I know d100s are shunned because they take too much time to roll, and 1d10+1percentile is really fun, but do they share the same probabilities? Is it better to use a (digital) d100?

Also, do d100s have 0? You can't roll a 0 with percentiles, I think.


7 Answers 7


Yes, a d100 is the same as 2d10 with one as the percentile.

A d100 goes 1–100, a d10 goes 0-9. Neither allows you to roll a 0, because of the way you count a percentile dice. (00 on the percentile and a 6 on the other dice forms 6, 00 on one and 0 on the other is 100, no option will result in 0.)

Do remember to use different colors of dice, else you will find things getting confusing quickly.

As an extra to the answer, it just occured to my why you asked the question. I guess you were wondering because 2d10 is not the same as a d20?

This is because (asides from one being a range of 2–20), there are multiple ways of getting the same result. If you roll 2d10, you can get a result of 7 by having: a 5 and a 2; a 6 and a 1, a 3 and a 4, etc. Because there is more than one way to get a ‘7’ result, you are more likely to get a 7 than a 2, which you can only get with 2× 1.

In a situation with a percentile dice and a ‘normal’ one, there is only one way to get every result (1 in a 100, to get 54 you need 5 on the percentile and 4 on the normal, no other result will work).

  • 6
    \$\begingroup\$ It's probably worth noting that the reckoning is different depending on whether or not you're using special percentile dice, where the first one is labeled 00-90 and thus 10 and 6 would be 16, not 6. \$\endgroup\$ Jun 10, 2016 at 1:04
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    \$\begingroup\$ By the way: You can substitute a D20 with a D10 and a coin (aka "D2"). When the coin lands heads-up, add 10 to the D10. \$\endgroup\$
    – Philipp
    Jul 28, 2016 at 15:18
  • 2
    \$\begingroup\$ 2d10 is standard notation for 1d10+1d10, very different from d100. Maybe express that semtence differently to avoid confusion. \$\endgroup\$ Mar 23, 2021 at 15:37

D100 and d%+d10 have exactly the same probabilities. If all 3 dice involved are fair, then they should come up with very similar distributions when rolled repeatedly. Obviously this isn't always the case as dice aren't consistent and there is a lot of randomness unless you roll a lot of times.

It seems there might be some confusion as to why d% doesn't have a bell shaped curve since it's two dice being rolled. It looks a lot like rolling 2d10 which is not the same as rolling d20 (or d19+1). 2d10 has a bell shaped curve because the results are added. there are the same number of permutations of 2d10 as there are of d%, but the number of possible results is 19 instead of 100.

To show this let's take 10 possible results from 2d10 and d%

\begin{array}{r|cc} & \rlap{\text{meaning as a...}}\\ \text{dice faces} & \text{2d10 result} & \text{d% result}\\ \hline 1,1 & 2 & 11\%\\ 1,2 & 3 & 12\%\\ 2,1 & 3 & 21\%\\ 2,2 & 4 & 22\%\\ 3,1 & 4 & 31\%\\ 3,2 & 5 & 32\%\\ 4,1 & 5 & 41\%\\ 4,2 & 6 & 42\%\\ 5,1 & 6 & 51\%\\ 5,2 & 7 & 52\%\\ 1,5 & 6 & 15\%\\ 2,5 & 7 & 25\%\\ \end{array}

As you can see there is no duplication of results between 2d10 and the d%. You can also observe that ordering of the dice is significant, this is why when we roll percentage dice we roll either two different colors, specifying which die is the 10s place and which is the 1s, or we use specially marked dice with 1-0 and 10-00.

As far as rolling 0%, no, neither a d100 nor a d% can roll a 0. They range from 1-100%.


No, they do not yield the same probabilities. While in theory they should be the same, in practice the current omnipresent d100 is the trademarked Zocchihedron, which is not perfectly symmetrical. An article from White Dwarf 85 (Jan 1987) ran some tests with the Zocchihedron and concluded it is not a fair die, with some numbers turning up with a much higher frequency than others after 5,164 rolls:

photo of the graph from the White Dwarf article, showing a bell-like curve with many frequency spikes

The first version of the die turned up numbers lower than 8 and higher than 92 much less frequently than other numbers. These were slightly addressed in later versions, where the high and low values were painted on different faces, but that only adjusted the statistical distribution of the bad rolls; it couldn’t fix the underlying problem.

The real problem is they are not perfectly symmetrical. The ends are slightly pointed, so numbers painted on those faces rarely turn up. They are not fair, they cannot be made fair, and they do not produce the same probabilities as rolling 2d10.

