The Dimensions of a Ten-Sided Die: Part II

The dimensions of a pentagonal trapezohedron can be expressed with two free variables. One of them determines the overall size of the polyhedron. The other determines the ratio of the distance between the polar vertices relative to the distance of the other vertices to the center.

Is there a canonical value for the ratio? We could insist all vertices lie in a sphere. This gives the die a round shape and is close to the ratio dice manufacturers use.

Since the vertices are now equidistant to the center we can impose the additional constraint $$n + \frac{h}{2} = \sqrt{\frac{h^2}{4} + m^2}$$ We arbitrarily set m to 1 and solve using Mathematica:

Solve[{k/n == m/(n + h),

   k/m == Cos[2*Pi/10],

   w/z == h/(n + h),

   y/(2*m) == Sin[2*Pi/10],

   z == Sqrt[m^2 + (n + h)^2],

   f == Sqrt[(z - w)^2 + (y/2)^2],

   g == Sqrt[w^2 + (y/2)^2],

   n + h/2 == Sqrt[h^2/4 + m^2],

   m == 1},

   {k, h, w, z, y, f, g, n, m}] // N

The values are $$k \approx 0.809017 \\ h \approx 0.212332 \\ w \approx 0.285586 \\
z \approx 1.49535 \\ y \approx 1.17557 \\ f \approx 1.345 \\ g \approx
   0.653491  \\ n \approx 0.899454 \\ m = 1 \\ \alpha \approx 51.8273^\circ \\ \beta = 90^\circ \\ \gamma \approx 128.173^\circ $$

The Dimensions of a Ten-Sided Die

TSR introduced the ten-sided die to gamers in 1981 but the idea of using a pentagonal trapezohedron as a die was first patented in 1906.

The pentagonal trapezohedron is face-transitive and each face is parallel to a face on the opposite side. Is the pentagonal trapezohedron the only ten-sided polyhedron with these properties? The pentagonal bipyramid is face-transitive but lacks parallel, opposite sides. The right, regular-octagonal prism has parallel, opposite sides but isn't face-transitive.

The pentagonal trapezohedron isn't vertex-transitive. Two of the vertices, which we can think of as the poles, touch 5 faces each. The remaining 10 vertices are arranged in two parallel pentagons and touch 3 faces each.

In the diagrams above a few labels are provided as an aid to figuring out the dimensions of the pentagonal trapezohedron. In the first diagram the dimensions n and h are along the axis connecting the two poles of the pentagonal trapezohedron. The left diagonal is the central spine of one of the kite-shaped faces, labeled z in the bottom two diagrams. If we rotate the trapezohedron around the axis 36°, we can bring the central spine of an adjacent face into the position of the right diagonal.  The non-polar vertices are in the planes perpendicular to the axis indicated by k or m.

If we rotate the polyhedron so that the axis is perpendicular to the plane of the page, then the second diagram shows how the non-polar vertices are arranged.

By similar triangles in the first diagram this holds $$ \frac{k}{n} = \frac{m}{n+h}$$ From the second diagram and trigonometry $$ \frac{k}{m} = \cos{\frac{2 \pi}{10}}$$ The remaining equations in the Mathematica code that follows can be justified by similar triangles, trigonometry, or the Pythagorean theorem:

  Solve[{k/n == m/(n + h), k/m == Cos[2*Pi/10], w/z == h/(n + h),
    y/(2*m) == Sin[2*Pi/10], z == Sqrt[m^2 + (n + h)^2],
    f == Sqrt[(z - w)^2 + (y/2)^2], g == Sqrt[w^2 + (y/2)^2]},
    {k, h, w, z, y, f, g}]

$$ k = \frac{1 + \sqrt{5}}{4}m \\ h = (2 + \sqrt{5})n \\ w = \frac{3\sqrt{m^2 + (6-2\sqrt{5})n^2}}{4} \\ z = \sqrt{m^2 +(6 - 2\sqrt{5})n^2}  \\ y = \sqrt{\frac{1}{2} \left(5-\sqrt{5}\right)} m \\  f = \sqrt{m^2 + n^2} \\ g = \sqrt{\left(9-4 \sqrt{5}\right) n^2-\frac{1}{2} \left(\sqrt{5}-3\right) m^2} \\ \alpha = 2 \sin ^{-1}\left(\frac{\sqrt{\frac{1}{2} \left(5-\sqrt{5}\right)} m}{2
   \sqrt{m^2+n^2}}\right) \\ \gamma = 2 \cos ^{-1}\left(-\frac{\left(\sqrt{5}-3\right) \sqrt{m^2-2 \left(\sqrt{5}-3\right) n^2}}{2
   \sqrt{4 \left(9-4 \sqrt{5}\right) n^2-2 \left(\sqrt{5}-3\right) m^2}}\right) \\ \beta = \frac{2 \pi - \alpha - \gamma}{2}$$We are free to choose m and n to be any positive values.

