Die from Two Dice

Download a (readable)  image_preview .pdf version.  A video demo here: YT

Alternative Dice
Need more dice? Want a D16, D24, D36 or D64? The good news is that you have more dice in your pocket than you think. Below is a system that gives you access to new dice that you (probably) didn’t know you had.

Picture this: You ask your players to roll a D96, offering them a D8 and D12 – a 5 and 10 appear on the pips. You sagaciously announce “Hrmmm … 70 huh, but is that enough …”. DM mystique leveled up, player mutiny probable!

d1Everyone knows that two D10s can be used to make a (‘composite’) D100 die. One D10 is the ‘biggie’ die (hereafter the ‘boss’ die), and the other D10 is the ‘smallie’ die (hereafter the ‘slave’ die). When the ‘boss’ die and ‘slave’ die are combined a random number from 1 to 100 is generated.

In the dice shown left/above a roll of 96 is obtained when the red die is the ‘boss’ die and the white die is the ‘slave’ die.

We are so used to the ‘D100 concept’ that we often don’t realize that we are actually using a simple mathematical procedure to work this result out.

That is, we are multiplying the 9 by 10 (i.e. 9 × 10 = 90) before adding the 6 to get 96 (i.e. 90 + 6 = 96). Obviously, if the white die were the ‘boss’ die, then the roll would have been 69 instead (i.e. (6 × 10) + 9 = 69).

So far, so good!

Plot Twist
There is nothing stopping us using the same (mathematical) procedure with any kinds of dice (i.e. not just with two D10s). By using any two dice you gain access to lots of new ‘composite’ dice (see the table below).

The only catch is that the maths is a bit more complicated than with the simple case of using two D10s. However, fear not, by using three simple rules, the procedure becomes relatively simple.

That said, if you prefer a non-mathematical option, see the ‘grid’ option at the end of this article.

Die Size (the ‘D’ number)
The size of the ‘composite’ die (the ‘D’ number) is set by multiplying the maximum possible values of both dice together. That is, two D10s give a ‘composite’ D100 die, so likewise, a D4 together with a D6 gives a ‘composite’ D24 die (i.e. 4 × 6 = 24), and a D6 and D10 give a D60 (i.e. 6 × 10 = 60). The table below gives the ‘composite’ dice that are available using standard polyhedral dice:

Composite Dice 1st Die
D2* D3* D4 D5* D6 D8 D10 D12       D20




 2nd Die

D2* D4 D6 D8 D10 D12 D16 D20 D24 D40
D3* D9 D12 D15 D18 D24 D30 D36 D60
D4 D16 D20 D24 D32 D40 D48 D80
D5* D25 D30 D40 D50 D60 D100
D6 D36 D48 D60 D72 D120
D8 D64 D80 D96 D160
D10 D100 D120 D200
D12 D144 D240
D20 D400


* = D2, D3 and D5 can be made from the repeating units found in larger dice. For example a D2 could be made by making the odd numbers on a D6 equate to 1, and the even numbers equate to 2. In a related way, a D3 can be obtained from a D6, and a D5 from a D10.

= polyhedral dice of this kind already exist, but this is an alternative option, that might be used to mess with your players. For example, a ‘composite’ D20 (D4 with a D5*) could be used to keep players from knowing if they have passed their saving throw/ability check, or not. This is more so if they don’t know which die you are using as the ‘boss’ die.

Please note, the above table does not include Dungeon Crawl Classics polyhedral dice, which also include D3, D5, D7, D9, D11, D14, D16, D18, D22, D24 and D30s, giving rise to still further possible composite dice.

Dashed cells are repeat combinations and have not been shown in the table for simplicity.

Three Simple Rules (Method 1)
As mentioned above, this system uses any two dice to make a ‘composite’ die (e.g. two D10s are used to make a ‘composite’ D100). Decide which die is the ‘boss’ die and which is the ‘slave’ die, and then roll the two dice:

  • Rule 1: Multiply the ‘boss’ die roll by the D-size of the ‘slave’ die
  • Rule 2: If the ‘boss’ die roll is the maximum value for that die (for a D6 that’s a 6), then the ‘boss’ roll equates to zero.
  • Rule 3: Add together the ‘boss’ die value from above to the ‘slave’ die roll

D10’s – Please note:  For the system to work properly the ‘0’ on a D10 should be treated as 10. This is because no other die has a ‘0’ on it.

