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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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You can't roll a 0 with 2d10 nor with a d100. –  Lucifer Feb 20 at 11:45
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The confusion may be over d10s typically having a "0" on the die with "00" being considered 100 despite "01" being 1. Personally, I'd argue that using a good set of d10s is better than trying to use a physical d100 because of the difficulty in balancing such a die properly. A little bit of damage can go a long way in skewing outcomes. –  Sean Duggan Feb 20 at 12:19

7 Answers 7

up vote 35 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 2x 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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Note:

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 at 14:29
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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 at 16:03

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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If you roll 2d10 as 1d100 you also have to fix which is the tens digit and which is the units digit before rolling the dice. The possibility of two ambiguous readings must be accounted for or there is inherent human bias in the die readout. For example, if I rolled a 3 and a 5, that could be 35 or 53 unless I already knew which dice was the tens dice and which was the units dice.

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No, they are not equivalent. A d100, trademarked as a Zocchihedron, is not perfectly symmetrical, and therefore it is not a perfectly fair die. Some numbers will 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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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, 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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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 (9 on the tens digit and 10 on the ones digit).

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