Actuarial Outpost Combinatoric Probability problem
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#21
09-11-2012, 11:13 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 1,366

Quote:

1560 = 4^6 - 4(3^6) + 6(2^6) - 4
#22
09-11-2012, 11:50 PM
 AAABBBCCC Member Join Date: Feb 2012 Posts: 1,329

Did you figure this out on your own? Or did you get it from a combinatorics book? I guess it is not that hard, but it is still pretty slick. My answer is the same, but yours is much better for obvious reasons.
#23
09-12-2012, 12:13 AM
 AAABBBCCC Member Join Date: Feb 2012 Posts: 1,329

In general, if you fix r toy choices and you have n trials, then you want to count the number of ways of placing n objects into r bins with each bin containing at least one element. This is just a sum of multinomials:

$\sum_{I} \frac{n!}{n_1!\cdots n_r!}$ with I the index set {(n_1,...,n_r):n_1+...+n_r=n, n_i>=1 for all i}.

You then multiply by r_2 choose r for the total number of ways of getting r unique toys in n trials given an initial set of r_2 unique toys.

acadeactuary, how did you get that closed form solution? did you use induction?

Last edited by AAABBBCCC; 09-12-2012 at 12:26 AM..
#24
09-12-2012, 12:38 AM
 AAABBBCCC Member Join Date: Feb 2012 Posts: 1,329

haha!

YOu see, if the bins are not labeled then you just get the stirling numbers:

http://en.wikipedia.org/wiki/Stirlin...he_second_kind

If they are labeled, then you just multiply the stirling numbers by r! (r bins) and you obtain the formula:

$\sum_{j=0}^r(-1)^{r-j}\binom{r}{j}j^n$
#25
09-12-2012, 10:01 AM
 Academic Actuary Member Join Date: Sep 2009 Posts: 1,366

Quote:
 Originally Posted by AAABBBCCC In general, if you fix r toy choices and you have n trials, then you want to count the number of ways of placing n objects into r bins with each bin containing at least one element. This is just a sum of multinomials: $\sum_{I} \frac{n!}{n_1!\cdots n_r!}$ with I the index set {(n_1,...,n_r):n_1+...+n_r=n, n_i>=1 for all i}. You then multiply by r_2 choose r for the total number of ways of getting r unique toys in n trials given an initial set of r_2 unique toys. acadeactuary, how did you get that closed form solution? did you use induction?
I just looked at a simple specific case and generalized.
#26
09-15-2012, 02:09 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 1,366

I did find the formula in a text entitled Elementary Probability by Stiraker. The general problem is an urn with n distinct balls and r selected with replacement. What is the probability each ball is selected at least once?
#27
09-15-2012, 02:17 PM
 jas66Kent Member SOA Join Date: May 2012 Location: Canterbury, UK Studying for ST2, 4, and 9. Favorite beer: Corona :) Posts: 7,737

This question is actually stochastic in nature and not combinatronic.

You can think of each box as a renewable set of probabilities.

If you open one box you get four probabilities with one choice. Then the next box offers you the same probabilities and one choice. Etc.. i.e. Each time you get a box the stochastic system renews itself. i.e. It's a renewal process
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