Non-standard bell curve? WARNING! CONTAINS MATH!

[Chimera245]

I know that multiple dice rolled at once can produce a bell curve effect, where the average values are more likely to come up than the extreme values. But is there a way to create Asymmetric bell curves with dice combinations some how?

I need a range of possible numbers that have the common, average values not in the middle.

Like, I need a bell curve that has more Above Average results than Below Average results, but still has the average come up most often.

Is there a way to manipulate the "peak" of the bell curve without moving the "feet"?

I never took any kind of advanced probability-based studies in school, so I'm hoping some kind of math-majored college graduate will chime in with "Oh, just do this simple thing which completely solves all your problems," but I'm not gonna hold my breath.

[Knaight]

Do you want to maintain the same average as in a symmetric bell curve, or are you on board with having a different, higher average where high results are also favored? If it's the first then this is going to get really convoluted. If it's the second, then it's easy - you can use either a roll and keep system (the 4d6 best 3 of D&D stat generation actually is an example of this), you can use a dice pool where more than half the results of the die count as a success (roll Xd10, count everything 3+), or any number of other things. I'm assuming the second, so this should be easy. Here's some AnyDice code for both of these situations, demonstrating curvature induced in both methods.

METHOD 1:
output [highest 3 of 3d6]
output [highest 3 of 4d6]
output [highest 3 of 5d6]
output [highest 3 of 6d6]
output [highest 3 of 7d6]

METHOD 2:
output 10d{0,0,1,1}
output 10d{0,0,1,1,1}
output 10d{0,0,1,1,1,1}
output 10d{0,0,1,1,1,1,1}
output 10d{0,0,1,1,1,1,1,1}

[GuzWaatensen]

...
Like, I need a bell curve that has more Above Average results than Below Average results, but still has the average come up most.
...

That is pretty much impossible by definition...

[Arctanaar]

That is pretty much impossible by definition...

d{-2,0,1,1} would have a mean of (-2)*0.25+0*0.25+1*0.5=0 with more above average results than below average ones.
I think that you are wrong.

[Yora]

The question is obviously about a non-bell shaped curve with the maximum at neither th center nor the ends.

One solution would be NdX, ignore results greater than Y.

An unrelated dice probability curiosity I just discovered: 1d6+1d4 leads to a curve with a perfectly flat top, with the same probably for 5, 6, and 7. (Seems to work with all combinations of two different dice.A greater difference in size leads to a wider flat line.)

[halfeye]

If the average moves to one side or the other, that's called sqewed:

https://en.wikipedia.org/wiki/Skewness

If there are two bumps on the distribution, that's bimodal:

https://en.wikipedia.org/wiki/Multimodal_distribution

If it's neither of those, it probably doesn't work in real statistics.

[GuzWaatensen]

That is pretty much impossible by definition...

...
I think that you are wrong.

Embarrassingly you are right, I made the age old mistake of mixing up average and median.

[Chimera245]

Skewed! Yes, that's exactly what I want. I've tried mixing die sizes and discovered the "flat top bell curve" thing. But I want to achieve it with the simplest possible math.

Specifically, I need a random number from 1 to 13, with 6 as the most common result. 2d7-1 puts the average at 7 instead, and 1d6+1d8-1 gives a flat-top bell with 6, 7, and 8 tied for most common.

I'll use that if I have to, but if there's an easy way to be wholly accurate, I'd rather do that.

[5a Violista]

I found a more complex rolling thing that gives you what you want (numbers 1 to 13, with the highest frequency at 6): http://anydice.com/program/937c. (The code I used was "output [highest 1 of 2d6]+[lowest 1 of 2d8]-1")
So you'd have to roll 4 dice, 2 d6 and 2 d8. You keep the best of the d6 and the worst of the d8.

The average result is 6.66, by the way.

[Mastikator]

Just take a regular bell curve and displace it by 50%

An easy example is 3d6, would have a distribution from 3 to 18, most common numbers 11 and 12. So just add 8, if above 18 subtract 16 (19 becomes 3, 20 becomes 4, etc).

