Simulate d616 on Anydice

anydicestatistics

In Marvel Multiverse Role-Playing Game, the players roll d616. It means that they roll 3d6, but:

One die has a different color, its 1 counts as a 6 and it's called the Marvel die, which would mean 1d{6,2,3,4,5,6};
The other two dice are normal d6s, which means they total 2d6;
I believe a normal roll would be 1d{6,2,3,4,5,6} + 2d6;

output 1d{6,2,3,4,5,6} + 2d6

The complicated part is when you add Edges and Troubles:

If you add an Edge, you reroll the lowest die and keep the highest result for that die between the roll and the reroll;
If you add a Trouble, you reroll the highest die and keep the lowest result for that die between the roll and the reroll;
When you reroll because of an Edge or a Trouble, it doesn't matter which die it is, it can be the Marvel die (d{6,2,3,4,5,6}) or any of the other two d6s;
Your Edges and Troubles are cumulative (e.g., you can roll with 2 Edges or 3 Troubles);
An Edge eliminates a Trouble (i.e., you cannot roll with Edges and Troubles at the same time, 2 Edges and 3 Troubles becomes 1 Trouble);

I've been trying to create a function where I can inform in a parameter if the roll gets Edges (1,2,3,…), Troubles (-1,-2,-3,…), or no modifier (0), and it gives me the probabilities for each result since last Friday, but I simply can't.

Can anyone help me get there?

EDITED FOR MORE CLARITY:

Edges and Troubles have priorities:

For Edges: If you have 2 lowest numbers, you reroll the Marvel Die (the most advantageous to reroll and get a higher number). If you rolled 3, 3, and 6, and one of the 3s is the Marvel Die, you reroll it to try and replace it with a higher number;
For Troubles: If you have 2 highest numbers, you reroll one of the normal dice (the lest advantageous to reroll and get a lower number). If you rolled 3, 3, and 6, and one of the 3s is the Marvel Die, you reroll the normal one to prevent the Marvel Die from getting lower that it already is;

Stacked Edges and Troubles work one at a time:

If you rolled 1, 3, and 6 and had two Edges, your first Edge would reroll the 1. Imagine you rerolled the 1 and got a 2, then you would have the sequence 2, 3, and 6. your second edge would reroll the 2.
If you rolled 1, 3, and 6 and had two Troubles, your first Trouble would reroll the 6. If you rerolled the 6 and got 2, you would then have 1, 2, and 3. Your second Trouble would reroll the 3.

Stacked Edged and Troubles with priorities:

If you rolled 1, 3, and 6 and had two Edges, your first Edge would reroll the 1. Imagine you rerolled the 1 and got a 3, then you would have the sequence 3, 3, and 6. If one of the 3s if the Marvel Die, you reroll it.
If you rolled 1, 3, and 6 and had two Troubles, your first Trouble would reroll the 6. If you rerolled the 6 and got 3, you would then have 1, 3, and 3. If one of the 3s is your Marvel Die, your second Trouble would reroll the 3 from the normal die.

Best Answer

DMARVEL: d{7, 2, 3, 4, 5, 6}

function: normal A:n marvel B:n normal C:n edge E:n {
 if E = 0 | (A = 6 & B = 7 & C = 6) {
  result: A + B + C - (B = 7)
 }
 L: [lowest of A and C]
 H: [highest of A and C]
 if E > 0 {
  if B <= L {
   result: [normal L marvel [highest of B and DMARVEL] normal H edge E-1]
  } else {
   result: [normal [highest of L and d6] marvel B normal H edge E-1]
  }
 } else {
  if B > H {
   result: [normal L marvel [lowest of B and DMARVEL] normal H edge E+1]
  } else {
   result: [normal L marvel B normal [lowest of H and d6] edge E+1]
  }
 }
}

output [normal d6 marvel DMARVEL normal d6 edge 1]

Link.

Fantastic rolls

Note that we temporarily assign the M result a value of 7 in order to distinguish it from a 6. This is needed to distinguish Ultimate Fantastic rolls of 6M6 which bypass any Trouble, and to prioritize rerolling M over a normal 6 when in Trouble. This also allows us to determine Fantastic rolls in the hundreds digit by replacing the base case

result: A + B + C - (B = 7)

with

result: A + B + C + (B = 7) * 99

though note that the reroll strategy strictly prioritizes the total over trying to get a Fantastic roll.

Related Solutions

[RPG] How to model Burning Wheel probabilities in AnyDice

AnyDice handles exploding dice in a peculiar fashion that's poorly-suited for roll-and-count dice pools. I have cobbled together a somewhat inelegant solution, but it appears to work correctly.

Here's the link to the program.