Maximally Separated Points on a Sphere

One of the designs for a seven-sided die is a sphere truncated at 7 points as far apart from each other as possible. The die doesn't have opposite and parallel sides so there isn't a face for writing the result on. Still the technique is general and could be used to create dice with anywhere between 5 and 100 sides.

How do we find points as far apart as possible from each other? For most values of n the problem is hard. One approach is to sum the pairwise distances of the points and use gradient ascent to find a maximum. The objective function is not convex and has local maxima, but if you think a local maximum is better than nothing, here is code find one. The code uses Euclidean distance, though I wonder if using the angle between two points would work better.  The partial derivatives are with respect to the θ and φ of spherical coordinates since we are constrained to the surface of the sphere.

Running the code:

  $ ./maximally_separated_points_on_sphere.py -n 7

  (0.139, -0.576, -0.805)
  (-0.191, 0.963, 0.191)
  (-0.608, 0.338, -0.718)
  (0.173, 0.142, 0.975)
  (0.884, 0.38, -0.273)
  (-0.855, -0.418, 0.308)
  (0.458, -0.828, 0.322)
  31.530926 

The output contains the coordinates of the 7 points found and the value of the objective function at those points.  A convex polytope can be defined in polymake using a set of points—in which case it finds the convex hull—or a set of half spaces—in which case it finds the intersection. Polymake requires homogeneous coordinates.

  $ polymake
  polytope > $p = new Polytope(POINTS => 
    [[1, 0.139, -0.576, -0.805],
     [1, -0.191, 0.963, 0.191],
     [1, -0.608, 0.338, -0.718],
     [1, 0.173, 0.142, 0.975],
     [1, 0.884, 0.38, -0.273],
     [1, -0.855, -0.418, 0.308],
     [1, 0.458, -0.828, 0.322]]);

  polytope > print $p->VERTICES_IN_FACETS;

  {1 2 5}
  {0 2 5}
  {0 2 4}
  {1 3 5}
  {1 3 4}
  {1 2 4}
  {3 4 6}
  {3 5 6}
  {0 5 6}
  {0 4 6}

Polymake tells us the polyhedron has ten faces. This website says there are 5 topologically different polyhedra with 7 vertices and 10 faces. Since polymake tells us which vertices belong to which faces, we can use Blender to visualize the polyhedron.

Following the instructions above yields this picture:

Our polyhedron is topologically equivalent to a pentagonal bipyramid. The figure is the dual of the 7-sided polyhedron we would like to make a die out of. The dual can be found in polymake using polarize:

  polytope > $dual = polarize($p);

  polytope > print convert_to<Float>(

    dehomogenize($dual->VERTICES));
  1.455537382 -0.7147719412 -0.1762669144
  1.083809015 0.804520317 0.8537214292
  -0.4415750134 -0.506366941 1.52830861
  1.059871785 -0.6049596605 -1.125593382
  -0.8893855532 -1.073711966 -0.7114555899
  -0.2646212759 -1.291044842 1.009076016
  -1.423698754 0.1131924589 -0.7895112253
  0.2939932965 0.9002156047 -1.208914314
  0.4841557507 1.557693116 0.211262627
  -1.274046573 0.7045065006 0.518151282

  polytope > print $dual->VERTICES_IN_FACETS;
  {1 2 8 9}
  {0 3 4 5}
  {0 1 2 5}
  {3 4 6 7}
  {2 4 5 6 9}
  {0 1 3 7 8}
  {6 7 8 9}

Polymake lists the vertices in a face in ascending order. Blender requires that we list the vertices in the order they occur as we travel around the polygon—otherwise the wireframe will have extra edges in the faces. Trial and error was used to find the correct order:

The polyhedron is topologically equivalent to a pentagonal prism.

Difficult Rolls

Here are the dice I keep behind the screen:

Making Do With a d20 and a d6

In a pinch I could get by with the two d20s and three d6s with pips. The d20s can be used for a percentile roll with the convention that pink is tens and white is ones.

The d20 can simulate a d4 by taking the result modulo 4, treating 0 as 4.

The d20 can simulate a d8 or a d12 with the help of a d6. Take the result of the d20 modulo 4, treating 0 as 4. Roll a d6 and add 4 if the result is 4, 5, or 6. Or add 4 if the result is 3, 4 and add 8 if the result is 5, 6.

This DM is a minimalist and only uses a single d6 and a d20 when running a game. Perhaps he uses the techniques above. Maybe he runs an old version of the game where everyone has d6 hit dice and does d6 damage. The DM could ask a player to make a roll.

Discrete Uniform Distributions

How many discrete uniform distribution are needed?