A d10 is a pentagonal trapezohedron, which is a symmetrical fair die. Even an imperfect d10 is far more fair than a d100 can be, due to the limits of design. And 2d10 can be rolled as the two digits representing the 10's and 1's digits to yield a perfectly symmetrical distribution of values from 00-99, giving a nice flat probability curve.

A digital d100 dice app will also yield a symmetrical distribution of values from 0-99%, but it won't inspire as much excitement in your players as when you trot out the giant golf-ball of randomness.

  • \$\begingroup\$ RE: "These were slightly addressed in later versions, where the high and low values were painted on different faces, but that only adjusted the statistical distribution of the bad rolls; it couldn’t fix the underlying problem." Really, in 30 years, no real effort's been made to make a fairer d100? That's pretty shocking, really. I mean, some guys made a fair d120—a fair d100 should be a snap by comparison. \$\endgroup\$ Feb 4, 2018 at 18:43
  • \$\begingroup\$ @HeyICanChan I expect it's merely due to infinitesimal demand compared to non-zero development cost. We already have means to roll d100 fairly, and the physical d100 novelty is nearly non-existent (as compared to the 90s) in these days of d120s and d13s and etc. \$\endgroup\$ Feb 4, 2018 at 22:48
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    \$\begingroup\$ @HeyICanChan , a d120 is fair because the geometric shape itself is a mathematically symmetrical shape. A d100 is not symmetrical - you can see this by examining it. The d100 has defined “ends” where the shapes or arrangement of faces are different than other faces on the dice. This is true of the Zocchihedron, and all the other shapes of d100s out there. Look at any of the closeup pictures of d100 on Amazon and you’ll see that they’re not symmetrical. An end might be hexagonal, but it’s right next to a five-sided blob shape. \$\endgroup\$ Jan 8, 2021 at 17:14
  • \$\begingroup\$ It might be wise to incorporate these (several-years-old!) comments into the answer. I mean, you visit so rarely and all. ;-) \$\endgroup\$ Jan 8, 2021 at 19:50

The standard method for 2d10-as-1d100 is to designate ahead of time which die will serve as the tens digit and which die will serve as the ones digit, treating a roll of 00 as equal to 100.

From a probabilistic standpoint (assuming fair dice), this is equal to 1d100. The key to understanding this is that you don't add the numbers up: each number is its own digit, independent of the others. This is what makes it work. There are two factors at play in that:

  • There are no outcomes on 2d10 which produce a result that couldn't be rolled on 1d100. Likewise, there are no result on a d100 that can't be produced on 2d10.
  • For every result that could be rolled on 1d100, there is exactly one 2d10 outcome which produces that result.

Let's compare this to, say, a claim that rolling 2d10 and summing the numbers (treating a 0 on each die as 10) is equivalent to rolling 1d20. The dice involved in the two claims are the same, but your claim is equivalent and mine is not.

Your claim fits the first point: I could name any result that a d100 could roll, and show how 2d10 could roll it. I could also do this backwards: if I name any result that 2d10 could roll, I can show how 1d100 could roll it. My claim fails this test, because I can roll a 1 on 1d20, but I can't roll it on 2d10.

Your claim also fits the second point: for any number I named that 2d10 could roll, I can show only one way to roll it. The real key here is that each result has the exactly the same number of ways to roll it: that number happens to be one, but that doesn't really matter here (though it makes the math easier). Because of this, all numbers on 2d10 are equally likely, just like on 1d100.

My claim is different. There's only one way to roll a 2 on 2d10-and-sum roll (1+1), and there's also only one way to roll a 20 (0+0): these two numbers are equally likely. But, for example, there are three ways to roll a 4 (1+3, 3+1, 2+2). This means that a 4 is more likely to come up than a 2. That's not like 1d20, where all results are equally likely, so they're not equivalent.

This is why 2d10-as-digits is equivalent to 1d100, even though 2d10-and-sum isn't equivalent to 1d20.


Rolling 2 d10s is the same as 1d100 for one simple fact:

They are rolling the digits independently as opposed to rolling 3d6 or 1d18 in which the results are added together and the lower cap for each set can be different (3 or 1 respectively).

When you roll 2 d10s for the purpose of replacing 1d100 the lowest result you can have is 1 (created by rolling 10 on tens digit and a 1 on the ones digit) and the highest you can roll is 100 (10 on the tens digit and 10 on the ones digit).


Yes. A 1d100 and a 2d10 share the same probability chances.

