since Jan, Mar, May, July, Aug, Oct, and Dec have 31, and you want a total of 18 birthdays in those months
and April, June, Sept, and Nov have 30, with 2+2 + 3 + 3 birthdays
And Feb has 28 days,
you ultimately wind up with:
28^2*31^18*30^10 out of 365^30 for the probability of pulling out days for those months... In that exact order (j,j,f,f,m,m,...)
However, there are 30! ways to rearrange the tiles you pulled out, so multiply that probability by 30!

And, crud. It's over 1. What'd I do wrong??
(I'm hitting reply so I can review my work in a larger window)


Darn darn darn darn... I'm missing where my logical explanation broke down.
I considered for a moment that multiplying by 30! was too much, (because Feb Feb isn't different from Feb Feb) However, I reconsidered an endless supply of tiles, in 1 second increments, and I keep the tiles as I pull them. Then, with minimal chance of duplicate tiles, there would be 30factorial ways to rearrange them. (and I could go to an even larger supply of tiles with microsecond increments...) 30! seems to be correct.
My individual probabilities for if the order mattered seem to be correct.

What on earth did I do wrong?! Forget about helping the OP, someone help ME please

 

[DrPizza]

Darn it! This problem is annoying me.

Okay, 12 sided die. Roll it. Write down the month.
Probability of J, J, F, F, M, M, . . .
in that order is 1/12^30

Number of different ways to rearrange those letters (well, we'll use different J's for June and July and Jan,...)
is
30!/(2!2!2!2!2!2!3!3!3!3!3!3!)
So, multiply that by 1/12^30
That sounds too low
3.7 * 10^-7

I give up, until someone finds and error in my previous post. It's really annoying me!
Regardless, I don't think the probability is going to be that high.
If as someone said, if you did it 11ty brazillian times, you'd expect it to average about 2.5 people per month.
But, 2 Jan, 3 Feb, 2 Mar, 3 Apr, 2 May, 2 June, 3 July, ...
would have the same probability (assuming each month is 1/12) as the problem in question. And, there are quite a few combinations of 6 months having 2 and 6 months having 3 and... I just had an idea...

Never mind, that idea gave me a headache.
After 2 minutes of attempting to figure out how many possible combinations of where to stick the birthdays,
starting with 30 in one month (12 different ways to do that)
then 29 in one month and 1 in another month (12 * 11)
then 28 in one month and 2 in another month (12 * 11)
etc.
Then 28 in a month and 1 in another month and 1 in yet another month
12*11*10
and 27 in a month and 2 in another month and 1 in yet another month
12 * 11 * 10
etc.

keep going,
then try to figure out how many ways there are to have 2 in each of 6 months, and 3 in each of the other 6 months.
choose 6 out of the 12 months to have 2 birthdays in:
there are 924 ways to have 2 birthdays in 6 months and 3 birthdays in the other 6 months.
So, if that were the *only* way they could be arranged (either 2 or 3 in a month), then the probability in question would be only 1/924. Obviously, it has to be less than that.

 

[DrPizza]

for what it's worth, I also liked djhuber's response