If we are using the above dice, it is sensible to stick to dice rolls in the ranges 1–4, 1–6, 1–8, 1–10, 1–12, 1–20, and 1–100. But if we could generate an integer randomly from any range, then any list could be used as a random generation table.

The Holmes rulebook lists 14 1st level magic-user spells, 18 2nd level magic-user spells, and 18 3rd level magic-user spells. What if we need to determine a spell randomly, say because a scroll is found or an NPC spell caster is encountered? Small wonder the spell lists were truncated to 12 per level in the Moldvay rulebook.

Integer Division and Round Up or Modular Arithmetic

Two techniques for getting a smaller roll from a larger roll

To get a d2 or a d3 we can use the d6 and apply integer division and rounding up. When rolling a d6 with pips I like to use ⌈d6/3⌉ or ⌈d6/2⌉.

For a d5, I use d20 % 5, replacing 0 with 5.

As discussed previously one could use modular arithmetic to get the d2 and d3 rolls. Why the inconsistency? I prefer modular arithmetic nowadays, but the group has a tradition going back to childhood for how a d2 or d3 is rolled.

d6 Helper

Getting a larger roll from two smaller rolls

Back when icosahedral dice were numbered 0 to 9 twice, grognards would roll them with a helper d6 to determine whether the result was in the range 1–10 or 11–20. Specifically the formula d10 × (d2 – 1) × 10 was used. The technique generalizes:

    d15: d5 + (d3 – 1) × 5

    d16: d8 + (d2 – 1) × 8
    d18: d6 × (d3 – 1) × 6
    d24: d12 × (d2 – 1) × 12
    d30: d10 × (d3 – 1) × 10
    d36: d12 × (d3 – 1) × 12
    d40: d20 × (d2 – 1) × 20
    d60: d20 × (d3 – 1) × 20

I use a d6 with pips as the helper. When rolling a d18 I combine it with the d6 with digits.

If High Then Low

Interpolating a roll in between a smaller and a larger roll

A truly uniform result for most of the other ranges is not possible, at least not with a single roll. We use the "if high, then low" technique to get a near uniform approximation. To get a d7, one rolls a d8 and a d6. If the d8 is too high the result on the d6 is used. In general we find the next higher and next lower ranges that we can roll exactly and use those.

To get a d17 a d16 and a d20 are rolled. I don't use a d16 and a d18 because both need a helper d6; there would be ambiguity if the dice are rolled at the same time.

Falling Distributions

Motiving the falling distribution

Gaming has gotten by with uniform (e.g. d20) and normal (e.g. 3d6) distributions. Both of these are symmetric distributions with expected values at the center of the distribution.

When world-building, an asymmetric distribution in which small values are more likely than large values is desirable. We should use a distribution whose probabilities fall in this way to determine the level of an NPC if we want low-level characters to be common and high-level characters to be rare. It could also be used to determine the number of humans in a settlement or the number of orcs in a war party, since populations tend to follow a Pareto distribution.

I know about exploding rolls and penetrating rolls, but they are close to uniform in practice.

One gets a geometric distribution by rolling repeatedly until a certain value is obtained and counting the number of rolls that is needed. Although strictly decreasing, it is a slow method which requires a lot of re-rolling.

Update: I've found a better way.

Absolute Value of the Difference of Two Rolls

A linear falling distribution with a single roll of two dice—almost

This interview mentions a promising technique, which is to roll the same dice twice and apply the absolute value function to the difference:

Because of the zero value, this isn't a perfect falling distribution, but if one is willing to re-roll until the result is not zero a falling distribution is achieved.

Tables

Falling distributions and in general any distribution using percentiles

Here is code for generating a d100 table which approximates an arbitrary distribution. It was used to make this table:

d100	level
01-33	1
34-55	2
56-70	3
71-80	4
81-86	5
87-91	6
92-94	7
95-96	8
97	9
98	10
99	11
00	12

An exponential distribution was used with the parameter chosen so 12 is the highest possible value. The distribution looks like this:

Testing Dice for Fairness: Part II

The power of the fair die fair test.

Previously a method was outlined for testing a die for fairness.

The test calls for rolling the die a large number of times, computing a p-value, and classifying the die as unfair if the p-value is below a significance threshold.

The example uses 100 trials and a significance threshold of .05. It doesn't mention why these numbers are chosen.

The Two Types of Error

To intelligently choose the numbers, we have to consider two types of mistakes. We might reject a fair die, or we might accept an unfair die.

The chance of making the first error—rejecting a fair die—is our significance threshold. To reduce the chance of making the first error we make the significance threshold low. A threshold of .05 is a bit high; our test has a 1 in 20 chance of rejecting a fair die!

The Power of a Test

The power of a test is the chance of not making the second error.  That is, the power is the chance of rejecting an unfair die.

The power of a test usually depends on how unfair the die is. The more lopsided the chances, the easier it is to detect the unfairness.

