Math challenge

I need some help trying to figure out how to solve this problem.
It is a dice problem, but for those of you who dislike craps discussions, it is not a craps problem, per se.

I have three 6 sided dice (A , B , C)
Assume that the dice are biased in the following manner

For 36 rolls of each die

Die	A	B	C
Face	cnt	cnt	cnt
1	5	7	5
2	6	6	6
3	6	6	6
4	6	6	6
5	6	6	6
6	7	5	7

So for an example on Die A you get this distribution
1 = 5/36
2 = 6/36
3 = 6/36
4 = 6/36
5 = 6/36
6 = 7/36

Now I start getting way over my head.

Randomly, I am going to select any 2 of the 3 dice.
It is easy for me to see that if I selected the combination of AC, that I will have more 12s than 2s, but I'm not sure exactly how to calculate that.

Finally I believe there are 3 combinations of Dice I can have,
33% of the time I will choose a pair of dice that have the 6-6 faces show up the most.
66% of the time, I will choose a pair of dice that have the 1-6 faces show up the most.

How do I build a table that would show me all of the expected results of the numbers 2 through 12, accounting for the bias above, AND the random selection of any 2 of the 3 dice available?

There are 3 possibilities for the 2 dice selection: AB, AC, BC
Caclulate for AB the 2 dice probs, ie
1,1 5/36 x 7/36 = 35/1296
1,2 5/36 x 6/36 = 30/1296
Do the same for AC and BC
For example for AC
1,1 5/36 x 5/36 = 25/1296
and For BC
1,1 7/36 x 5/36 = 35/1296

Then get the average of the 3 possibilities
For 1,1 (35+25+35)/3 = 31,66/1296

Quote: RaleighCraps

How do I build a table that would show me all of the expected results of the numbers 2 through 12, accounting for the bias above, AND the random selection of any 2 of the 3 dice available?

Think of it like you have three "meta-dices" X = AB, Y=AC, and Z=BC. First calculate the expected results obtaining 2-12 from X, Y and Z separately. Then, as you always choose X, Y, or Z at random (with probability 1/3 each), average the results for 2-12 over X, Y, and Z.

It's easier if you just plug the numbers into a spreadsheet since there are only 216 lines (or 648 if you triplicate the logic for each perm of two die).

Going down have ({A}1,{B}1,{C}1) (1,1,2)...(6,6,6)
Going across start with Dice A, Dice B, Dice C. Then add in the chances of the dice showing their number (via a lookup OFFSET). Then (a) multiply chances A and chances B to get the chances of that (b) add values on A and B to get total. Send across to a column (e.g. IF $E2=O$1 then $D2 else 0) under the total. Repeat for AC and BC (probably easier by adding more rows or using another sheet).

I find some gambling problems are easiest to solve by running the perms down a spreadsheet. For instance two cards AA AK AQ..A2 KK KQ.... 22, is easy to program (since there's only one carry). Similarly all the perms of six cards (ignoring their suit and order) runs to about 18564 lines. Personally I created that list using programming (do I = 1 to 13, do J = I to 13...., write I J K L M N).

I would do it like this, but I'm evil like that.

Quote: miplet

I would do it like this, but I'm evil like that.

Thanks miplet. I need to study your ss a bit more to understand how you have this laid out, but this is exactly what I was looking for.

I think I can modify from here to create more data.

I was surprised to see 7 was still 16.7%, but quickly realized that would probably be correct because of the numbers I chose for the bias.
But this is interesting, because it shows that the 7 would occur as often as expected, yet the other numbers must deviate from expectation of dice with no bias