Odds question concerning dice

[Wad]

You have ten dice and get one roll.

What are the odds of rolling 8 of a kind, 7 of a kind, or 6 of a kind on a single roll?

[bigpooch]

Here, we assume fair dice with 6 faces.

These kinds of problems are solved in this way:

If you roll an m of a kind (where m=8, 7 or 6 as you asked
in the original post), there are C(10,m) ways of picking
the specific combinations of the dice for which the die roll
matches the rank of the m of a kind. Then you just multiply
by the probabilities of the individual independent results.
Then multiply by the 6, the number of different m of a kinds
that can result.

Note: C(n,m) = combinations of n objects taken m at a time;
thus, the number of draw poker hands is C(52,5). C(n,m) is
just n!/((n-m)!m!) where k! = 1 x 2 x ... x k.

Thus, for

m=8: 6 x C(10,8)(1/6)^8(5/6)^2 = 0.000111633

m=7: 6 x C(10,7)(1/6)^7(5/6)^3 = 0.001488435

m=6: 6 x C(10,6)(1/6)^6(5/6)^4 = 0.01302301

[Mano]

Let's try a counting approach. I am going to use an ordered approach.

10 dice, each has 6 possible outcomes so 6^10 = 60466176 different configurations for the ten dice. For any subset of 8 dice there are 6 configurations such that all 8 are the same number, and for the remaining 2 dice there are 5*5 = 25 configurations where neither die is the same as the first 8. There are C(10,8) = 45 different ways to choose a subset of 8 dice from the original 10. So our probability is (6*25*45)/6^10 = 6750/6^10 = .000116 or 8957:1 against.

For 7 of a kind we would get (6*5*5*5)*C(10,7)/6^10 = 90000/6^10 = .001488 or about 671:1 against.

For 6 of a kind (6*5*5*5*5)*C(10,6)/6^10 = 787500/6^10 = .01302 or about 76:1 against.