This question stems from Probability of 3 of a kind with 7 dice as a means to check my understanding. I'm not sure whether this is general practice on this website, so I'm sorry if creating a new topic is frowned upon. I figured the question is large enough on its own to warrant a new topic. (Note: I intend to go back later to choose "best answer" on the previous one. It's a tough choice because I've used both extensively.)
Now without further ado:
With a set of three desired from 9 dice, there are
a) $ \binom{6}{1}\binom{9}{3}5^6 $
possible outcomes which contain one set of three.
These outcomes also contain other undesirables such as 111222222, 111222223, 111222333, etc. which must be accounted for.
Number of possibilities which contains a set of six in the remaining dice:
b) $ \binom{6}{2}\binom{9}{6}4^0 $
Number of possibilities which contains a set of five in the remaining dice:
c) $ \binom{6}{2}\binom{9}{5}4^1 $
Number of possibilities which contains a set of four in the remaining dice:
d) $ \binom{6}{2}\binom{9}{4}4^2 $
Number of possibilities which contains a set of three in the remaining dice:
e) $ \binom{6}{2}\binom{9}{3}\binom{6}{3}4^3 $
Number of possibilities which contain two sets of three in the remaining dice (three sets of three in the nine):
f) $ \binom{6}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}3^0 $
Now, this last set of three option (e) (ie. 111222345) may contain something that looks like 111222333 due to the randomness of the three remaining dice, so we need to subtract (f) from (e) in order to find the number of options which contain only 2 sets of three.
g) e - f: $ \binom{6}{2}\binom{9}{3}\binom{6}{3}4^3 - \binom{6}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}3^0$
Because (a) computes all combinations where there is one set of three, it will pull certain things twice, and certain other things even more. ex. 111222344 is the same as 222111344 AND 111222333 ~ 222111333 ~ 333222111: there are 3! different arrangements of this latter scenario.
I was initially planning to write out everything as I understood it, but as I try, I find myself going in circles and getting massively confused. To get the final solution of "how many ways to get exactly one set of 3 in 9 dice," there should be something like the following unless I'm mistaken:
$ a - b - c - d - ??? $
How many e's, f'g, and/or g's need to be taken out to account for both the initial duplicates counted in (a) (as described two paragraphs up) AND how many e's, f's, and/or g's need to be removed to ensure only one set of three remains confuses me.
Is what I have so far along the right path? If not, why?
If so, or even if just the overall idea is okay but the specifics are off, is there an easy/managable way to keep track of duplicates and items that need to be excluded in situations like this one?
I was also wondering if there's a good way to make use of the knowledge of that 111222333 ~ 333222111 is really just abc ~ cba; or 3! as opposed to working out more specifically what to include or exclude. Any ideas?
As always,
Many thanks.
PS
I got pulled away too many times while writing thanks to a dog who wanted to run around the house with a bloody toe, so I hope it all flows well.
Solution: I'm sad to say I basically gave up. I thought I had it, I can figure out specific scenarios, but in my chart the probabilities weren't adding up to 1 when all scenarios should have been accounted for. Therefore something someplace went horribly wrong. I took programmer approach and wrote a short script to count what I wanted out of a million rolls and filled in the chart that way (no, not exact, but close enough for what I want from it). I need a bath.
On the bright side all this grueling let me know why, when the program turned out unexpected results, it was right and could be explained.