An Approximation

Because of the dependence on the amount of unfairness, we can't represent the power as a single percentage. It is natural to characterize an unfair die as a categorical distribution with n - 1 free parameters where n is the number of faces. Even in the case of a d6 we can't plot the power as a function of the free parameters: there are too many.

For this reason let's only consider unfair dice characterized by a single parameter which is the probability of rolling the most likely face. All other faces are assumed to be equally likely.

The Power of the .05 Significance Test

Now we can plot the power of a test, given the probability of the mostly likely face.

If we increase the number of trials, the chance of detecting an unfair die goes up. Even if we only do 100 trials, the chance of detecting an unfair die where one of the faces is twice as likely as expected is about 90%.

I would like the percentage to be higher, but when I think the about the effort of making 200 trials instead, the percentage seems acceptable.

The code used for calculating the power of a test is online. It uses the Monte Carlo method as discussed previously. The number of Monte Carlo trials was 10,000 which gives a 95% confidence our estimate is within 1%.

The Power of the .01 Significance Test

What if we decide to use a .01 significance level instead?

The chart doesn't change much. The chance of detecting the die where one of the faces is twice as likely as expected is now 77%.

Cloud Giant Damage

A general method for converting range notation to dice notation.

My first copy of the game was a Holmes edition with geomorphs.

The rules have staying power. This summer I met a Holmes player born after they went out of print!

And so the questions we asked thirty-five years ago are still getting asked. A classic is how to roll the 6–63 damage that cloud giants deal out:

The experienced referee makes this roll with a 3d20+3, but—and this is another example of the ambiguity of range notation—a 19d4-13 will also work.

One wonders whether there are other ways to make the roll. Also, is there an easy way to find those ways?

If we use a single type of die, then we must use a d20 or a d4 as above. If we allow more than one type of die in the roll, then 6d10+d4-1 works.

In fact, if we use more than one type of die, there are 164 ways to roll cloud giant damage.

To find them all, observe a fact about the difference between the largest and smallest value of each die; a d6 produces results in the range 1–6 and has a difference of 5. When multiple dice are summed, the difference of the sum is the sum of the differences of each die used; 2d6 has a range of 2–12 and a difference of 10; 3d6 has a range of 3–18 and a difference of 15.

The rule holds if we use more than one type of die: a d4 has a range of 1–4 and a difference of 3. A d4+d6 roll has a range of 2–10 and a difference of 3 + 5 = 8.

To find out how many ways there are to roll 6–63, we ask ourselves how many ways there are to sum to the difference of 57 given the differences we have available: 3 (d4), 5 (d6), 7 (d8), 9 (d10), 11 (d12), and 19 (d20). This is the unbounded knapsack problem where we are only interested in solutions which fill the "knapsack" completely.

The complete list of ways to roll cloud giant damage is here.

Code for finding ways to roll an arbitrary range is here.

Binomial Parameter Estimates

More on estimating class qualification probabilities.

Choosing a Character Attribute Generation Method

On p. 11 of the DMG four alternatives for generating character attributes are given. How should a referee decide which one to use?

The referee might have a sense for how easy it should be to roll up a prestige class. Does he want the players all playing paladins and rangers instead of fighters? If so, he should consider the chances listed in the previous post. The 4d6kh3 with re-arrangement method gives the best chances overall.

Did the Designers Know the Odds?

If you asked the designers forty years ago to guess the chance of rolling a paladin, how close would they have been? The calculation requires writing code and running it on a modern computer, so the exact chance was probably unknown in the 1970s. Still, the designers could have made a reasonable guess just by rolling lots of characters.

Monte Carlo Method

To estimate the chance of rolling a class, the designers could have rolled 100 characters and counted how many of them qualified. If 35 of them qualified, the estimated chance is 35%. This is the Monte Carlo method applied to the binomial parameter estimation problem.

The Monte Carlo method is slick! It is easy to understand. Complicated mathematics is avoided.

Monte Carlo can be used to verify complicated mathematics. After writing code to do the exact calculation, I wrote code which uses Monte Carlo to make sure the exact code is in the right ballpark.

Often it isn't enough to get an estimate; we also have specify how accurate it is. We can increase our accuracy by doing more rolls. For example, we will get a better estimate of the chance of a paladin if we roll 1000 characters instead of 100 characters. It is natural to ask how many characters must we roll to get an answer which is accurate enough for our purpose.

Normal Approximation Interval

One way to represent the accuracy of an estimate is with a confidence interval. For binomial parameter estimation, one way to get a confidence interval is with the normal approximation:
$$p \pm z_{1 - \frac{α}{2}} \sqrt{\frac{p \cdot (1 - p)}{n}}$$Here n is the number of characters generated, p is the observed fraction, and α is the chance of being outside the confidence interval. When α is 5%, z_1-α/2 is 1.96.  In R one can compute z_1-α/2 with

  qnorm(1-alpha/2)

The normal approximation suggests that if we increase the number of simulations 10-fold, the length of the confidence interval will shrink to about 0.32 of its previous size.

