@BillMisc,
If the probability of getting a hit (block) is p, then:
- with 1 die, the probability of getting 0 hits (blocks) is 1-p
- with 1 die, the probability of getting 1 hit (block) is p
- with 2 dice, the probability of getting 0 hits (blocks) is (1-p)^2
- with 2 dice, the probability of getting 1 hit (block) is 2p(1-p)
- with 2 dice, the probability of getting 2 hits (blocks) is p^2
- with 3 dice, the probability of getting 0 hits (blocks) is (1-p)^3
- with 3 dice, the probability of getting 1 hit (block) is 3p(1-p)^2
- with 3 dice, the probability of getting 2 hits (blocks) is 3p^2(1-p)
- with 3 dice, the probability of getting 3 hits (blocks) is p^3
For instance, with 2 red dice:
- the probability of getting 0 hits is 1/36
- the probability of getting 1 hit is 10/36
- the probability of getting 2 hits is 25/36
And with 3 yellow dice:
- the probability of getting 0 blocks is 1/8
- the probability of getting 1 block is 3/8
- the probability of getting 2 blocks is 3/8
- the probability of getting 3 blocks is 1/8
To determine the probabilities of the net number of hits with 2 red vs. 3 yellow, compute the products of each of the red and yellow outcomes:
...........................2 red
3 yellow....0 hits.......1 hit.........2 hits
0 blocks...1/36*1/8..10/36*1/8..25/36*1/8
1 block....1/36*3/8..10/36*3/8..25/36*3/8
2 blocks...1/36*3/8..10/36*3/8..25/36*3/8
3 blocks...1/36*1/8..10/36*1/8..25/36*1/8
2 red nets 2 hits against 3 yellow with probability 25/36*1/8
2 red nets 1 hit against 3 yellow with probability 25/36*3/8 + 10/36*1/8
2 red nets 0 hits against 3 yellow with probability 25/36*3/8 + 10/36*3/8 + 1/36*1/8
2 red nets -1 hits against 3 yellow with probability 25/36*1/8 + 10/36*3/8 + 1/36*3/8
2 red nets -2 hits against 3 yellow with probability 10/36*1/8 + 1/36*3/8
2 red nets -3 hits against 3 yellow with probability 1/36*1/8
Note that the above are sums of diagonals starting in the upper right corner.
Based on your post, I assume that 0, -1, -2, and -3 net hits are all equivalent to 0 hits.