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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.

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8 Answers 8

up vote 47 down vote accepted

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

A d100 goes 1–100, a d10 goes 1–10. Neither allows you to roll a 0, because of the way you count a percentile dice. (10 on the percentile and a 6 on the other dice forms 6, 10 on one and 10 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).

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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..14, not 2..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 (AdB and XdY) have an exact mapping if and only if max(B^A,Y^X) % 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 eachother.

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 % 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 4 3's 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.

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Poor and confusing answer. BTW 1d4+1d2 averages to 4 while a d8 averages to 4.5 – Simanos Feb 26 '14 at 14:29
Simanos, this is not about summing dice. Rather this is about simulating dice (which is what the question is about). I recognize that this topic is more advanced, and would probably not come up in play except in simple circumstances. (I have used 1d4 + 1d6 to simulate a d8 before - take the 1d4 result and add 4 if the 1d6 is odd, for instance. I would not have a need for simulating a d640. As for what I mean by +, you should read my opening paragraph where I cover the distinction. So your comment tells me "This was a confusing answer because I didn't actually read it." – corsiKa Feb 26 '14 at 16:03
This is a confusing answer because it overcomplicates using 2d10 as d% – Adeptus Sep 15 at 4:41
You shouldn't use plus as notation if you're talking about combining dice rolls in any way other than addition. It makes your answer unnecessarily confusing. – okeefe Sep 15 at 19:16
Thanks. The comma notation works, and should be less confusing. – corsiKa Sep 16 at 17:23

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 (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.

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"for any number I named that 2d10 could roll, I can show only one way to roll it." This is incorrect: the probabilities of arriving at a particular number need to be equal, not the number of ways of arriving there. I.e. if there were six non-intersecting ways of rolling the value of 4 with one set of dice, each with a probability of, say 0.02, that set of dice gives P(X=4)=0.12. If a second set of dice has exactly one way of rolling 4, but that one way has P(X=4)=0.12, then the two sets of dice have equivalent probabilities for the value of 4. If this is true of all values, you are golden. – Lexible Sep 15 at 6:58
Given that each result for each die has an equal probability (since we are assuming fair dice), then the probabilities of arriving at a particular number are directly proportional to the number of ways of arriving there ie. if there are 2 ways of getting a number, it will be twice as likely as if there were 1 way. Thus The Spooniest is correct given the context of the question, although you are also correct and more explicit. – GreySage Sep 15 at 18:23

No, they are not equivalent. A d100, trademarked as a Zocchihedron, is not perfectly symmetrical, and it is not a fair die. Some numbers turn up with a higher frequency than others.

A d10, however, is a pentagonal trapezohedron, which is a symmetrical fair die. And 2d10 can be rolled as the two digits representing the 10's and 1's digits to yield a perfectly symmetrical value from 00-99.

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Note that the question asks for a comparison with a digital d100, putting the focus on ideal probabilities rather than dice imperfections. – SevenSidedDie Sep 16 at 4:33
@SevenSidedDie, the questioner explicitly mentioned both the "slow" and "fun" of rolling d100s and d10s, implying comparisons of the physical objects. I answered the question asked, which was if the physical d100 shares the same probabilities as rolling two physical d10 dice, which it does not. In his second question, "is it better to use a (digital) d100?" he used the word digital only in parenthesis, de-emphasizing it. And it only asks about the subjective "better", without saying what outcome he wants: a "fun" experience, or a "statistically perfect outcome". – John Deters Sep 16 at 4:54
Everyone else seems to have notice the probability question inherent here. It is interesting to know that a physical d100 is not actually a perfect d100, but it's a half-answer without tackling the probabilities issue. – SevenSidedDie Sep 16 at 4:56

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.

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From a purely mathematical standpoint; these are not equal concepts. Assuming fair forms of dice, a d100 is a pure set where each face equals its face value in probability. So, 55 on a d100 equals 55% chance of hitting 55 or lower (which is the primary point of % methods - "at or lower").

2d10's has two main schools of thought: one holds that 90 on the tens and 0 on the ones = 100, and the other holds that 00 on the tens and 0 on the ones equals 100. Neither method equals the probability of a 100% fair die (if one has issue with the physical die due to physical flaws, compare to a digital 100 sided die - e.g. a random number generator from 1-100).

2d10's does not equal a 100 sided probability because 2d10's coupled in this manner (that is, not summed) functions under the same rules as dual test coin toss.