Instead of using dice in the usual manner, I've created several functions that, given parameters of a die, return its output, i.e. the number of successes it has produced.

function: roll ROLL:n threshold T {
  if ROLL >=T {result: 1}
  result: 0
}

function: rollexploding ROLL:n threshold T{
  if ROLL =6 {result: 1+[rollexploding 1d6 threshold T]}
  result: [roll ROLL threshold T]
}

function: rerollfailures ROLL:n threshold T{
  if ROLL < T {result: [roll 1d6 threshold T]}
  result: [roll ROLL threshold T]
}

function: rerollthenexplode ROLL: n threshold T {
  if ROLL < T {result: [rollexploding 1d6 threshold T]}
  result: [rollexploding ROLL threshold T]
}

Then I made a wrapper function that would figure out which of these functions to call, and handle requests for multiple dice being rolled:

function: wrapper DICE:n threshold T explode E reroll R{
  RES:0
  loop N over {1..DICE} {
    if E & R {RES: RES+[rerollthenexplode 1d6 threshold T]}
    else {if E {RES:RES+[rollexploding 1d6 threshold T]}
      else {if R {RES:RES+[rerollfailures 1d6 threshold T]}
        else {RES:RES+[roll 1d6 threshold T]}
      }
    }
  }
  result:RES
}

Finally, I made a bunch of functions that simplify input by providing pre-determined combinations of parameters. I'll provide only 3 of them here:

function: b DICE:n{
  result:[wrapper DICE threshold 4 explode 0 reroll 0]
}

function: be DICE:n{
  result:[wrapper DICE threshold 4 explode 1 reroll 0]
}

function: gr DICE:n{
  result:[wrapper DICE threshold 3 explode 0 reroll 1]
}

To use it, type things like output [b 4]

[RPG] Rerolling and replacing the lowest result in Anydice

Doppelgreener's answer is good if the player must always reroll their lowest die, no matter what they originally rolled. However, if using the ability is optional, the player will most likely choose not to use it if they roll, say, two twelves on their first 2d12 roll.

In general, it's hard to model such optional decision-making processes mathematically, since the rationally optimal decision may depend on what the player's specific goal is (not to mention that players are human, and thus often don't act rationally!). However, in this case, a fairly reasonable class of decision-making rules to consider are those where the player rerolls the lowest die only when it's less than some fixed threshold. In fact, if the player's goal is simply to maximize the expected average result of their roll, their optimal strategy is to reroll a die only when the original value of that die is less than the expected average of the reroll (which, for a d12, is (1+12)/2 = 6.5).

Here's a basic AnyDice script to model that decision-making strategy:

function: reroll lowest of ROLL:s as REROLL:d if less than MIN:n {
  LOWEST: (#ROLL)@ROLL       \ the lowest die is sorted last \
  if LOWEST >= MIN {
    result: ROLL + 0
  } else {
    REST: {1..#ROLL-1}@ROLL  \ all but the lowest die \
    result: REST + REROLL
  }
}

output [reroll lowest of 2d12 as d12 if less than 7] named "2d12 replace lowest if < 7"

Note that the function in the code above is generic enough to allow arbitrary initial dice pool sizes (although only the lowest die is ever rerolled) and thresholds, and even provides the option for rerolling with a different die than the original pool had, should that be desired.

Looking at the output of the script, we can see that this "reroll lowest if less than 7" strategy significantly outperforms both "always reroll" and "never reroll":

Of course, we could also consider thresholds other than 7 (≈ 6.5). However, the summary statistics do reveal that, at least as far as the expected average outcome is concerned, 7 is indeed the optimal threshold for rerolling a d12.

All that said, other decision-making rules can still do even better in specific circumstances. For example, if the player is trying to roll to meet or exceed a particular target number, the natural and likely optimal rule is simply to reroll if the sum of their original roll is less than the target, and let the original roll stand otherwise.

Of course, we can model that in AnyDice as well:

function: roll ROLL:s vs TARGET:n with optional REROLL:d reroll {
  SUM: ROLL + 0  \ force the sequence to be summed into a single number! \
  if SUM >= TARGET {
    \ no need to reroll, since we've already met the target \
    result: 1
  } else {
    \ discard and reroll the lowest die \
    REST: {1..#ROLL-1}@ROLL
    result: REST + REROLL >= TARGET
  }
}

loop TARGET over {2 .. 24} {
  output 2d12 >= TARGET named "2d12 vs [TARGET]"
  output [roll 2d12 vs TARGET with optional d12 reroll] named "2d12 vs [TARGET] with optional reroll"
}

In this case, we unfortunately don't get such nice graphs out, since each comparison against the target number just outputs 0 if the roll fails and 1 if it succeeds. Still, looking at the "transposed" view (which I've linked directly to above), we can see that allowing the reroll is slightly better than granting the player +3 to their roll. For example, an unmodified 2d12 roll has a 61.81% chance of meeting a target of 12, whereas 2d12 with an optional reroll has a 64.53% chance of meeting a target of 15, and a 56.71% chance of meeting a target of 16.

Best Answer

Fantastic rolls

Related Solutions

[RPG] How to model Burning Wheel probabilities in AnyDice

[RPG] Rerolling and replacing the lowest result in Anydice

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