"Ulysses used a freely available online tool to find the formula for the averages."
No, not really. I used a freely available tool to expand the generating functions. The entire problem here is that you can't figure out pairs by averages on the dice. The expected value on 2 dice is 7 and tells you nada about the odds of pairs.
Counting dice can be a pain, as I am sure you know, and while the dice game created here looks a lot like a formv of Yahtzee (descended from existing games) the odds are not the same. One of the easiest ways to get around this is to write a program to roll 10,000 dice, check against success, and figure that this is a good representation of the odds. When you brute force it like that it is sometimes called the "Monte Carlo" method for pretty obvious reasons. It's also nice and straight-forward, if not trivial, and can make a CPU run a little hot.
(Note: Pascal's triangle, the basis for modern combinatorics, was in fact created for a dice game as legend has it.)
I thought about Monte Carlo but I was obsessed with finding the mathematical way of doing it. Statistic books had sections on counting using combinatorics but that covered essentially permutations (n^6) and combinations in total, and not of sub-sets with in. So it didn't provide a way to count pairs.
Eventually, thanks to my brother and the Great Google (Ceremonial robe recommended but not required.), I found discrete math and generating functions:http://stats.stackexchange.com/questions/3614/how-to-easily-determine-the-results-distribution-for-multiple-dice
You can get a formula that represents each number of times a specific combination is found.
So if you take, say. (A+B+C+D+E+F)^2 and expand it (which you can do on Wolfram Alpha per the link above) you get something like:
a^2 + 2 a b + 2 a c+2 a d+2 a e+2 a f+b^2+2 b c+2 b d+2 b e+2 b f+c^2+2 c d+2 c e+2 c f+d^2+2 d e+2 d f+e^2+2 e f+f^2
Each term is a combination of dice, but using letters for each face (A=1, B=2, etc). So [2 a b] means a 1 and a 2. The coefficient represents the number of times each combination occurs. So [2 a b] means there are two ways to roll a 1 and a 2 on two dice i.e. 1,2 or 2,1.. Everything to ^2 is a pair.
This method expands out to N-dice. The downside is that there are 3003 combinations for 10 dice, and over 60m permutations. So while this is more manageable it's still a LOT of work. But since I know ^2 is a pair and ^3 is a triple, etc, it's possible to grab those terms using what amounts to a find function and then add up the coefficients using...well, Excel, because I wanted an answer more than I want to do the programming.
I wasn't questioning either his numbers or the wisdom of hiring a stats guy (thank you for that, getting it right does matter, and that's why we pay people with Masters.) (Hey, cool! I was right! I am genuinely a little surprised.....)
Here is what I worked out if you're curious. Alas, the expansion I embedded in the sheets didn't carry through, but I can always redo it to include them.https://docs.google.com/spreadsheet/ccc?key=0AggHPTvQaNPydFMtN0w2eVg3OGZZVkxZU2N3S1JhaEE#gid=1
I confess, I worked all this out and was here to see if my numbers matched other peoples findings and then I find out there was a table...in the book...and the folks thinking about running it missed it too.....
I'd feel foolish but as I got the right answer and learned a very valuable tool I'm just gonna be happy, And finish my beer.