The normal approximation doesn't work well when the observed fraction p is close to zero or 1.  For example, when n is 100 and p is .01, the 95% confidence interval is [-.95%, 2.95%] even though we know a negative value is not possible. This is a consequence of using the normal distribution as an approximation for a distribution with non-negative support.

In the extreme cases where p is exactly 0 or 1, the normal approximation gives an obviously invalid interval with zero length. The usual advice is to never use the normal approximation when there are fewer than 5 successes or 5 failures. That is, there should be at least 5 characters which qualify as a paladin and 5 characters which don't.

Wilson Score Interval

At the cost of some extra complexity, there is a formula which works better when p is near zero or one:
$$\frac{1}{1 + \frac{1}{n}z_{1-α/2}^2} \bigg[p + \frac{1}{2n} z_{1-α/2}^2 \pm z \sqrt{\frac{1}{n} p (1 - p) + \frac{1}{4n^2} z_{1-α/2}^2}  \bigg]$$

Beta Distribution

We can look for a distribution which describes the possibilities for the parameter we are estimating. This is more informative than a confidence interval. Bayesian statistics gives us a way to do this, but we must start with an initial distribution, the prior, which is somewhat arbitrarily chosen. We use the data we observe and the prior to compute the posterior distribution, which is the distribution we seek.

When estimating a binomial parameter, the beta distribution is convenient because it makes computation of the posterior easy. The beta distribution has two parameters. If the prior has parameters α and β and we observe m successes in n trials, then the posterior is a beta distribution with parameters α + m and β + n - m. We can think of the parameters as tallies of the number of successes and failures we observe.

How should α and β be chosen for the prior? If we set them both to 1, then the prior is identical to the uniform distribution. That is, we are saying that before we look at the data we think each possible value for the parameter is equally likely. For those who know information theory, the beta distribution with parameters (1, 1) is the beta distribution with highest entropy. Such a prior, incidentally, was how Laplace solved the sunrise problem.

Suppose we decide to start with a beta (1, 1) prior. We roll 100 characters and 35 of them qualify. Then the posterior parameters are (36, 66). We can use R to compute the chance the actual probability is within any interval. For example that chance that the true probability is inside [.32, .38] is 47.4%:

  > pbeta(.38, 36, 66) - pbeta(.32, 36, 66)
  [1] 0.4739478

If we think we already know the mean and variance for the distribution, we can use them to set the parameters of the prior. Here is how to use Mathematica and the formulas for the mean and variance to solve for the parameters:

  Solve[{μ == α/(α+β),
         σ^2 == α β/((α+β)^2 * (α+β+1))},
        {α,β}]

Here are the formulas for the parameters, given mean μ and variance σ²:
$$α = \frac{μ^2 - μ^3 - μ σ^2}{σ^2}$$ $$β  = \frac{(-1 + μ) (-μ + μ^2 + σ^2)}{σ^2}$$

Hoeffding's Inequality

Hoeffding's Inequality is invaluable when doing Monte Carlo estimates. It directly tells us how much data we need to ensure our estimate is near the true value with a given probability:

$$ \mathrm{P}\big(\big|p - \mathrm{E}[p]\big| \geq t\big) \leq 2 e^{-2nt^2}$$

Here p is the estimate, E[p] is the true value, t is half the length of our confidence interval, and n is the number of observations. If we know the probability and interval we want, we recast the inequality as:
$$ \frac{\mathrm{ln} \frac{2}{\mathrm{P}(|p - \mathrm{E}[p]| \geq t)}}{2 t^2} \leq n$$ If we want to be 95% certain the estimate is within a tenth of a percent, we get this lower bound on n:
$$ \frac{\mathrm{ln} \frac{2}{.05}}{2(0.001)^2} \approx 9.22 \times 10^5 \leq n$$ The number might seem discouragingly large, but generating a million characters is feasible on modern computers; just write code to do it!

Chance of Rolling a Class

e.g. how hard is it to roll up a paladin?

It's an obvious question. What is the chance of rolling a character that qualifies for a given class?

Minimum Attributes

The chance depends on the minimum attributes for the class. Let's use the first edition values:

minimum attributes per Players Handbook (1978)

            Str Int Wis Dex Con Char
cleric:       3   3   9   3   3   3
druid:        3   3  12   3   3  15
fighter:      9   3   3   3   7   3
paladin:     12   9  13   3   9  17
ranger:      13  13  14   3  14   3
magic-user:   3   9   3   6   3   3
illusionist:  3  15   3  16   3   3
thief:        3   3   3   9   3   3
assassin:    12  11   3  12   3   3
monk:        15   3  15  15  11   3
bard:        15  12  15  15  10  15 

Odds by Class: 3d6

The original method of rolling up attributes is to use 3d6 for each attribute. The attributes are generated in the order of strength, intelligence, wisdom, dexterity, constitution, and charisma. No rearrangement is allowed.

