Please help me compute outcomes

I believe that order is not important. If you rolls 366, 636 or 663 are equivalent because they all result in the same 3 ways to place these dice.

You also cant just multiply by six because replace a 2 with 2 is not really a new combination.

Also this is not really any different then a pool of four white dice.

I will think about this one and try to get you an answer.

There are four six-sided dice, so there are 6^4=1296 possible rolls.
For the white dice, there are 6^3=216 permutations (where order counts) but only 56 combinations (distinct outcomes ignoring order).
Adding the black die, 56*6=336 states.
There are precisely 6 outcomes where the black die is irrelevant: all four rolled the same number.
I think this means 336-6=330 distinct states.
The black die is going to map the 330 states into one of the 56 combinations of three white dice, but that depends on player decisions that I can't model without more knowledge of the game.

Edit: Just saw the note above that order is important. So you have 1296-6=1290 distinct states that map into the original 216 permutations according to player decisions.

Well here is what I came up with:

I think that you roll all the dice and place the three white ones and then optionally replace one of the white ones with the black. Which I equated to rolling four dice and picking three. For four dice there are 126 unique combinations that we can break into 5 cases (I did some SQL programming to do this part. I can share the results and or code if desired):

Case #1:

In case number 1 there are 4 choices for the first slot then 3 choices for the second slot and 2 choices for the third slot. Which is 24 possible combinations for each of the 15 cases. This means a total 360 possibilities for Case #1

Case #2:

Lets represent of these numbers as A, B, C, and C. Then there are 12 possible combinations:

for 12 combinations for each of the 60 cases is 720 total possibilities.

Case #3:

Lets think of these numbers as A, B, B and B. Then there are 4 possible combinations:

For 4 combinations for each of the 30 cases is 120 total possibilities.

Case #4:

Lets think of these numbers as A, A,B and B. Then there are 6 possible combinations:

For 6 combinations for each of the 15 cases is 90 total possibilities.

Case #5:

There is one way to arrange these dice for each of these six occurrences. For a total of 6 possibilities.

360+720+120+90+6=1296

I think this is correct, but please feel free to double check.

@FrankM if you can explain this more clearly please do so.

The combinations that I came up with for each case, did not consider the possibility that this combination would have already exist in another case. For example 1234 would be a case one combination and 1233 would be case 2 combination. In my previous post I counted 123 from 1234 and 123 from 1233 as distinct possibilities, which obviously they are not. I will correct this in this post, but I think there is simpler way to answer this question.

Case 1 (ABCD): Which is all of the unique numbers is fine. Subtotal of 360.

Case 2 (ABCC): All of the arrangements that have 3 different numbers already exist in Case 1. Which reduces to the following combinations:

for 6 combinations for each of the 60 cases is 360 total possibilities.

Case 3 (ABBB): All of the possibilities using 2 numbers already exist from Case 2. Which means the only new combinations is BBB. Seems that I over counted this my original post as well. Since there are 6 different possible combos of BBB this adds another 6 possibilities.

Case 4 (AABB): All of the possibilities are covered in case 2. 0 new possibilities added.

Case 5 (AAAA):The possibility are covered in case 3. 0 new possibilities added.

360+360+6+0+0=726

Like FrankM points out some of these possibilities may not make strategic sense in your game. An extreme example would be picking a one over a six. So the effective possibilities are certainly less.

@FrankM thanks Frank you are usually better at explaining the math than I am.