I have an idea for a roleplaying game that uses continuous probability distributions (like the normal distribution) to decide random outcomes (instead of discrete distributions that comes from using dices). I am wondering if this already exists?


Jill has 50 Dex and has +10 in firing bows. She tries to fire an arrow some distance. Her skill gives her normal distribution function a mean value of 60 and the properties of using a bow gives this action a standard deviation of 15. because of the difficulty of the shot needs to "roll" a 70 or higher.

Then you use any normal distribution random generator N(60,15) (like Matlab, internet or some app on smart phones) to roll and see if you get higher or lower than 70. if she gets 90 or higher it's a critical hit, if she gets 30 or lower it's an epic fail.

(Note that I haven't thought these numbers through at all and I can see like five holes in these numbers while writing them, the example is for the demonstration of how the mean value and the standard deviation would be decided and how the distribution should be used.)

Does this exist anywhere?

  • 4
    \$\begingroup\$ The edits made to this question have brought it to a point where it has nothing to do with the title "continuous probability" nor anything to do with the existing answers. If you discover from the answers to your question that you actually have a different question (or simply asked the wrong one) please ask a new question. We can have two great questions with great answers instead of one mediocre one. I'd suggest reverting some of the edits and separating them into a new question. \$\endgroup\$
    – Cirdec
    Apr 11, 2015 at 16:53
  • \$\begingroup\$ You may also want to un-accept my answer, as you've said it doesn't solve your problem. \$\endgroup\$ Apr 11, 2015 at 18:52
  • \$\begingroup\$ I will do that. I accepted the answer since it answered the question "does it exist?" Which was the original question. \$\endgroup\$
    – marbri
    Apr 11, 2015 at 18:54

5 Answers 5


Yes, this exists... sort of.

I say "sort of" because you're assuming that continuous probabilities are possible to generate, when it is physically impossible to do so. (All random number generators, both digital and physical, can only produce discrete numbers, even if there are so many that the curve looks smooth; and the results from physically-contnuous randomisers are inherently limited by the discrete resolution of your measurement tool.)

However, the motive that you appear to have underlying this desire is, indeed, a goal that some roleplaying games have been written to satisfy. Rather than using summed dice, a small but persistently popular selection of RPGs are percentile systems.

In a percentile dice system, the probability of an outcome is modelled with percentages, as in your example, and success/failure — including critical successes and epic failures — can be determined by comparing a roll of percentile dice to the probability: any number equal or lower is a success.

These dice have rolled an 18 result, which in "roll under" percentile systems would be a clear success against a probability of 70%:

A "tens" and "ones" pair of percentile dice showing '10' and '8': a result of 18
Image © IanWatson, licensed under CC-BY-SA 3.0

These dice do not generate a bell curve, however — in such systems, the location of the bell curve is in the part of the system that determines the percentage odd you need to roll under, and the dice only select a percentage from the always-flat flat curve that 0%–100% always has, to compare against the already-generated bell curve marked with percentiles.

Depending on the system, that 18 result might also be a critical success. Few systems use standard deviations by name, but some do use static numerical modifiers to the probability which end up being equivalent (e.g., a task that is easy and gets a +20 to the roll, which is the same as if the bell curve was shifted 20 points to the right.)

There are a number of games that use percentile dice systems to generate transparent probabilities like this. The very first one in RPG publishing history is and was Chaosium's RuneQuest (1978), which they eventually turned into a "house system" used in many other of their games and even other publishers' games. That original percentile game system is still available (with some small changes) in either Chaosium's universal-genre RPG Basic RolePlaying (2008), Chaosium's fantasy-specific Magic World (and its free Quick-Start edition), or a recent RuneQuest edition and a number of the retroclones of its older editions.

There are lots of percentile systems out there. I can't recommend any single one, but all of them satisfy the desire you and many other roleplayers have had for clear, real-looking probabilities in a roleplaying game. And as a bonus, their dice-based resolution is quite a bit faster than trying to use Matlab during a roleplaying session!

