I don't have a lot of dice yet, just a basic set of polyhedral dice:

  • 1d4
  • 1d6
  • 1d8
  • 2d10
  • 1d12
  • 1d20

Sometimes an ability or save calls for a die that I don't have.

Strike, Unarmed



  • A Medium character deals 1d3 points of nonlethal damage with an unarmed strike.
  • A Small character deals 1d2 points of nonlethal damage.


Source: http://www.d20pfsrd.com/equipment---final/weapons/weapon-descriptions/strike-unarmed


Type poison (ingested); Save Fortitude DC 14

Onset 10 minutes; Frequency 1/minute for 6 minutes

Effect 1d2 Str damage, target can attempt one save to cure a lycanthropy affliction contracted in the past hour; Cure 1 save

Source: http://www.d20pfsrd.com/gamemastering/afflictions/poison/belladonna

Is it okay to substitute a d2 for a d4, or a d3 for a d6? I'm worried that maybe I'm overlooking some of the math behind the probability distribution.

When I'm substituting a d2 for a d4, I rule that rolling a 1 or 2 equals a 1, and rolling a 3 or 4 equals a 2. I use the same principle for d3 substitutes or any other substitute I have to make.


8 Answers 8


The d3 is a rare die, and the d2 is a coin.

Substituting a D3 can be done with any die whose total number of faces can be divided by 3. These include in your case the d6 and the d12. The easiest way to do this is to use a d6 and say the following:

1 → 1, 2
2 → 3, 4
3 → 5, 6

Or in other words, divide by 2, rounded up. You can also work in cycles, with one number per facing before working in the next cycle:

1 → 1, 4
2 → 2, 5
3 → 3, 6

But this is less intuitive to use. The d12 can be used as a d3 as well:

1 → 1,  2,  3,  4
2 → 5,  6,  7,  8
3 → 9, 10, 11, 12

But this creates more possible results than a d6, and it's easier to use the smallest die you have. You could work in cycles here as well, but that makes things needlessly complex and I advice against using this:

1 → 1, 4, 7, 10
2 → 2, 5, 8, 11
3 → 3, 6, 9, 12

The d2 problem can be solved by either flipping a coin or rolling any die with an even number of faces. Either divide the results in half to determine the outcome:

1 → 1, 2, 3
2 → 4, 5, 6

Or make the even/odd split:

1 → 1, 3, 5
2 → 2, 4, 6

You can try using a d20 on the even/odd roll with all numbers above 10 being 2:

1 →  1,  2,  3,  4,  5,  6,  7,  8,  9, 10
2 → 11, 12, 13, 14, 15, 16, 17, 18, 19, 20

Or you can use the d6 or another die, whatever you like best.

There is no mechanical difference to doing it with a bigger die than using a d2 or a d3 unless the die you are using as a substitute is crooked.

  • 2
    \$\begingroup\$ You can also simulate a d3 or d2 by rolling any size die and re-rolling if it gets a result that couldn't occur on the size of die you're trying to simulate. This can occasionally lead to an annoying number of re-rolls, however (especially if rolling a very large die), so the divide-by-x method is generally better. \$\endgroup\$
    – GMJoe
    Commented Jun 16, 2017 at 9:35

Yes you can by choosing a die that is multiple of the one your are trying to simulate and then divide, as explained by some of the others.

If you don't want to perform any calculations, you can also use any die with more sides than the one your are trying to simulate, and reroll when you roll higher than the die you are trying to simulate. So, you can use a d10 to roll a d9 by rerolling on a 10.

Of course simulating a d3 using for example a d20 might take a lot of rerolls, so you will probably want to choose a die that is close to the one you are trying to simulate.

Why this works

Lets assume that you are trying to roll a d3 using a d4 and reroll on 4. It is clear that the only possible outcomes are 1, 2 and 3. Rolling any of these on the first roll has equal probabilities (each 25%). When you roll a 4 (which has a 25% probabilitiy) you reroll. Again each of the scores 1, 2 and 3 has an equal probability of being rolled (again 25%). There is no reason why it would be more likely to roll a 1 on the second roll than a 2. This remains so for each subsequent reroll.

Therefore, the only possible outcomes are 1, 2 and 3, and each of these have an equal probability which has to be 1/3.

  • \$\begingroup\$ Continuing this math, the sum of 0.25 + 0.25^2 + 0.25^3 + . . . + 0.25^N = 1/3 in the limit where N goes to infinity, which is exactly what you want for 3 answers of equal probability. \$\endgroup\$
    – gto
    Commented Feb 28, 2021 at 9:00

Yes, you can and are intended to do that, division has no other effect on the probability distribution. You can use any other die with sides divisible by your target die size (so you can use a d6 both as a d3 and d2 safely, for example).


Substituting by multiplication and division is easy. What you can't do is substitute by adding and subtracting. Others have gone into the basic explanation here.