A 1d100 die, and 2d10 both have equal opportunities to represent numbers from 1-100. The 1d100 can display numbers 1-100, and the 2d10 can display the numbers 00-99. (In some games, a 0 on both dice is interpreted as "100" instead of "0". In this interpretation, the 2d10 displays 1-100, just like the d100.)

Both dice combinations have 100 possible results, and each of the 100 results are unique, which results an an equal probability that either has a chance to land on any specific number.



the following only applies when dice rolls are treated as unique. In most cases, we don't treat dice like we do below. Normally, 1d6 + 1d8 would usually have a range of \$[2\mathrel{{.}\,{.}}14]\$, not \$[2\mathrel{{.}\,{.}}48]\$ because in most cases we sum dice instead of counting each individual combination as unique. Below, I use the notation 1d6 + 1d8 to mean a single roll of these and we are counting the unique possibilities the dice could provide. So 2d6 treats \$[1,3]\$ as a different roll than \$[3,1]\$ even though when we roll for damage, they're both just 4 (a pretty crappy roll, honestly...). So keep that in mind when reading this post!!

Moving on...

Mathematically, two rolls (\$A\text{d}B\$ and \$X\text{d}Y\$) have an exact mapping if and only if \$max(B^A,Y^X)\mod min(B^A,Y^X) = 0\$. That is to say if you take for each roll the number of sides raised to a power equal to the number of dice, and the lower value divides into the higher value, there will exist an exact mapping.

In your case, not only do they divide each other but they're equal (\$100^1 = 10^2)\$ so they have a 1:1 mapping. This means you can apply any 1:1 mapping you want. You could have \$[5,2]\$ on your d10s map to 59 on the d% if you really felt like it and had a methodology for tracking what results map to each other.

You can try this with any combination: 5d2 ≉ 2d6 because \$2^5 = 32\$ and \$6^2 = 36\$. However, you could come very close to approximating your 5 coins with a 2d6 roll and just mapping 4 values to reroll.

You can even combine dice, like so:

  • 1d4, 1d2 ≈ 1d8 because \$4✕2 = 8\$
  • 2d5, 1d4 ≈ 2d10 because \$5^2 ✕ 4^1 = 25 ✕ 4 = 10^2 = 100\$
  • 2d8, 1d10 ≈ 3d6, 1d12, 1d20 because \$8^2 = 64 ✕ 10 = 640\$ and \$6^3 = 216 ✕ 12 ✕ 20 = 51840\$ and \$51840 \mod 640 = 0\$

When you add dice rolls, you multiply their power results together. So that last roll means that if you had needed to simulate [2d8, 1d10] but you only had [3d6, 1d12, 1d20], you could still do it with a proper mapping. (For those keeping track, it's an 81:1 mapping because \$51840 / 640 = 81\$. This means that you can do anything you want to eliminate 81 (\$3^4\$) from the dice you do have to make it work. There's only four 3s in the dice you have, so you have to map your 3d6 to 3d2 (using 1,2,3 vs 4,5,6 or odd vs even) and then mapping your 1d12 to 1d4. Now you have mapped both sides to 640 combinations, from which you can apply your garden variety 1:1 mapping.)

  • \$\begingroup\$ What is an exact mapping? I know what injections, surjections and bijections are. \$\endgroup\$
    – Tommi
    May 10, 2017 at 8:19
  • \$\begingroup\$ @Thanuir For example, you can make an exact mapping of a regular six sided die to a coin (a two sided die) by mapping 1,2,3 to heads and 4,5,6 to tails. You cannot make an exact mapping of a six sided die to a four sided die because there would be leftover elements. So while bijective is one-to-one, the exact mapping is X-to-one where no element in set A points to multiple elements in B. If you grouped the the elements in A into sets, those sets would be bijective onto B. \$\endgroup\$
    – corsiKa
    May 10, 2017 at 16:15
  • \$\begingroup\$ Okay, so a mapping where the pre-image of every element has the same cardinality (and so a surjection, in particular, unless one considers the map from the empty set to a non-empty set). Is this standard terminology in number theory, algebra, or in some other field? Anyway, you might want to clarify this in the post itself, as I highly doubt I am the only reader who did not immediately understand the terminology. \$\endgroup\$
    – Tommi
    May 11, 2017 at 6:30
  • \$\begingroup\$ As I researched the topic I was unable to find a standard term to describe the phenomenon. Perhaps a question for math stack? \$\endgroup\$
    – corsiKa
    May 14, 2017 at 17:09
  • \$\begingroup\$ I am reasonably familiar with mathematics terminology, which is why I asked - this would have been something I should know if it was established terminology. \$\endgroup\$
    – Tommi
    May 15, 2017 at 10:44

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