(Method 2)
Method 2 is the same as Method 1, only Rule 2 is different:

  • Alternative Rule 2: the maximum value on ANY die equates to ZERO, unless both dice roll their maximums, then the maximum values should be taken instead.

In the traditional two D10s method, there are no “10’s” shown on the dice, they are replaced with zeros. It is only when you get two zeros that the zeros flip over and become 10s, and so give 100! Alternative Rule 2 replicates this by ‘replacing’ the maximum values on the dice with zeros.

Therefore, Method 2 works exactly the same as the traditional two D10s system, except the maths is a bit more awkward (because you need to flip maximums to zeros most of the time). Method 2 might work best if you simply paint over the maximum values on the dice, and perhaps paint a zero or add a wild-card ‘star’ symbol. Kickstarter anyone?

Worked examples (using Method 1 only)
:: Example 1:  D24

d2A D24 is rolled (i.e. using a D4 and D6). It is decided to make the D6 the ‘boss’ die and the D4 the ‘slave’ die.

3 (on the ‘boss’ die) and 4 (on the ‘slave’ die) are rolled.

Rule 1:  3 × D4 = 12
Rule 2:  This rule does not apply
Rule 3:  12 + 4 = 16 rolled on the D24

:: Example 2:  D24 (revisited)

Please note, if in Example 1 the D4 were the ‘boss’ die and the D6 were the ‘slave’ die, then a different result would have been obtained, i.e.:

Rule 1:  The ‘boss’ die roll of 4 is the maximum roll possible on the D4. Therefore, Rule 2 applies!
Rule 2:  The boss die result is deemed to be 0
Rule 3:  0 + 3 = 3 is rolled on the D24

So why is Example 1 different to Example 2? Think about a D100 where one die roll is a 9 and the other die roll is a 6. Depending on which you decide is the ‘boss/slave’ die you get either 69 or 96. Therefore, it is very important to decide which is the ‘boss’ die and ‘slave’ die before you roll. Or have a system in place like the smallest die is always the ‘boss’ die (the maths is usually a bit simpler if the ‘boss’ die is the smaller die), or the when the dice are rolled the leftmost die is the ‘boss’ die, etc.

:: Example 3:  D60

d3A D60 is rolled (i.e. using a D6 and D10). It is decided to make the D6 the ‘boss’ die and the D10 the ‘slave’ die.

2 (on the ‘boss’ die) and 6 (on the ‘slave’ die) are rolled.

Rule 1:  2 × D10 = 20
Rule 2:  This rule does not apply
Rule 3:  20 + 6 = 26 is rolled on the D60

If the ‘boss’ and ‘slave’ dice were reversed then the result would be (6 × 6) + 2 = 38.

Combining Examples 1 and 3 (i.e. D24:D60) we can generate a random time of day i.e. 16:26.

Grid Option
Not everyone will enjoy doing this kind of mental gymnastics. Thankfully, if that applies to you, then the maths can be eliminated by making a simple reference grid as explained below:

On one axis (e.g. the horizontal axis) write out the numbers of the first die. On the other axis (e.g. the vertical axis) write out the numbers of the second die. Make a grid. Then simply write 1 up to the ‘D’ number in the grid formed.

For example, for a D24 make a 6 by 4 grid, and write the numbers 1 to 24 in the boxes formed:

D24 1st Die roll (i.e. the D6)
1 2 3 4 5 6

2nd Die


(i.e. the D4)

1 1 2 3 4 5 6
2 7 8 9 10 11 12
3 13 14 15 16 17 18
4 19 20 21 22 23 24

d4Using this D24 reference grid as a tool, the result of a D24 can be easily determined when the two dice are rolled.

For example, if the D4 rolls a 4 and the D6 rolls a 5, then by simply cross‑referencing the rolls on the grid, a value of 23 is obtained (i.e. the red cell). Likewise, rolling two 3s would give 15 (not shaded in this case). Simples!

No doubt, a DM type screen with such reference grids could be made up. Also, I’m sure a program/app could be whipped up to do the same.

I suspect, you’ll either like this idea or hate it.

– – –

c AntMe on DriveThruDriveThru; at the moment I’m mainly pimping my procedural adventure ‘Carapace‘ about a giant ant colony.