You'll have a probability distribution that looks flipped vertically. The most common numbers would not be 11 and 12, they'd be 3 and 18.

[Knaight]

I found a more complex rolling thing that gives you what you want (numbers 1 to 13, with the highest frequency at 6): http://anydice.com/program/937c. (The code I used was "output [highest 1 of 2d6]+[lowest 1 of 2d8]-1")
So you'd have to roll 4 dice, 2 d6 and 2 d8. You keep the best of the d6 and the worst of the d8.

The average result is 6.66, by the way.

That seems a bit convoluted. A dice pool using 12d10 where 7-10 is a success goes from 0-12, mode of 5. Add 1 to that, and you get 1 to 13, mode of 6.
output 1+12d{0,0,0,0,0,0,1,1,1,1}

[MrStabby]

OK, so we can rule out summing dice together. The is a bit of maths called the standard limit theorem that rules that out.

Now pushing the limits towards the edges of the distribution is easy if we know what effects we are looking for. Do we want the most extreme values to be the most common? Roll 100 pairs of D6s and the result is the least common output? Cumbersome and impractical without a computer but you get the idea. In principal a sufficiently close approximation to anything you want should be possible.

2d12s take the highest if sum is odd, take lowest if sum is even will give you a bit more of a spread. You can add as many dice as you want to this depending on how big you want the spread to be and how unlikely the value in the middle.

[Knaight]

OK, so we can rule out summing dice together. The is a bit of maths called the standard limit theorem that rules that out.

That's not accurate at all, as the workable method upthread involving summing dice together demonstrates. Adding two uniform distributions together isn't going to work, but as long as the distributions can be made nonuniform we're good.

[5a Violista]

That seems a bit convoluted. A dice pool using 12d10 where 7-10 is a success goes from 0-12, mode of 5. Add 1 to that, and you get 1 to 13, mode of 6.
output 1+12d{0,0,0,0,0,0,1,1,1,1}

Yes, it is pretty convoluted. I agree. Your method also works and is simpler, but requires rolling three times as many dice. While mine's convoluted, I also think it's pretty elegant in how it fulfills the requirements.

So, I picked "Roll 2d6 take best, roll 2d8 pick worst, subtract 1" because it required the least amount of dice to roll out of the ones I could find that fits the criteria. Definitely not the simplest (I believe yours fits that bill), but does require fewer dice. (Well, I guess you could just make a single custom 13-sided die that's weighted towards 6, but that would take a lot of design work.)

Actually, I think it would be pretty neat to graph out the distributions of all the suggestions in this thread so far. As long as I'm procrastinating, I might as well do it myself. Expect me to edit in pictures of these distributions made from some spreadsheet program or something.

Spoiler: Only the ones that give 1 to 13

Show

Only two of the suggested methods actually fit both criteria, so those ones have a high saturation color.

[Chimera245]

Wow, now that I see it actually charted out, that is pretty much exactly the curve I wanted. Now I just need to remember how to program those kinds of operations into Maptool macros...

Just for my own curiosity, how is it that the probabilities come up those ways from those dice formulas? I've got a pretty good head for math, but never really explored probability very deeply...

[Knaight]

Wow, now that I see it actually charted out, that is pretty much exactly the curve I wanted. Now I just need to remember how to program those kinds of operations into Maptool macros...

Just for my own curiosity, how is it that the probabilities come up those ways from those dice formulas? I've got a pretty good head for math, but never really explored probability very deeply...

It depends on the formula. For mine, I knew that dice pools used a binomial distribution that can be made arbitrarily wide, hence the 12dX. I also know that binomial distributions can be skewed one way or the other by making the success chance something other than 50, to skew low I lowered the chance. For reasons that boil down to sheer laziness, I did this by making a pseudo d6, finding it didn't work, then using a pseudo d10 and fiddling with the numbers until they worked out way.

