# Is this the right formula for the total permutations of a set of dice? [closed]

I know the total number of permutations (binomial) on tossing $n$ fair coins is given by:

$$\mathrm{Total} = 2^n$$

What about dice? If I grab some dice and roll $4\mathrm{d}2 + 4\mathrm{d}4 + 6\mathrm{d}3$, is the total number of permutations given by this formula?

$$\mathrm{Total} = 4^2 \times 4^4 \times 6^3$$

• This might be better as a Math.SE question. – Ian Miller May 6 '17 at 2:05
• did you mean 4 to the power of 2, and 6 to the power of 3? would that not be rolling 2d4 and 3d6? – Tritium21 May 6 '17 at 8:07
• Did you mean 4d2 + 4d4 + 6d3, or did you actually mean to ask about 2d4 + 4d4 + 3d6? (Few people I know could “grab some dice” and end up with any d2s or d3s, let alone many of them.) – SevenSidedDie May 8 '17 at 15:28
• I'm voting to close this question as off-topic because it's a statistics and dice question with no direct relation to RPGs or RPG design. – WrongOnTheInternet Jul 14 '17 at 19:06
• @WrongOnTheInternet I think it's on-topic because understanding dice probabilities and how to calculate them is understanding how to use dice, which is understanding how to use an RPG tool, which is on-topic. There's an ongoing meta on this, though, so I'll hold off on reopen votes. – Please stop being evil Jul 15 '17 at 5:40

Dice permutations are calculated as "Permutations with repetition/replacement," which means, intuitively speaking, that if you roll 2d6, getting a 6 on the first die does not prevent you from getting a 6 on the second die. (As opposed to "Permutations without repetition/replacement." An example of that is putting putting six numbered slips of paper into an urn and drawing out two of them, literally without replacing the first slip before you draw out the second-- now, getting a 6 on the first slip does prevent you from getting a 6 on the second slip.)

Calculating permutations with repetition is very easy. I'll build up the intuition in a few short steps:

1. Consider rolling 1d6. The number of permutations is, trivially, 6:

\begin{array}{r|llllll} \text{Die One} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} \\ \hline \end{array}

I'll note in passing that $6^1$ happens to be $6$ and that we could use this for any single roll of some die with Y faces: $$\text{1dY} \rightarrow Y^1 = Y$$

1. Consider rolling 2d6:

\begin{array}{r|llllll} \text{Die Two\Die One} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} \\ \hline 1 & \text{1,1} & \text{1,2} & \text{1,3} & \text{1,4} & \text{1,5} & \text{1,6}\\ 2 & \text{2,1} & \text{2,2} & \text{2,3} & \text{2,4} & \text{2,5} & \text{2,6}\\ 3 & \text{3,1} & \text{3,2} & \text{3,3} & \text{3,4} & \text{3,5} & \text{3,6}\\ 4 & \text{4,1} & \text{4,2} & \text{4,3} & \text{4,4} & \text{4,5} & \text{4,6}\\ 5 & \text{5,1} & \text{5,2} & \text{5,3} & \text{5,4} & \text{5,5} & \text{5,6}\\ 6 & \text{6,1} & \text{6,2} & \text{6,3} & \text{6,4} & \text{6,5} & \text{6,6}\\ \end{array}

You can count the entries in the table and come up with 36. But you can also think of this as extending the table for a 1d6 roll: Each of the six entries in the first table is turned into its own new, separate list with six unique entries. This is because no matter what we roll on the first die, we can get any of the six values on the second die. So we can simply calculate 6 times 6 entries = 36. If we generalize that to two dice, each with Y sides, that leaves us with:

$$\text{2dY} \rightarrow Y^2$$

1. Consider rolling 3d6. I'm not going to draw the big table, but extending what we did above, we turn each of the 36 table entries into, again, its own unique list of six more entries. (Because again, no matter what we roll on the first two dice, we can roll any of the six values for the third one.)

For the specific case of 3d6 we would have 36 times 6 = 216 entries.

This basic thought process holds for any die with Y sides that we toss X times, i.e., $$\text{XdY} \rightarrow Y^X$$

Note that the X and the Y swap places on either side of the expression! This is not a mistake.

1. But what about something weird, like 1d6 + 1d4? The same basic process:

\begin{array}{r|llllll} \text{Die Two\Die One} & \text{1} & \text{2} & \text{3} & \text{4} & \text{5} & \text{6} \\ \hline 1 & \text{1,1} & \text{1,2} & \text{1,3} & \text{1,4} & \text{1,5} & \text{1,6}\\ 2 & \text{2,1} & \text{2,2} & \text{2,3} & \text{2,4} & \text{2,5} & \text{2,6}\\ 3 & \text{3,1} & \text{3,2} & \text{3,3} & \text{3,4} & \text{3,5} & \text{3,6}\\ 4 & \text{4,1} & \text{4,2} & \text{4,3} & \text{4,4} & \text{4,5} & \text{4,6}\\ \end{array}

Just note here that each of the six entries from the first die is only matched with FOUR entries from the second die. In the expression below, I'm using Y values with subscripts

$$\text{1dY}_1 + \text{1dY}_2 \rightarrow Y_1^1 \times Y_2^1$$

1. It doesn't matter how weird we get from there, as long as we're talking about simple dice, we just keep multiplying by the number of faces on the new die we've just added. The full formula in all its generality becomes:

$$\text{X}_1\text{dY}_1 + \text{X}_2\text{dY}_2 + \dots + \text{X}_N\text{dY}_N \rightarrow Y_1^{X_1} \times Y_2^{X_2} \dots Y_N^{X_N}$$

1. As a final note, the addition of a constant (e.g., 3d6 + 6) changes nothing, and the constant can be ignored for finding the number of permutations. Technically, one might consider it "a single sided die" and multiply by one, but that's a bit precious.

For your specific example, there are 2,985,984 permutations:

\begin{array}{r|llll} \ N & X_N & Y_N & {Y_N ^ {X_N}} \\ \hline 1 & \text{4} & \text{2} & \text{16} \\ 2 & \text{4} & \text{4} & \text{256} \\ 3 & \text{6} & \text{3} & \text{729}\\ \hline \text{product} & \text{ } & \text{ } & \text{2985984}\\ \end{array}

• Addition of MathJax to site: +1, Novak approves. – Novak May 6 '17 at 21:00

Each die is an independent event. The correct number of permutations is $2^4 4^4 3^6$.