  • \$\begingroup\$ I understand what you mean. What I dont like with those systems is that advancement is most often linear which means that if you advance with 10 then its 10 percentile units easier to succeed with the percentile system, but with the normal distribution system, the change in probability from a 10 point advancement would depend on how close you already are to succeeding. Also, some things we do are more random and have less to do with skill(larger std) than others. Like firing an arrow versus picking a lock, both require skill but firing an arrow has more random variables affecting it. \$\endgroup\$
    – marbri
    Apr 9, 2015 at 19:18
  • 2
    \$\begingroup\$ @Martin You have to consider what standard deviation actually matters though: for an arrow, it doesn't matter how complex the inputs are, when the output is binary "hit or miss". Even when you add more outputs (crit pass/fail), you still have discrete outputs that don't care how complex the inputs are. In the end more random things and less random things both have an single X% probability to do Y outcome. As for advancement: most d% games make advancement slower the higher your skill is, so you get larger increases over time at the low end and smaller increases over time at the high end. \$\endgroup\$ Apr 9, 2015 at 23:12

Ultimately, your “normal-distribution random number generator” is neither random nor normally-distributed. It’s also not actually continuous; all numbers in a computer are always going to be discrete.

Any system that rolls a number of equally-sized dice and sums them up is more-or-less approximately normal (see also: central limit theorem). More dice produces a smoother result, and the results of your computer’s algorithm is mathematically identical to just rolling a very-large number of very-many-sided dice. And all that really accomplishes is a smoother approximation of a normal distribution than, say, 3d6.

To a certain extent, the smoothness matters. Certainly, I can look at the probability distribution of 3d6 graphed from 3 to 18, versus a graph of what Matlab or Wolfram Alpha call a normal distribution with the same mean and standard deviation, and see a striking difference. But that difference isn’t really that big; I tend to doubt that players will really notice a difference.

  • \$\begingroup\$ Thanks fo the feedback. The main difference im thinking of is that its complicated to vary the std if you can only use a handfull of dice(more would be difficult and exhausting). And if one would go through all the trouble to find out a good way to vary the std with a handfull of dice chances are it would result in a not so simple formula that is a hassle to use everytime you need to. I realize that a computer also uses a discrete distribution, but the "steps" are of the size 10^-16 (in matlab). I feel like its closer to the real world since some things are more widely distributed than others. \$\endgroup\$
    – marbri
    Apr 9, 2015 at 17:21
  • \$\begingroup\$ @Martin You should clarify your question, then, to say that. The short answer is, no, I don’t think that’s a good idea. \$\endgroup\$
    – KRyan
    Apr 9, 2015 at 17:30

It's very hard to detect the difference. You can just use 3d6 or something similarly simple, because you won't really notice a difference

Let's pretend you have two ways of determining the result of tests like the one Jill is participating in. One of them generates results from an ideal normal distribution. The other is a small handful of dice, say 3d6. You choose one to use for the game and hide it behind a GM screen so the players don't know which one you are using. We're going to choose the 3d6. How long will it take the players to detect that you are using the 3d6 instead of the ideal normal distribution? We'll work through the Jill example.

Jill's player knows that Jill will shoot that distance if she rolls over \$70\$ on a normal distribution with mean \$60\$ and standard deviation \$15\$. Looking this up in a table, she expects that she will succeed \$0.2524925\$ of the time.

Behind the screen we calculate that Jill's player needs to hit \$\dfrac 2 3 = \dfrac{70 - 60}{15}\$ of the standard deviation above the mean to shoot the arrow that distance. We have 3d6 with mean \$10.5\$ and standard deviation \$2.95804\$. She needs to hit \$ 12.472 = 10.5 + \frac 2 3 \times 2.95804 \$, so she will shoot the distance with a roll of 13 or higher. She only makes the shot \$0.199012\$ of the time.

Jill's player thinks that you might have 3d6 hidden behind the GM screen. To find out if you do, she is going to attempt this shot repeatedly and keep track of how many times it is successful or not. Asking for help on cross-validated, she finds out that if she wants suggestive evidence (\$68\%\$ sure you are using the wrong dice or \$1\$ standard deviation) she will need to repeat the shot this many times:

$$ 244 = \left( \dfrac{1}{ \arcsin (\sqrt{0.2524925}) - \arcsin (\sqrt{0.199012}) } \right)^2 $$

If she wants to convince herself (96% chance she's right or 2 standard deviations) she will need to make this many shots:

$$ 975 = \left( \dfrac{2}{ \arcsin (\sqrt{0.2524925}) - \arcsin (\sqrt{0.199012}) } \right)^2 $$

Jill's player can't even suspect that you aren't really using an ideal normal distribution until she's made 244 shots and can't be convinced of it until she's made 975 dice rolls.