The odds of a character qualifying for a given class by this method are:

  assassin:      7.031%
  bard:          0.002%
  cleric:       74.074%
  druid:         3.472%
  fighter:      67.215%
  illusionist:   0.429%
  magic-user:   70.645%
  monk:          0.040%
  paladin:       0.099%
  ranger:        0.176%
  thief:        74.074%

We knew that rolling up a paladin was a long shot, but rolling up a monk is harder still. Your chance of rolling up a bard are 1 in 58,140.

Odds by Class: DMG Methods

Some prestige classes might seem pointless, given how unlikely it is to roll a character that qualifies to be one. However, the DMG allows, at referee discretion, one of four methods for rolling attributes, each producing higher attributes on average. For example, in method I the player rolls the six attributes with 4d6kh3 and then rearranges the attributes as desired.

The chances of rolling up a character according to these four methods are:

  (METHOD I) 4d6kh3 PLAYER-ARRANGED:
  assassin:      93.616%
  bard:           1.581%
  cleric:       100.000%
  druid:         78.245%
  fighter:      100.000%
  illusionist:   35.782%
  magic-user:   100.000%
  monk:          13.641%
  paladin:       25.169%
  ranger:        30.470%
  thief:        100.000%

  (METHOD II)  3d6 BEST-OF-12 PLAYER-ARRANGED:
  assassin:      96.180%
  bard:           1.887%
  cleric:       100.000%
  druid:         68.206%
  fighter:      100.000%
  illusionist:   24.302%
  magic-user:   100.000%
  monk:           9.231%
  paladin:       18.704%
  ranger:        33.984%
  thief:        100.000%

  (METHOD III) 3d6 BEST-OF-6 EACH ATTRIBUTE:
  assassin:      87.053%
  bard:           3.572%
  cleric:        99.970%
  druid:         41.544%
  fighter:       99.970%
  illusionist:   10.936%
  magic-user:    99.970%
  monk:           8.487%
  paladin:        8.324%
  ranger:        29.788%
  thief:         99.970%

  (METHOD IV) BEST-OF-12-CHARACTERS 3d6:
  assassin:      58.309%
  bard:           0.021%
  cleric:       100.000%
  druid:         34.562%
  fighter:      100.000%
  illusionist:    5.024%
  magic-user:   100.000%
  monk:           0.475%
  paladin:        1.179%
  ranger:         2.097%
  thief:        100.000%

Overall, method I is the best, though method II is best for an assassin or ranger and method III is best for a bard.

Code

The code for calculating the probabilities is on GitHub.

The Arch-Mage Casts a Fireball

Fireball damage: its calculation and its distribution.

The spell components are ready:

This is what the game is about. I cast the spell:

Look at all the damage! But how much damage, exactly? My enthusiasm is tempered by the prospect of adding up the pips.

Fortunately, as a wizard I am familiar with the arcane rule of tens. Prestidigitation ensues and the hidden number is revealed:

Apropos of nothing, I timed myself on a few more rolls:

• 65 hit points: 22.2s
• 63 hit points: 19.6s
• 56 hit points: 28.3s
• 63 hit points: 28.3s
• 62 hit points: 24.9s

An average of 24.7s spent to get a number that was usually within a couple hit points of 63.

A Shortcut?

When younger I was tempted to roll a single d6, multiply it by 18, and be done with it. That is, use a d6 × 18 roll in place of the 18d6. However, even then I couldn't escape the feeling that this was not an equivalent roll. I could see that the latter method always produces a multiple of 18, but I wondered if it wasn't close enough all the same.

Both methods cause 63 hit points of damage on average, so in that respect they are the same. But if the mean is all we care about, why roll dice at all?

Plotting the probability mass distributions is a good way to show how different rolls are. Here the 18d6 distribution is in red and the d6 × 18 distribution is in blue:

The standard deviations are 7.2 and 30.7, a large difference.

Chebyshev's Inequality

At this point, I'm going digress in an attempt to make the standard deviation seem useful.

According to Chebyshev's inequality, the chance of being more than k standard deviations from the mean can never be greater than 1/k². Thus, in the case of the 18d6 distribution, the chance the value is less than 49 or more than 77 is no more than 25%. The same probability for the d6 × 18 distribution is 66⅔%.

Chebyshev's inequality is a rough upper bound that works for any distribution for which we know the mean and standard deviation. If we know the distribution, we can do a summation or integration to get the exact probability; for the 18d6 distribution the exact probability is less than 5%.

A Better Shortcut?

