How can I simulate decreasing randomness even though randomness is increasing?

Question

I am seeking a way to decrease variance as character skill advances. I've considered various mechanisms for dice tests. I have one in particular that does not accomplish this by default, but I like the mechanism for other reasons.

I want to know if there is a way I can simulate my goal while otherwise retaining the mechanism in its current form.

Count wins in [SKILL]d6

• As character level advances, various SKILLS advance.
• When attempting actions using SKILLS they are better at, characters roll more dice, increasing the chance to roll wins, thereby increasing "accuracy"
• To the best of my (not math oriented) understanding, rolling more dice means more opportunities for random and extreme results, thereby reducing "precision".

Is there a way I could affect a reduction in variance (increase in precision) by using an auxiliary die, some sort of reference table, or some other mechanism?

This is where I begin to struggle to mentally visualize the details necessary for understanding, but so far I have considered a sort of "luck die" that is rolled at the same time as a dice test. The luck die determines some modification to the dice test results (perhaps increasing the range of numbers that count as a win, allowing to turn a certain amount of "fails" into wins, or something else).

How can I visualize these details better to tweak the system, or what are some mechanisms that achieve my goal of decreasing variance with SKILL advancement while keeping a d6 dice pool mechanism?

Answer 1

Sort of.

As the number of dice increases, the range of potential outcomes increases.

As the number of dice increases, the likelihood of an extreme outcome decreases.

To show this, let's assume a WIN is 50%: 4,5,6 on a d6. If you roll 1d6, the likelihood of the max result (1) is 50% and the likelihood of the minimum result (0) is also 50%. It's very "swingy" within that range. With two dice, it's 25% minimum, 50% middle, and 25% maximum. As you add more dice, those results cluster more and more toward the middle. With enough dice the result smooths out to something that looks like a normal distribution (it isn't, but it looks a lot like one).

So, yes, the result has a higher potential range with more dice, but no, it actually becomes more predictable, not less.

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You can find the odds yourself fairly easily. The formula looks complex, but it's actually something you can do on the back of an envelope for up to 10 or so dice (and trivially in a spreadsheet). I'll state it, walk briefly through the how it works, and show you how to generalize it:

For k dice with d sides and w winning faces, the probability of exactly n wins is: [( n! / (k! * (n-k)! ) )] * [w/d]^[n] * [(d-w)/d]^[k-n]

First, each die is independent. Think of rolling them one at a time; the way each die lands does not affect the way the next die lands. So you have a giant binary decision tree: with two dice, the results could be:

• Loss then Loss (LL)
• Loss then Win (LW)
• Win then Loss (WL)
• Win then Win (WW)

Let's change the earlier assumption and say instead that a WIN is only when a die shows six. so to get the first result, you have a (5/6) event, then another (5/6) event: 25/36 two-dice rolls will end showing Loss-Loss.

Similarly, wins are (1/6), so WW is only 1/36.

LW is (5/6) then (1/6), which is 5/36. WL is (1/6) then (5/6) which is also 5/6.

Collectively, that's 36 out of 36 possible outcomes.

You can build a similar tree for any number of dice, but it's easier to generalize: the likelihood of each discrete outcome there is [probability of win]^[number of wins] * [probability of loss]^[number of losses]. (Anything raised to the power 0 is 1). So:

• WWWW would be (1/6)^4 * (5/6)^0 = (1/1296) * (1) = 1/1296
• WLWWWL would be (1/6)^4 * (5/6)^2 = (25/36) * (1) = 25/46656
• LLWWWW would be (1/6)^4 * (5/6)^2 = (25/36) * (1) = 25/46656
• LWWWWL would be (1/6)^4 * (5/6)^2 = (25/36) * (1) = 25/46656

(last three examples show that order doesn't matter, just the totals.)

Counting all the ways you could have 2 losses and four wins is hard. Fortunately, there are two ways to to figure that out. First is the "binomial coefficient": for k items, how many ways can you choose n of those items? Or more relevant here, for for k dice rolled, how many ways can n of those dice show wins?

You could calculate it using a formula: ( n! / (k! * (n-k)! ) ) If you're doing that, plug it into a spreadsheet instead.

Or you can use a little math trick to generate the coefficients: Pascal's Triangle. It's faster up to about 5 dice, and still good to about 10:

          1
        1   1
      1   2   1
    1   3   3   1
  1   4   6   4   1
1   5   10  10  5   1

Notice how on each line, the outside numbers are 1, then each other number is the sum of the two numbers diagonally up from it. So, going from the fourth line (1 3 3 1) to the fifth line ( 1 4 6 4 1):

• 4 is the sum of 1 and 3
• 6 is the sum of 3 and 3
• 4 is the sum of 3 and 1

...and you can extend that on down infinitely (next line would be 1,6,15,20,15,6,1).

The numbers on each row are the binomial coefficients for that number of dice, starting with 0. So, from the top:

• There is only one way to get a result with 0 dice.
• With 1 die, there is one way to get 1 win and one way to get 1 loss.
• With 2 dice, there is one way to get 2W, two ways to get 1W 1L, and one way to get 2L.

