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3

Anydice is unfortunately a bit tricky in that it only allows us to return a single number from a function, so we have to use these workarounds if you want to get more information out of a single roll. But that's entirely doable, in this case, since the introspection you want to do isn't very complicated. Here's a program which implements your mechanic: ...


1

Just answering: How to make Anydice see if two values are the same among 3 dice and keep them? Because the explanation after that makes no sense to me. This outputs the middle number if it matches either the highest or the lowest of 3 dice and 0 if it doesn’t. function: matches DICE:s { result: (1@DICE=2@DICE)*2@DICE + (1@DICE!=2@DICE)*(2@DICE=3@DICE)*2@...


2

Using output 2@3d20 should provide the second result from the top. If two (or all three) results are equal, that also 'spits out' that same result.


11

Here's a fairly efficient solution: TARGET_DIST: [highest of 1@3d6 and [highest of 1@2d8 and 1@1d12]] output TARGET_DIST named "highest of 3d6, 2d8 and 1d12" function: roll versus TARGET:n { P: d6 > TARGET Q: d8 > TARGET R: d12 > TARGET result: 2dP + 1dQ + 1dR } output [roll versus TARGET_DIST] named "2d6, 1d8 and 1d12 vs. ...


3

Do it? Sure. Do it well? Not so much The easiest way to do mixed dice pools in AnyDice is to cast them as seperate sequence inputs into a function, and then stitch the sequences together. Note that you'll have to build the function with a maximum different dice possible, but it is possible to give a 0 or {} if the function has more inputs than your pool. We ...


1

Over a series of rolls Arithmetic coding requires on average the fewest real rolls per emulated roll. The basic idea is that every possible outcome from the real rolls as well as every possible outcome from the emulated rolls are mapped to a range of real numbers between 0 and 1; with the width of each range being equal to that outcome's probability. Note ...


0

disclaimer: I am not a math genius whatsoever and this is not a mathematically correct method. So my system is simple, say you want to replicate a d4, if you roll your d20 from 1 to 4 you don't have to do anything and you apply the roll as normal. However if the roll is greater than 4 you subract 16 from it. And if the difference is negative you roll again ...


2

A d20 roll has more information than a d6 roll (or any other of the smaller dice). Thus it should be possible to average less than one d20 roll for a d6 roll. To be more precise, log(6)/log(20)=0.5981 rolls of the d20 should suffice. How can it be less than one? Imagine emulating a d2 with a d4. Each d4 roll is worth two d2 rolls (take the d4 result divided ...


1

To be able to emulate all of the required numbers, you need to be able to emulate d2, d3, and d5. For instance, to emulate a d10, you roll a d2 and a d5, and if the d2 comes out as 2, you add 5 to the d5 result. As all of the basic dice are multiples of 2, 3 and 5, with no other prime factors, this is what you require. Similarly, to create d6, simply roll d2 ...


13

By putting the d20 in your pocket You don't need a die to emulate a die. Have the GM secretly choose a whole number. Have the player rolling declare some other whole number. Add the two together and then take the modulo of that sum by the size of the die to be emulated. I've done this for years on account of being poor and not always having internet until ...


8

Summary: Rolling and discarding averages fewer than 2 d20 per any other die, even d12. This is good for time and the stated criterion of number of rolls. Divide and round up is easier than actually working through your formula, and maybe easier than doing the modulo-like operation that's equivalent to your formula, except for d10 where you just take the ...


16

To directly implement a d6 or a d12, you can increase your d20 to a d60, by using the fact that Every face of a standard d20 has 3 distinctive vertices. Looking at an example face, it is seen that the number defines the 'up' direction, as well as 'left' and 'right': Therefore, the three vertices of a face can be distinctively described as "upper vertex&...


11

In theory, you can simulate any die if you Think of your d20 as a compass. The idea is to split \$360^\circ\$ into N even parts and let the "compass" select a part. (Similarly to a Twister spinner). To simulate a dN, you roll the die and ignore the value of the number rolled. Instead you look where the "upper vertex"1 points to. Then, ...


33

Emulating a d6 using a fixed number of d20 rolls is impossible. Of course, as the other answer points out, it is possible to do so if we take into account more than just the rolled result, such as its orientation, but let's ignore that for a moment. You rightly say that we can emulate a d6 with a d20 by mapping 1-18 to the d6 and rerolling on 19 or 20. This ...


25

An Icosahedron has 12 vertices. Since an icosahedron has 12 vertices, all we need is a method for identifying the vertex which corresponds to a particular result. Orientation Each face of a d20 is a triangle, so we can easily determine which vertex is the result based on its orientation relative to us. Simply take the vertex that is the furthest away or &...


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