If Jill's player is particularly clever, she might decide to find out what test in the game will allow her to, as quickly as possible, demonstrate that you are using 3d6 instead of an ideal normal distribution. She carefully considers a graph of the cumulative distribution functions for both 3d6 and the ideal normal distribution with mean \$10.5\$ and standard deviation \$2.95804\$. These are the probabilities that her roll will be less than or equal to the target number.

3d6 compared to normal distribution

She realizes that the error is biggest at the integer values, which are exactly what you can roll. The errors not measured on the left side of the graph are exactly what's measured on the right. She writes down the values of the CDFs for each of the possible dice rolls. For each pair of probabilities she calculates how many checks will be required to be concerned (\$68\%\$ chance or \$1\$ standard deviation) or convinced (\$96\%\$ chance or \$2\$ standard deviations) that you are using the wrong dice.

\begin{array}{lllll} \text{Target} & \text{3d6 <= Target} & \text{Normal} & \text{N (Z=1)} & \text{N (Z=2)} \\ \hline 3 & 0.00462 & 0.00561 & 20948 & 83792 \\ 4 & 0.0185 & 0.013 & 3113 & 12451 \\ 5 & 0.0462 & 0.031 & 676 & 2704 \\ 6 & 0.0925 & 0.064 & 353 & 1412 \\ 7 & 0.162 & 0.118 & 252 & 1006 \\ 8 & 0.259 & 0.199 & 194 & 776 \\ 9 & 0.375 & 0.306 & 189 & 754 \\ 10 & 0.5 & 0.432 & 221 & 883 \\ 11 & 0.625 & 0.567 & 288 & 1149 \\ 12 & 0.740 & 0.693 & 370 & 1480 \\ 13 & 0.837 & 0.800 & 432 & 1728 \\ 14 & 0.907 & 0.881 & 567 & 2266 \\ 15 & 0.953 & 0.935 & 655 & 2618 \\ 16 & 0.981 & 0.968 & 570 & 2280 \\ 17 & 0.9953 & 0.9860 & 393 & 1570 \\ 18 & 1.0 & 0.994 & 178 & 712 \end{array}

Jill can't suspect that you are using the wrong dice until making 178 dice rolls, and she can't be sure until she makes 712. The difference between 3d6 and an ideal normal distribution won't be detectable except in very long campaigns.

The following summarizes how long it would take players to detect that an ideal normal distribution isn't being used for their checks for various popular dice systems.

\begin{array}{lll} \text{Dice} & \text{N (Z=1)} & \text{N (Z=2)} \\ \hline \text{d6} & 14 & 55 \\ \text{2d6} & 52 & 207 \\ \text{3d6} & 178 & 712 \\ \text{4d6} \text{ (opposed 2d6)} & 266 & 1063 \\ \text{6d6} \text{ (opposed 3d6)} & 409 & 1636 \\ \text{d20} & 20 & 80 \\ \text{4d3} \text{ (fudge dice)} & 72 & 287 \\ \text{8d3} \text{ (opposed fudge dice)} & 136 & 541 \end{array}