So then that brings us down to the final piece: to get the probability of exactly n wins on k dice with w winning faces, you calculate:

[binomial coefficient] * [probability of win]^[number of wins] * [probability of loss]^[number of losses]

Which is:

[binomial coefficient] * [w/6]^[n] * [(6-w)/6]^[k-n]

...which still seems complicated, but is actually pretty quick. For example, getting 2 wins on 3 dice, with 2 winning sides is:

[3] * [2/6]^[2] * [4/6] = 3*(2^2)*4 / (6^2 * 6)

The only difficulty there is 6^3 = 216 (which will be the denominator for everything with 3 dice), so with those inputs the probability of two wins is 48/216.

Answer 2

Dice pools are actually excellent at producing the sorts of results you seem to want. Extreme results are possible but increasingly unlikely. Here is something to help you out with your math...

https://anydice.com/program/31fc3

This little program runs on the assumption that you roll [X]d6, and count up how many 5s and 6s you get as your "Wins."

In event of link rot, the program is very simple:

output 4d(1d6 > 4)

Replace that first '4' with however many dice you want to simulate. Just to give a few examples:

• If you roll 1 die, you get 0 successes 66.67% of the time, and 1 success 33.33% of the time
• If you roll 2 dice, you get 0 successes 44.44% of the time, 1 success 44.44% of the time, and 2 successes 11.11% of the time.
• 3 dice gives 0 successes 29.63% of the time, 1 44.44% of the time, 2 22.22% of the time, and 3 3.7% of the time

Now let's make some big jumps up...

8 dice gives you...

• 0 a mere 3.9% of the time
• 1 success 15.61%
• 2 successes 27.31%
• 3 successes 27.31%
• 4 successes 17.07%
• 5 successes 6.83%
• 6 successes 1.71%
• 7 successes 0.24%
• 8 successes 0.02%

In summary, you have an 87.3% chance to get 1, 2, 3, or 4 successes. So any other outcome is a significant outlier.

And so it continues. The basic result is that you get a very strongly normed distribution that centers around 1/3 of how many dice you rolled (assuming a 5/6 to succeed). So, dice pool systems (such as Shadowrun, Warhammer 40K Wrath and Glory, World of Darkness, and Exalted) tend to give you successes in a very reliable range as you increase your total dice count.

Even if you throw 20 d6s, you have a roughly 85% of getting between 4 and 10 successes.

Answer 3

As the other answers have explained, yes, the more dice you roll, the more possible different outcomes you can get. As you are only counting successes, the more dice, the more possible successes. However, the extremes of the success range become relatively less and less likely the more dice you use.

To help with the visualization aspect of this, here is another simple anydice program that loops over various numbers of dice, and shows you the distribution of successes.

The output looks like this (using 5 or 6 as a success on a d6):

As you can see, the extreme number of course increases to match the number of dice you used, but these extremes become relatively less likely: from 33% with just one die, down to just 0.14% with six. At the same time, the share of middling numbers of successes takes a larger and larger share. Another way too look at this is to plot the expected number of successes as a graph:

You can see how the bulk of the successes is increasingly in the middle of the pack as the number of dice increases.

Answer 4

These methods yield progressivey larger average and smaller variance, with a maximum of 120: 6d20, 10d12, 12d10, 20d6, 30d4.

Something similar can be achieved with a single type of die. Make the player roll d6 5 times, put them in ascending order, and before summing them, let the player replace the lowest few dice (how many depends on the player's skill level) with the first non-replaced die. Example: 1 2 2 3 5 with the lowest 2 replaced becomes 2 2 2 3 5.

See other answers for counting the average and the variance.

Answer 5

Don't mess with n. Mess with p.

This is some advanced math; I would recommend a college-level course on probability if you want to develop some good "probability instincts" instead of having to manually verify the impact of your changes.

The relevant concept here is binomial distribution; variance for the binomial distribution is n (number of attempts) * p (chance of success) * q (chance of failure). In cases where you either succeed or fail and q equals 1 - p, that's n(p - p^2). Holding n constant, the slope of that curve is 1 - 2p with an inflection point at p = 1/2, and since the slope's negative after that, 1/2 is the local maximum.

As probabilities grow toward 1/2, variance increases; as they grow beyond 1/2, variance decreases. So maybe... roll some number of d10s, d12s, or d20s determined by situation, and try to get under your skill rating, which starts at 50% odds?

Answer 6

If I understand your question correctly, what you're asking for is a method that has the following requirements?

1. Lets the player roll multiple d6's and count how many of those are WINs
2. Higher SKILL means a higher average of WINs
3. Higher SKILL means more consistent results (in other words a smaller difference between the min and max number of WINs you can get)(in other words the skill would feel more reliable to the player, because it's less likely that the actual result is much lower than the expected result)

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If so, this method might suit your needs (assuming max SKILL level is 10, can be any arbitrary number):

1. Players roll a number of green D6's equal to their SKILL, and add a number of red D6's to that to make a total of 10d6.
2. Red D6's count 1-3 as a loss, 4-5 as a single win and 6 as a double win.
3. Green D6's count a 1-4 as a single win and 5-6 as a double win.

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This method would have players always roll 10d6, but when their SKILL increases they can replace that many red dice with green dice. This means statisticly:

• The average number of wins increases, as the green dice have more chances of a single win and more chances of a double win (higher skills feel stronger)
• The minimum number of wins increases, as the green dice have no options of zero wins but the red dice do (higher skills feel more reliable, less chance of catastrophic failures)
• The maximum number of wins stays the same, as both dice have the possibility of giving double wins (even the lowest skilled character can get super lucky, but this would be extremely rare) (if you don't like this you can replace the red dice double win with a single win)