Help Calculating Dice Odds for a Unique d6 Dice Pool

Question

I'm attempting to design a unique d6 dice pool system, and I want to try programming the probabilities in AnyDice but I'm not sure how. Here's how the system works:

• Dice pools are between 2 and 10 dice
• A roll of 5 or 6 is a success
• Any dice that sum to 5 or higher also count as successes

For example, a roll of (6,3,2,3,5,4,3,1) would be 5 successes (1 from 6, 1 from 5, 1 from 3+2, 1 from 4+1, and 1 from 3+3).

What kind of program in AnyDice or Excel would I use to calculate average odds for each dice pool up to 10 dice?

Answer 1

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.

Answer 2

This is actually pretty simple to calculate. First, there is a 1/3 chance of each dice being a 5 or 6. So you will get (dice pool size) / 3 successes from that to start.

After that, you need to calculate how many successes you get from adding other dice together. Since every result that isn't a 5 or 6 is going to be a 1, 2, 3, or 4, that means that your non-success rolls will average out to 2.5. This means that, on average, you will get one success for every two non-successful dice. Since 2/3 dice are not successful, this means that we get, on average, (dice pool size) / 3 successes from this as well. (2/3 divided by 2 is 1/3)

So, in the end, you will get an average of 2 successes for every 3 dice rolled.

Answer 3

While @AgentPaper has provided a very nice formula, reality is slightly different.

I used a small python script to calculate the average number of results over 100000 tries.

import random, math

rep = 100000
data =[]
for n in range(1,11):
    result = []
    for _ in range(rep):
        R = sorted([random.randint(1,6) for __ in range(n)])
        S = R.count(6)+R.count(5)
        R = R[:-S]
        while len(R)>1:
            # Take highest number in R
            x = R.pop(-1)           
            try:
                # Try to find lowest complement to 5.
                R.pop(next(R.index(i) for i in R if x+i>=5))
                S+=1
            except StopIteration:
                # No complement found? Put (highest+lowest) back into list.
                R.append(x+R.pop(0))                   
        result.append(S)

    avg = sum(result)/float(rep)
    dev = math.sqrt(sum([(result[i]-avg)**2 for i in range(rep)])/(rep-1))
    print('{}: {:.2f}+-{:.2f}'.format(n,avg,dev))

The result is the green line, complete with standard deviation (of the single result, not the mean) the red line shows 2/3 #dice for comparison.

While this seems odd, as AgentPaper is not incorrect in his calculations, I believe the deviation is caused by left-over dice that can't be added up to 5.

As pointed out by Chris, instead of randomly rolling 100000 times, you can iterate over every single possible result. The number of such results is 6^n, which means this method takes quite some time for high n. To do so, you change the above code to use the following:

import random, math
from itertools import product

data =[]
for n in range(1,11):
    rep = 6**n
    result = []
    for r in product([1,2,3,4,5,6],repeat=n):
        R = sorted(list(r))
        S = R.count(6)+R.count(5)
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