You can also do it with exponentiation and roots, if you give each die its own meaning. For this to work, the number of sides on the large die has to be a power of the number of sides on the small die: for example, 8 is a power of 2, so you can susbstitute 3d2 for 1d8 (because 8 is 2^3).

Line your dice up in a row, and roll them one by one. Treat the highest possible result as though it were zero (d10s already do this). Each time you roll one, you multiply by the number of sides on the smaller die, to the power of the number of times you've already rolled, so the first die gets used as-is (multiplied by 1, since anything to the zero power is 1). The second die gets multiplied by the number of sides you're using (to the first power), so in the 3d2 example, you multiply the second die by 2. The third die gets multiplied by 4 (2 squared). You keep rolling down the line, increasing your multiplier each time, and when you're done, you sum the results. This gives the same number of outcomes as the larger die, so the math comes out clean again.

Why on Earth would anyone do this? Because it's how you substitute 2d10 for 1d100. This is probably not the way you think about it -you're just assigning a die to each digit- but that's how Arabic numberals work. If you're comfortable calculating numbers in other bases, you can make some of them almost as easy: substituting 3d2 for 1d8, as above, makes a lot more sense if you think about it in binary (base-2). You could substitute 2d4 (or 4d2) for 1d16, or 2d6 (but not 6d2) for 1d36, in the same way if you really wanted to.


There is no such thing as a d3 or a d2, except for very specialised dice. What you're supposed to do with these dice is take a d4 or d6, and use half the result of each side, rounding up. (Or for a d2, just flip a coin.) It is the same with a d5, substituting a d10.

  • 3
    \$\begingroup\$ I think there is even a d1, which is basically a Möbius Strip. \$\endgroup\$ Commented May 15, 2015 at 8:47
  • 2
    \$\begingroup\$ You can also use a coin for a D2. \$\endgroup\$
    – Tashio
    Commented May 15, 2015 at 8:51
  • \$\begingroup\$ A coin also works for a d2, yes. \$\endgroup\$
    – xanderh
    Commented May 15, 2015 at 9:07
  • \$\begingroup\$ There sure is such a thing as a d3: boardgamegeek.com/image/1002239/powerboats \$\endgroup\$
    – kasperd
    Commented May 15, 2015 at 12:03
  • 5
    \$\begingroup\$ @MarcDingena No, a d1 is a marble. \$\endgroup\$
    – Bacon Bits
    Commented May 15, 2015 at 12:27

Yes, it is perfectly acceptable and possible to make a d3 from a d6. If the desired dice is a divisor of the actual dice, you can apply the modulo function to it to get the desired dice.

For example, rolling a d6 % 3 will give you a d3.

  • Roll a 1: 1 % 3 = 1 -- thats a 1.
  • Roll a 2: 2 % 3 = 2 -- thats a 2.
  • Roll a 3: 3 % 3 = 0 -- thats a 3. Just bear with me on that one.
  • Roll a 4: 4 % 3 = 1 -- thats a 1.
  • Roll a 5: 5 % 3 = 2 -- thats a 2.
  • Roll a 6: 6 % 3 = 0 -- thats a 3. Again, just bear with me on that.

This also works for d5 from a d10.

As you can see, each option is still there and equally represented on the dice.

One of the dice games I've played in the past is Button Men and it has the situation where other oddball dice sizes are needed from time to time... like, how about a d9? There's an entire page of of these special dice at Dice Math for the game.

Its possible (as mentioned) to do a d9 by rolling a d10 and discarding a roll of a 10, but you can also do it by rolling two d6 dice that are specially set up:

d9 = d6(1,1,2,2,3,3) + d6(0,0,3,3,6,6)

This works the same way a d100 can be rolled using 2d10.

If you aren't adverse to making your own dice (say you dabble in woodworking or have a laser cutter), it is also possible to make barrel dice where one can make an N sided prism with the faces etched on it.


You can absolutely do that. Cut the value in half (round up) or take the value mod 3 and add one.

There is another option, however. In this case, it's odious, but it can be a useful tool: you can always simulate a small die with a larger die. If, for example, you found yourself in need of a d3, rolling a d4 and rerolling 4s will do fine. Rerolling can be a chore, but when you need a strange and physically unfeasible die (randomly select one of 17 peasants), it's a good option.


Your intuition is correct - you can use a d6 to simulate a d3 or d2 by assigning the sides of the d6 to the numbers 1 and 2, or 1, 2 and 3.

If you find yourself needing a d2 or d3 a lot, and don't feel 100% comfortable with making the substitution, you can simply take a d6 and repaint its sides. For example, paint the 6 to show 1, the 5 to show 2, and the 4 to show 3, and you have a d3.

Of course, the paint might upset the die's balance, so you can always simply buy d2 and d3 dice, that are cubic but with the numbers appearing twice or three times. Here is one place where you can buy such a thing, and a search for "six sided d3" will turn up many others.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .