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# Please help me compute outcomes

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questccg
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Joined: 04/16/2011

I know we've been down this "road" before... But I've changed how the dice are used and I'd like to have some "computation" done with regards to the probability.

So here is my problem that I need solved:

Roll 3d6 (White) + 1d6 (Black) = Compute ALL possible outcomes.

I don't want a listing, just the total number of possible outcomes. Where I am stuck is with the "Black" (1d6).

For that die, can replace any of the other "White (3d6)... If so desired. Usually you would do this to replace one of the lower dice. But at the same time IF you can drop one point and gain a better formula, you might also consider dropping to a lower value.

Anyone have the time to give me the details - would be much appreciated.

Before I was using 1d6 (Blue), 1d6 (Red), 1d6 (Green) and 1d6 (Black). And the values were fixed with the exception that the "Black" die could replace one of the other dice.

But instead of FIXING the dice colors and values, I figured I'd make them all "White" and one "Black" die. This gives players MORE "Outcomes" and possible "plays" (which is what I need help calculating...)

Many thanks!

questccg
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Joined: 04/16/2011
Okay so let me start by getting the ball rolling...

Since there are THREE (3) White (3d6), and order is important (because each die matches a "resource"), we have:

6 x 6 x 6 = 216 possible outcomes (not too bad!)

Now where I have no clue... Is how to introduce the ONE (1) Black (1d6)...

It's always rolled and then the players decide IF they want to SWAP.

Need HELP --> Here <-- !!!

let-off studios
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Joined: 02/07/2011
Maybe 1,296?

questccg wrote:
6 x 6 x 6 = 216 possible outcomes (not too bad!)

Now where I have no clue... Is how to introduce the ONE (1) Black (1d6)...

Can you just multiply that value by 6 to determine the number of possible outcomes?

Either that or I'm confused by your original question.

Fri
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Joined: 09/06/2017
I think that order is not important

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.

questccg
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Joined: 04/16/2011
Fri wrote:I believe that

Fri wrote:
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.

Well choosing "366" means:

• "3" for Bloodlust
• "6" for Treasure
• "6" for Food.

So order is important. "636" or "663" is not the same either. But "666" would be the same... Really not sure anymore.

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

Yeah see my example above with the three (3) "6"s.

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

Well the fourth (4th) dice is optional. Are you saying the color is irrelevant or the die is identical to the others - in computation terms?

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

Thank you ... Sorry it's not a clear problem (I guess).

FrankM
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Joined: 01/27/2017
Quick 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.

questccg
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Joined: 04/16/2011
Not 100% certain...

FrankM wrote:
...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.

Can you explain what you mean by "MAP"? You're saying 1290 outcomes... How do they "map" to 216 permutations??? What does that mean? How does this affect the outcomes??

I'm still not 100% convinced. Why? Look at this example:

• Die #1 = 6
• Die #2 = 6
• Die #3 = 3

Combinations are "663", "366", and "636". BUT there is something to consider:

• Die #1 (6), Die #2 (6) and Die #3 (3)
• Die #2 (6), Die #1 (6) and Die #3 (3)

Are both IDENTICAL where the order is not 100% important because two (2) combinations are the same.

Does this figure into your equation? Or are there simply MORE cases besides "6" to be subtracted from 1296 (6 x 6 x 6 x 6).

BTW I may be incorrect in this formulation... I just see that there SEEM to be other combinations that are the same.

Correct me if I am wrong - no worries...

Fri
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Joined: 09/06/2017
Well here is what I came up

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 number combination description Occurrences
Case #1 4 unique numbers each 1X 15
Case #2 3 numbers one repeated 2X 60
Case #3 2 numbers one repeated 3X 30
Case #4 2 numbers each repeated 2X 15
Case #5 1 number repeated 4X 6

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:

Slot 1 Slot2 Slot3
A B C
A C B
A C C
B A C
B C A
B C C
C A B
C A C
C B A
C B C
C C A
C C B

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:

Slot 1 Slot2 Slot3
A B B
B A B
B B A
B B B

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:

Slot 1 Slot2 Slot3
A A B
A B A
A A B
B A A
B A B
B B A

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.

FrankM
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Joined: 01/27/2017
Fri did a good job there

Fri did a good job of explaining what the 1296 rolls were and how the player could map them into the 216 effective assignments.

If there was a consistent way players decide (black die replaces the lowest white one), then it would be possible to work out the probabilities of those 216 assignments. But it sounds like there might be other tactical considerations.

Fri
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Joined: 09/06/2017
I found mistakes in my last post

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:

Slot 1 Slot2 Slot3
A C C
B C C
C A C
C B C
C C A
C C B

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.

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