[RPG] How to get the highest or lowest values of an irregular dice pool in AnyDice

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I want to calculate irregular dice pools, such as (2d10, 1d8) in AnyDice, but I'm having difficulties.

What I mean to do is out of a roll of 2d10 and a 1d8, take the highest (or lowest) two values of the three dice and sum them (total 2-20 with highest & total 2-18 with lowest). I.e. d10 => 7, d10 => 4, d8 => 5 results in either 4 & 5 for 9 total (keep lowest) or 7 & 5 for 12 total (keep highest).

I can do this in Troll, by putting sum largest 2 {2d10,1d8}. I would like to translate this to AnyDice, since I prefer it due to the overlapping graph when comparing probabilities, but am having difficulties since the odds don't match when doing output [highest of 2d10 and 1d8].

Best Answer

Take into account that when you pass 2d10 to Anydice, it automatically sums it. Trying to simulate what you wanted the easiest way I found, here's the code:

function: highestsum A:n B:n C:n {
 result: {1,2} @ [sort {A, B, C}]
}

function: lowestsum A:n B:n C:n {
 result: {2,3} @ [sort {A, B, C}]
}

output [highestsum 1d10 1d10 1d8]
output [lowestsum 1d10 1d10 1d8]

You can test it here.

What both functions do, is sum the three dices and then take out the highest or the lowest. Take into account that it's important to use three dices in this code, splitting the two 2d10 in 1d10 and 1d10.

Edit: Improved thanks to Ilmari Karonen

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[RPG] Help Calculating Dice Odds for a Unique d6 Dice Pool

An exhaustive generation of the possibilities can be done.

It is not 100% clear what you mean by "dice totalling at least 5", I've chosen to interpret this as "sum up to 5 dice" and that means one would have at least one success with a pool of 5 or more dice (and at least 2 successes with a pool of 10 dice).

In principle, the algorithm is simple:

# interpret "roll a die" as "loop from 1 to 6 and for each..."
declare a tabulation function
declare a "generate all rolls" function, taking dice to roll and rolls so far
  if we should roll only one more dice:
    roll it, add it to the rolls so far, pass that to the tabulator
  otherwise:
    roll a die, add it to the rolls so far and call ourselves, generating one less roll

Then you can tabulate all the number of different successes. So far, I have the results for pools of size 2-9 in and can edit the entry with the results.

[RPG] How to calculate the sum of 2 random dice out of a 3d6 pool in AnyDice

~~The accepted solution disregards the core purpose of the actual question (how does user choice affect the outcome), so I had to try and figure it out...~~

I'm new to anydice, so I welcome corrections/suggestions. Here, if the player rolls their desired outcome on 2/3 dice, that becomes their choice, otherwise, output a random choice from the 3....

function: choice of A:n and B:n and C:n target T:n {
  if A=T { result: A }
  if B=T { result: B }
  if C=T { result: C }
  result: 1d{A, B, C}
}
function: combo of A:n and B:n and C:n target T {
  result: [choice of A+B and B+C and A+C target T]
}

loop T over {1..12} {
  output [combo of 1d6 and 1d6 and 1d6 target T] named "2/3d6 [T]"
}

Here, we can see the huge difference that player choice makes

Chance of getting a 7:

2d6: ~17%

(2/3)d6: ~42%

This matches the mathematical answer:

The probability that a pair of the three dice has sum '7' is $$\frac{3 \cdot 6 \cdot 4 + 3 \cdot 6}{6^3} = \frac{90}{216} = \frac{5}{12} ≈ 42/100 $$ Src: https://math.stackexchange.com/questions/3283971/probability-that-2-dice-selected-from-3-rolled-dice-will-have-sum-7

Best Answer

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[RPG] How to calculate the sum of 2 random dice out of a 3d6 pool in AnyDice

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