Back to rolling fireball damage.

What if we allow two rolls of a d6? I wrote some code which searches for an expression with the same mean and the nearest standard deviation. To keep things simple the first d6 always gets multiplied by a non-negative integer and the second d6 is taken as is. With those constraints here are the best approximations:

   1d6: d6 × 0 + d6 +  0
   2d6: d6 × 1 + d6 +  0
   3d6: d6 × 0 + d6 +  7
   4d6: d6 × 1 + d6 +  7
   5d6: d6 × 2 + d6 +  7
   6d6: d6 × 1 + d6 + 14
   7d6: d6 × 2 + d6 + 14
   8d6: d6 × 3 + d6 + 14
   9d6: d6 × 2 + d6 + 21
  10d6: d6 × 3 + d6 + 21
  11d6: d6 × 2 + d6 + 28
  12d6: d6 × 3 + d6 + 28
  13d6: d6 × 4 + d6 + 28
  14d6: d6 × 3 + d6 + 35
  15d6: d6 × 4 + d6 + 35
  16d6: d6 × 3 + d6 + 42
  17d6: d6 × 4 + d6 + 42
  18d6: d6 × 3 + d6 + 49
  19d6: d6 × 4 + d6 + 49
  20d6: d6 × 5 + d6 + 49
  21d6: d6 × 4 + d6 + 56
  22d6: d6 × 5 + d6 + 56
  23d6: d6 × 4 + d6 + 63
  24d6: d6 × 5 + d6 + 63

Here is how well the approximation works for the arch-mage; the 18d6 distribution is in red and the d6×3 + d6 + 49 in blue:

Dice Notation

What our group sees as the essentials of dice notation.

Here are examples of dice notation: d4, d6, d8, d10, d12, and d20.

The notation refers to a die, but more often a roll of a die. The rules might say that a weapon causes d8 of damage, which is equivalent to saying it does 1–8 hit points of damage.

The notation has been extended in a couple of ways: 3d6 means to roll three 6-sided dice and sum them, generating a number in the range 3–18. 1d4+1 means to roll one 4-sided die and add one to it, generating a number in the range 2–5.

The Introduction of Dice Notation

The notation isn't used in the original box set or the Monster Manual.

The Players Handbook (June 1978) was the first TSR publication to use it. Jon Peterson suggests the PHB was written as if players were already familiar with the notation, but the occurrences I've found are used in a parenthetical manner. For example, consider this spell description:

A fireball is an explosive burst of flame, which detonates with a low roar, and delivers damage proportionate to the level of the magic-user who cast it, i.e. 1 six-sided die (d6) for each level of experience of the spell caster. Exception: Magic fireball wands deliver 6 die fireballs (6d6), magic staves with this capability deliver 8 die fireballs, and scroll spells of this type deliver a fireball of from 5 to 10 dice (d6 + 4) of damage.

The November 1978 printing of the Holmes rulebook appends the following text:

In some places the reader will note an abbreviated notation for the type of die has been used. The first number is the number of dice used, the letter "d" appears, and the last number is the type of dice used. Thus, "2d4" would mean that two 4-sided dice would be thrown (or one 4-sided would be thrown twice); "3d12" would indicate that three 12-sided dice are used, and so on.

The blurb suggests that dice notation is used elsewhere in the Holmes rulebook, but it isn't!

The Dungeon Masters Guide (August 1979) also explains the notation:

Before any further discussion takes place, let us define the accepted abbreviations for the various dice. A die is symbolized by "d", and its number of sides is shown immediately thereafter. A six-sided die is therefore "d6", d8 is an eight-sided die, and so on. Two four-sided dice are expressed by 2d4, five eight-side dice are 5d8, etc. Any additions to or subtractions from the die or dice are expressed after the identification, thus: d8 + 8 means a linear number grouping between 9 and 16, while 3d6 - 2 means a bell-shaped progression from 1 to 16, with the greatest probability group in the middle (8, 9). This latter progression has the same median numbers as 2d6, but it has higher and lower ends and a greater probability of a median number than if 2d12 were used. When percentage dice are to be used, this is indicated by d%.

As Jon Peterson has discussed, essentially the same notation, albeit with a capital D, was being used in the fanzines for several years before TSR embraced it. It appears, amazingly, in the first issue of Alarums & Excursions from 1975.

The Old Notation Isn't Good Enough

Looking through the older texts, you can see a couple of different ways for specifying dice rolls: "5 + 1", "3-8 sided", and "2–24".

This notation is inferior in various ways. The first doesn't make clear which die to use: a 5d6+1 roll is intended. Of course, ambiguity can sometimes be advantageous. The way hit dice are specified in the Monster Manual might have made the book more appealing to players still using d6 hit dice for monsters.

