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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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up vote 52 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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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. – SirTechSpec Jun 10 at 1:04
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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. – Philipp 14 hours ago

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%

\begin{array}{rrr} \text{roll} & \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%.

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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 (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 '15 at 6:58
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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 '15 at 18:23

No, they do not yield the same probabilities. 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. 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 they are still not fair, and do not produce the same probabilities as 2d10.

A d10 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 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.

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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 '15 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 '15 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 '15 at 4:56

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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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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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\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.)

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protected by Oblivious Sage Sep 16 '15 at 0:38

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