If you take a coin, you have a 50% chance of getting heads or tails. If you make a rule that you need to hit heads on two consecutive tests, then your probability isn't 50% (1/2), it's 1/2*1/2 = 1/4 = 25% chance.

This is akin to how 2d10's as "percentile" work. If your target number is 55, then you need your tens throw to be 5 or less before you can consider your ones throw valid. Your one's throw is this considered at 5 or less.

The ones is dependent on the tens die and both need to succeed to grant a success.

Depending on which way you treat the die (90|0=100, or 00|0=100), your probabilities vary.

Take the above example. In 90|0 = 100: Your tens die hitting 50 = 60% chance This is because your values at or less than 50 are, 50, 40, 30, 20, 10, 00. 6/10.

Your 5 on the ones has a 50% chance, however. This is because 90|0 treats the ones as 1, 2, 3, 4, 5, 6, 7, 8, 9, 0 (10).

Just as we multiply on the coin toss we multiply the d10 results because the second test (ones) is dependent on the first test (tens).

So that becomes 6/10*5/10 = 3/5*1/2 = 3/10 = 30%

In the 00|0 = 100% Your tens hitting 50 = 60% as well for the same reason as the 90|0. However, your ones, unlike the 90|0 set, are also the same probability as your tens. This is because in this set, the ones are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 As such, 5 is 6/10.

This becomes 6/10*6/10 = 3/5*3/5 = 9/25 = 36%

This is, again, because the total test is two tests whereby the second test is dependent upon the first test (just like two heads in a row on a coin).

This creates a system for either method which looks something like a car's RPMs through gears. On a note, given the two, I'd rather use 90|0 method as there's a greater chance of 100 than on 00|0 - in 00|0, you have a 1% chance of accomplishing a 100%, whereas on the 90|0 (due to the nature of how it considers the 0 to be a 10 on the ones die) you have a 100% of hitting 100 or lower; a concept not as easily applicable to the 00|0 due to the 00 being re-used as both a 00 for tens and if combined with a 0 from the ones, only then equaling a 100. Basically, on the 00|0 method, 99 is your 90|0 equivalent "100 or below" in terms of probability. Plus; you don't have to change the rule of what the 00 refers to dependent on what the ones die is doing.

Either way, however, neither are equal to a percentile die system. The closest die to a d100 in terms of probability is a d20, which shares the exact same probability curve as a d100, but within a shorter interval (intervals of 5%, rather than intervals of 1%).

Note: graph represents the given value or lower as a two-test system at each value presented, always including the consideration of the ones position. 2d10 vs d100

Due to comments continually being erased, I'll try to help in this answer as best as I can to summarize the issues raised and discussed in the comments.

Some raised concern by pointing out that 1/10*1/10 is exactly equal to 1/100, or 1%. This is true. However, with 2d10 or d100, we aren't asking whether or not a single face value will be achieved. Instead, we are asking if a given value or below will be achieved, which is a very different question. It is in this form of the question of result that 2d10 radically departs from d100 probabilities.

It was also raised as a point that the above graph doesn't seem too good at representing odds of a given number or below, which is true; it is a general graph which always includes the ones and shows each value's range.

To plot what it looks like for a given value or below, we have to pick a given number and show that value set in both 90|0 format and 00|0 format. Here is what 50 and 55 as a target number looks like in probabilities on both forms of 2d10 as percentile.

The values on the left are the tens, while the values on the right are the ones. The ones are only of need for consideration when the tens on the left hits the 10% results respectively.

90|0 and 00|0 target 50 and 55

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Reminder: comments are for clarifying content, not discussion. Please take any discussion to Role-playing Games Chat. The prior 3̶8̶ 44 (!) comments have been purged. – SevenSidedDie Sep 16 at 1:45
I created a chat channel for this answer here: – MichaelS Sep 16 at 7:08
I would love to chat about this, but the room requires 20 reputation points to write in, so I cannot. – TheStumps Sep 17 at 1:32
@TheStumps The non-Q&A features of the site are gated by first submitting answers or questions that are well-received by the community; that's by design, in order to enforce with new users the site's laser-focus on productive Q&A. If you'd like to use chat, first you'll have to go beyond this single-issue interest, and contribute elsewhere on the site. All you need is an answer with two upvotes, or a question with four, to get the requisite 20 reputation. – SevenSidedDie Sep 17 at 17:35

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).

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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 lets take 10 possible results from 2d10 and d%

 roll     2d10 result    d% result
 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%

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%.

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protected by Oblivious Sage Sep 16 at 0:38

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