17 thoughts on “Die from Two Dice

  1. Random visitor

    I must be super dumb, but I can’t see how to aply these rules to a D16 for example. I have a D4 boss, and a D4 slave. Boss rolls 1, and slave rolls 1 too. So 1*D4 = 4. 4+1 = 5. If would use a grid crossing 1 and 1 should be equal to 1 so… what am I missing?


    1. Goblin's Henchman Post author

      Let’s flip this.

      If you roll two D10s and roll 1 and 1, the result is 11 and not 1. But, using a grid it would be 1.

      To get 1 on two D10s, you need to roll a 10 (i.e. a zero; D10s don’t have 10s on them) and 1. To get 1 with two D4s you need to roll a 4 and a 1, making use of the full rule structure.

      Essentially, the ‘rules’ work if followed. For two D4s, write out the 16 possible results using the rules, and check you get results 1 to 16. You should.

      Indeed, the grid doesn’t even need to be filled out sequentially – as long as 1 to X is in the table.

      So, it’s a good question. It’s just that the ‘rules’ and ‘grid’ operate differently. What’s important is that two D4s return 16 equally likely results no matter which system is used.

      Hope that helps



      1. RV for Random visitor

        I see. I should have known there’s no direct relations between the two methods. Anyway it’s an interesting solution. I was looking for something like this since I had an idea about measuring damages in my own game by using ranges of numbers, but I couldn’t because of dice limitations. But I’m afraid even with the grid method many players would find this system somehow akward, kinda… slow?

        Maybe it could work if one dice per pair had letters instead of numbers, so the grid system could feel more apealing. I don’t know.

        Thanks for the answer and very, very cool post. Keep it up. I will revisit your blog.


  2. Goblin's Henchman Post author


    Bespoke dice are a good idea. Some ‘D100’s have one die with the base added to it already e.g. 0, 10, 20, 30, 40 … 90.

    So there is no reason why the same couldn’t be done with you D16, one die has 0, 4, 8 and 12 which is added to the normal D4.



  3. sycarion

    I think I am doing something wrong with Method 1. With a D4 as a boss die and a D6 as a slave die, a roll of 3 (boss) and slave (6), I get 6 * D4 = 24 + 3 = 27. Method 2 gives me a 3 6 is max, so 0 + 3 = 3. Am I doing that right?


    1. Goblin's Henchman Post author

      So Sycarion (updated):

      Method 1 (where D4 is boss):
      (3 x D6) = 18 + 6 = 24 (i.e. the max)

      Method 1 (where D6 is the boss):
      Since 6 is max, it becomes 0, so:

      0 x D4 = 0 + 3 = 3.

      So, I think you accidentally flipped the boss and slave die over, and didn’t make the 6 a zero. So you ended up with:
      6 x D4 = 24 + 3 = 27!!

      For completeness If using Method 2
      (3 x D6) = 18 + 0 (6 on D6 = 0) = 18


  4. kevin

    Don’t you want to subtract 1 from the boss value in method 1? Otherwise it’s impossible to generate the number 1 and the other numbers between 1 and the size of the “slave” die.

    Liked by 1 person

  5. rorschachhamster

    For some reason I find it easier to subtract 1 from the boss die bevor multipliying with the max amount of the slave die and then adding the slave die. Works as well. 1 = 0 means slave die only, 2 = 1 times max slave die + slave die etc… That means, that max on both dices means max, and you don’t have to flip around for that one time.

    Liked by 1 person

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  8. Random Dicer

    There is an easier way to create any random number with 2 dice, even with more than 2.

    Example, d4 and d6 for d24
    If d4=1, then result = d6 (+0)
    If d4=2, then result = d6 +6
    If d4=3, then result = d6 +12
    If d4=4, then result = d6 +18

    This way you get a random number from 1 to 24 with equal chance of each number to appear. And it’s easier to keep in mind.

    And if you want it on a formula:

    If dA=n, then result = dB + (n-1)*dB(max)

    d2 and d12 to create d24:
    If d2=1, the result = d12 + (1-1)*12 => d12 + 0
    if d2=2, the result = d12 + (2-1)+12 => d12+12


    1. Goblin's Henchman Post author

      Thanks for the comment. I think I sort of covered this idea (albeit from a slightly different direction). For example you could have the D4 printed with 0, 6, 12 and 18 and simply add the D6 to it.



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