source code

  • \$\begingroup\$ Thanks, this was very useful information. This wasn't the main point of the question in mind, although i realize i posed it poorely. The main point is not the way to determine the outcome(i.e. dice or computer) but the way to determine the odds. A better way of showing my idea would be: to succeed with anything you need to roll 100, Juno has a skill level that gives the mean, say 70, and Juno wants to pick a lock. Picking lock is more about skill than luck, which gives a low std=15. so the odds would be P(x>=100) on N(70,15). \$\endgroup\$
    – marbri
    Apr 11, 2015 at 12:25
  • \$\begingroup\$ If she instead would have wanted to shoot an arrow, which is more about luck than lock picking, the std would have been higher, std=35. Then, if she has a skill of 70 in archery, the odds would be P(x>=100) on N(70,35). This reduces the problem(in writing the system) to finding out the luck-to-skill ratio of actions, there are methods to do this(at least with baseball). \$\endgroup\$
    – marbri
    Apr 11, 2015 at 12:27
  • \$\begingroup\$ @Martin I think that deserves a separate question which has two very simple answers. The first is throw more dice. This is easy to do if you have dice marked with an average of 0 like fudge dice (-1 0 1 on a d6) or -2 -1 0 1 2 on a d10; you don't need to adjust the mean to throw more dice which increases the variance. If your dice don't have a mean of 0 you need to in increase the target number by the mean of the dice. If you add 2dN of variance (throw 2 more dice) you need to increase the target number by N+1. \$\endgroup\$
    – Cirdec
    Apr 11, 2015 at 16:35
  • \$\begingroup\$ @Martin The other answer is change the target. Increasing the variance makes it easier to do hard things and harder to do easy things. To increase the variance, change the target towards the player's mean outcome. If Jill's performance has a median of 60 and she attempts something with a difficulty of 70 that has a high variance in her performance move the target towards 60, say 67. If she instead attempts something that should be easy for her (a difficulty of 50) but has a high variance, more the target towards 60, say 53. \$\endgroup\$
    – Cirdec
    Apr 11, 2015 at 16:46

If what you're wanting is more granularity than, say, 3d6 provides, then use whatever distribution curve you like, express the probability of success p as a decimal between 0 and 1, and success is defined as rolling a fraction smaller than p using d10 (or d20 ignoring tens digit) as many rounds as necessary to either reach the stated precision or know that further rounds are not needed to determine the outcome.

Example: You have a .3141592653589... chance of hitting a particular target with your arrow.

If your first roll is 0-2, you've hit the target. If it's 4-9, you've missed. In either case, no further rolls are required. Only if you hit 3 exactly do you roll again.

If your second roll is 0, you hit. If it's 2-9, you missed. If it's 1 exactly, you roll again.

If your third roll is 0-3, you hit. If it's 5-9, you missed. If it's 4 exactly, you roll again.

This repeats until you either stop hitting the exact target number or run out of digits.

It should be immediately apparent that 90% of the time you will roll once, 9% of the time twice, 0.9% three times, 0.09% four... The extra dice rolls are only needed when the previous rolls are exactly at the threshold of success.

It should take less than a minute for people to grasp how this system works.

  • \$\begingroup\$ Very smart solution! One that i will likely use \$\endgroup\$
    – marbri
    Apr 9, 2015 at 21:03

Unfortunately, much of our universe appears to be quantized. One could imagine a system wherein, instead of rolling dice, you precisely measured the net magnetic spin of some ideal gas on non-interacting magnetic moments each of either spin up (1) or spin down (-1). Theoretically such a property, when measured many times, should form a gaussian distribution (a bell curve) with an exceedingly small standard deviation. Unfortunately, this is not quite right. In actuality only certain discrete values of net spin are possible, based on the number of moments in the system. For example, if there is an odd number of moments, it is impossible for the net spin to be exactly 0. Also, obviously, it is impossible for the net spin to ever be any number that is not an integer or to be larger than the total number of particles. All of these concerns have a negligible impact on the behavior of large collections of particles, but they mean that, at a fundamental level, you don't have a continuous distribution.

The example deals with a very idealized situation, but really a great many physical quantities have been shown to be (very probably) quantized by modern physics. The question, then, requires us to find something that isn't.

Oddly, position is currently not believed to be precisely quantized, though this is disputed by some physicists. In the standard model of quantum mechanics, when one puts a single electron in the ground state in an infinite square well, a truly continuous probability distribution (the Schrodinger Equation) describes the likelihood thereafter of finding the electron to be at any given point in space within the box. Thus, with some sophisticated and expensive experimental physics equipment, one could actually maybe be measuring a continuous probability distribution. But here we encounter the second, and more fundamental issue with your plan:

The Arabic numeral system is inherently discrete. Most measuring devices give their outputs in numbers (citation needed). Lets take the number 5.5563, for example. If this is the number our instrument reports, the next possible number is 5.5564. The instrument can't report 5.55631, or any other number between the two possible results. In order to have a continuous distribution we would need an infinite degree of precision and an infinite number of digits. This is a problem.

