My interpretation of the question and a solution are given below.
If 3 faces display 1 and the other three faces each display 0, -1, 5, then we can make up a table to show all 36 outcomes of throwing 2 dice:
0 0 |
0 -1 |
0 5 |
0 1 |
0 1 |
0 1 |
-1 0 |
-1 -1 |
-1 5 |
-1 1 |
-1 1 |
-1 1 |
5 0 |
5 -1 |
5 5 |
5 1 |
5 1 |
5 1 |
1 0 |
1 -1 |
1 5 |
1 1 |
1 1 |
1 1 |
1 0 |
1 -1 |
1 5 |
1 1 |
1 1 |
1 1 |
1 0 |
1 -1 |
1 5 |
1 1 |
1 1 |
1 1 |
It's clear that there are duplicates. Now we can consider permutations or combinations when assessing the probability. If the dice are thrown together, we assume it's the combination, not the order (permutation), which matters. So, using the table we can list the outcomes without duplication and just count the boxes in the table, divide by 36, and make another table which gives the probabilities directly:
|
-1 |
0 |
1 |
5 |
-1 |
1/36 |
1/18 |
1/6 |
1/18 |
0 |
1/18 |
1/36 |
1/6 |
1/18 |
1 |
1/6 |
1/6 |
1/4 |
1/6 |
5 |
1/18 |
1/18 |
1/6 |
1/36 |
The red numbers are those on the dice faces for each of the two dice. In the table body there's the probability for the relevant combination of the two dice (16 combinations in all). For example, the combination 0, 5 appears twice in the first table so the probability is 2/36=1/18. When both dice display 1, the combination appears 9 times in the first table so the probability is 9/36=1/4. The second table demonstrates the law of distribution. Note the symmetry.