The third example, range notation, looks like a concise way to specify rolls, but it also can be ambiguous. For example, 3–12 can be either d10+2 or 3d4. The first method is uniform, whereas the second starts to approximate a bell curve.  If you roll it the first way, the chance of getting a 3 is 10%; if you roll it the second way the chance of getting a 3 is 1 in 64 or about 1.6%.

3–12 is the smallest range which can be ambiguous, and it is used in the PHB! A bardiche inflicts 3–12 hit points of damage on large opponents.

The New Notation Isn't Good Enough

On p. 10 the DMG explains dice notation and on the following page it describes a method for rolling attribute scores which can't be expressed with that dice notation: rolling 4d6 and dropping the lowest die!

The site roll20.net has some notation for this. The roll can be written as 4d6d1 to indicate the lowest die is dropped, or 4d6k3 to indicate the highest three dice are kept.

There is alternate notation which make it explicit that the lowest die is dropped: 4d6dl1 and the highest three dice are kept: 4d6kh3.  One could call for the highest die to be dropped: 4d6dh1 or the lowest three dice are kept: 4d6kl3.

The lowercase L seems like an opportunity for confusion with a numeral one, so we just use 4d6kh3 for a 3d6 with negative skew and 4d6dh1 for a 3d6 with positive skew.

Code

We don't like laptops, ipads, or even phones at the table. Nevertheless it was convenient to implement a command line tool which understands dice notation—including the "keep high" and "drop high" extensions:

  $ roll 6d6
  16
  
  $ for i in $(seq 1 6); do roll 4d6kh3; done 
  14
  8
  7
  14
  15
  10

The code is on GitHub.

Factor Rolls

One can use the dice to generate other ranges of integers:

1–2  ⌈d6/3⌉
1–3  ⌈d6/2⌉
1–5  ⌈d20/4⌉
1–10 ⌈d20/2⌉

In case the notation on the right is not clear, one rolls the indicated die, divides by the following number, and then rounds up. The most practical notation is d2, d3, d5, d10. I'm not aware of a standard term for this type of roll; we've been calling them factor rolls.

Positional Rolls

Percentile dice are an example of what we've been calling a positional roll. We could use two d6 to create a d36, for example. This is not multiplication, but more like working with base 6 numbers. The percentile dice make the process easier by using zeros and distinguishing the tens die from the ones die. One formula for getting a range 1–36 is

    6×(d6 - 1) + d6

Dice of different colors are needed for a positional roll. Our convention is to use a white die for the ones die. If dice of different colors are not available, a single roll is not possible; roll the most significant die first.

The dice do not have to have the same number of faces. If you wanted to use the 30 sided dice gaming tables published by the armory, you could generate d30 rolls with a d6 and a d10.

If factor rolls and positional rolls are allowed, then the only numbers we cannot generate are ones containing a prime larger than 5 as a factor.

Re-rolling

Thus there is no way to generate d7 uniformly using a single roll. One could roll a d8 and re-roll if the result is 8.  Simply writing d7 is the best notation.

Seven sided dice have been manufactured. One design is a pentagonal prism, and another "is based on spacing points as equally as possible on a sphere and then cutting planar slices perpendicular to those directions."  It would be interesting to test these dice and see whether the distribution is uniform.

Testing Dice for Fairness

Are Chessex dice fair; are Koplow dice fair; how to test dice.

Testing dice for fairness is tedious but simple: roll the dice a large number of times and tally how often each face comes up.

Here is an example where a set of Chessex dice are rolled 100 times each:

We don't expect the numbers to be exactly the same, even if the die is fair. Any outcome is possible from a fair die, though some outcomes, such as seeing a 6 a hundred times and the other faces not at all, are vanishingly unlikely. How do we recognize implausible results from a fair die?

My time in the statistics department at Ohio State acquainted me with a test statistic which can be used to answer the question.

Let n be the number of sides the die has. Let O_i be the number of times we observe the i-face to come up. Let E_i as the number of times we expect the i-face to come up, assuming the die is fair. The test statistic is:

$$ χ^2 = \sum_{i=1}^n \frac{(O_i - E_i)^2}{E_i} $$
The test statistic has a Chi-squared distribution with n - 1 degrees of freedom. It is used to assign a p-value to the result, which is the chance that a fair die would produce results as or more extreme than what we observed. If you are interested, here is some code for making the calculation. The closer the p-value is to zero, the stronger the evidence that the die is not fair.

I took a set of Koplow dice and a set of Chessex dice and rolled each die 100 times. The only die which had a p-value less than .05 was the Koplow d6. However, if we compute p-values for 5 fair dice, there is a 22.6% chance that one of the dice will have a p-value less than .05. To account for this, I applied the Bonferroni correction, which raised the p-value of the Koplow d6 to 0.11.

So as far as I can tell, my Koplow dice and my Chessex dice are fair. It doesn't mean yours are too, but you can test them.