Clearly the solution is to use a measuring device that doesn't report numbers. Unfortunately, unless it its measurement is based off of gravitational interaction, the act of measurement itself now creates problems for us. All force interactions other than gravity possess a good amount of evidence supporting the existence of a mediating particle. Such mediating particles result in quantization of the interaction, so that only interactions involving an integer number of such mediating particles (quanta) are possible. While people like to speculate about gravitons, there really isn't much evidence yet for their existence and a far bit of evidence to the contrary, so you're actually on pretty solid ground if the measurement device is solely using gravity to measure the position of the electron in the box and reporting that position by means of deflection in its own position (the uncertainty in the latter is of no concern to us; indeed it is helpful to you).

So now we have a continuous probability distribution measured and reported in a continuous manner. So we're good right? Not quite.

Gravity is very weak, and electrons are very small. Unfortunately, large (i.e. massive) objects have an uncertainty in position that is astronomically small. So yes, this works, but you can't see it, which kinda defeats the purpose of a measuring device in the first place. So, for all intents and purposes, it is not currently technologically feasible to generate a random number from a truly continuous probability distribution. In order to do so we'd need something like a macroscopic atomic (i.e. indivisible) object with the mass of an electron, or some such.

And even then quantization would interfere if you were telling where it was by looking at it, feeling it, smelling it, or otherwise measuring it in a physical way via biology.

So, you're pretty much screwed.

But wait! There's hope! The human mind is a powerful thing, and evidence suggests it can create and emulate a continuous probability distribution. You can experience the spiritual and mental directly, rather than indirectly as with the physical, and you can control directly the physical and mental as you cannot the spiritual. So then, if you were to conceptualize a continuous probability distribution and by virtue of the free will granted to you by God you selected a truly random point in the distribution you could have a number of the data type you need to make this work. Do any systems currently existing not only have you do that but also ask you to do math on it? No. Not really.

There are systems that ask you to do this, though. In Amber Diceless, the GM is supposed to (if I understand it correctly) come up what exact thing is retrieved by quick Logrus summoning via a random result with a normal distribution centered about the desired item and the standard deviation directly proportionate to the reciprocal of the time use to search for the thing. This is the only example I am aware of.

  • 1
    \$\begingroup\$ The lack of continuity and randomness in the world really isn't a problem. Its more about varying the probability curve than it is about the continuity. \$\endgroup\$
    – marbri
    Apr 9, 2015 at 19:47
  • 1
    \$\begingroup\$ @thedarkwanderer The obvious objection to dice is that simulating a normal distribution by rolling a large number of dice and adding them up really breaks the flow. \$\endgroup\$ Apr 9, 2015 at 20:09
  • \$\begingroup\$ Where did you find evidence suggesting that the human brain can do anything close to a normal distribution, let alone a continuous normal distribution. For example, plotting numbers chosen "at random" by a group of people whose native cultures all use decimal numerical systems will result in huge spikes on those integers with 3s and 7s in them, especially those with 3 or 7 in the least significant digit. \$\endgroup\$ Apr 13, 2015 at 11:19
  • \$\begingroup\$ @MatthewNajmon 1) you aren't pulling from a random sample of people. You are creating and then pulling from a normal-distribution construct. 2) Presumably you are aren't even thinking in numbers since, as mentioned, that would break continuity. Depending on your beliefs about the human mind this may or may not be possible; there is about equal evidence either way, in my opinion. Evidence for the ability to do this would include the exceptional nature of consciousness as a phenomena and the perception of being able to do this. Evidence against it would include the quantization of most stuff. \$\endgroup\$ Apr 13, 2015 at 14:34
  • \$\begingroup\$ @MatthewNajmon Throughout this answer I consistently take the most favorable position for the querent. I find it odd that you object specifically to the consciousness argument as that is by far the best supported (scientific evidence, which is what I suppose you to be looking for, is equal-ish, but the philosophical evidence is overwhelmingly in the querent's favor; c.f. philosophy zombies, critiques of determinism, objections to materialism if you're interested) as, for example, supposing that there isn't a graviton is far less reasonable. \$\endgroup\$ Apr 13, 2015 at 14:43

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