The other formula is a bit trickier. The notation used in the graph is 2d6b1+2d8w1-1, getting 1-13 with an average of 6. Consider 1d6+1d8-1, which also gets 1-13 with a flat average as a baseline. Using 2d6b1 skews the d6 upwards a bit, using 2d8w1 skews the d8 downwards a bit. Because the d8 has more sides than the d6 the magnitude of the skew is higher, which pushes the probability of the 6 up a bit relative to the 7. Essentially it involves adding two probability curves together, and shifting the peak in the process.

These differences also carry through to the rest of the distribution. Binomial distributions tend to be very middle heavy, which is why the probability of 6 on that one is a bit higher, and why the probabilities of 11-13 are tiny. When adding two offset distributions the edges are usually only really affected by one of them, and with a d6 and d8 being fairly similar you see almost symmetrical behavior on the ends for roll and keep.

EDIT: This does suggest an obvious third method that minimized dice while retaining a simple dicepool structure - a trinomial distribution. For 1-13 there's basically a pair of mathematically identical ways to do it for each set of 6 usable dice, one of which involves each die having a result of 0-2 (weighted) and adding 1, and the other of which involves each die having a result of -1 to 1 (weighted the same way) and adding 7. Here's three different ways to do it, using either d6 or d10.

output 7+6d{-1,-1,0,0,0,1}
output 7+6d{-1,-1,0,0,0,0,0,0,0,1}
output 7+6d{-1,-1,-1,0,0,0,0,0,1,1}

That essentially works out to d6, 1-2 is -1 and 6 is +1, d10, 1-2 is -1 and 10 is +1, and d10, 1-3 is -1 and 9-10 is +1. All are skewed, all have 6 as the most probable result, and in the case of the d6 dicepool the mean is 6.00 (the other two both hit 6.40). So there's another three options. A d4 pool would not work, I didn't check d8, and d12 and d20 are both guaranteed to work (I didn't check them either, but the d6 and d10 results can be duplicated). If you feel like doing something weird with the dice used, the pools of unusual die types work perfectly.

[CharonsHelper]

An easy example is 3d6, would have a distribution from 3 to 18, most common numbers 11 and 12.

You mean that 10 & 11 are the most common.

[Algeh]

Math teacher here: If you want some not-easy-to-get-with-dice distribution, one other option is not to use dice. :)

For an in-person game, for example, you might use cards/pieces of paper drawn out of a hat. You could visually "make" your initial distribution by setting up your cards with say, one "1" card, two "2" cards, twenty-seven "3" cards, five "4" cards, ten "5" cards, or whatever until it looked right, then just put all the resulting cards in a bag and draw one out instead of rolling a die each time. This lets you have any mix of probabilities you want since you're not doing any adding together so you won't produce a bell curve unless that's the distribution you chose with the mix of cards in the first place. (Make sure to put the card you just drew back in each time unless you want an effect where numbers get less likely when they've happened already.)

You can also do this by re-visiting the old D&D Percentile-based Random Encounter tables model (I have no idea if these are still a thing - the last D&D edition I played regularly was 2nd edition AD&D). Assign a bigger "chunk" of the table to the results you want to have be more common and a smaller "chunk" to the results you want to have be less common. Each number is equally likely to be produced, but entire clusters of numbers map to the same outcome and you distribute the size of the clusters to make things more or less likely as appropriate for the distribution you actually want.

You could also make a custom spinner with regions of sizes that mapped to how often you wanted those outcomes, but I'm not sure I'd want to use spinners in an ongoing game since it can be hard to get ones that work very well.

Anyway, if using a system where you're adding dice together and then doing a bunch of fiddling doesn't get you the distribution you want or seems to have so much overhead that it's a major pain, you might try a completely different random generator instead.

[enderlord99]

Xd(YdZ) might work, or it might be the exact opposite; I'm not sure which.

[nedz]

The question is obviously about a non-bell shaped curve with the maximum at neither th center nor the ends.

One solution would be NdX, ignore results greater than Y.

An unrelated dice probability curiosity I just discovered: 1d6+1d4 leads to a curve with a perfectly flat top, with the same probably for 5, 6, and 7. (Seems to work with all combinations of two different dice.A greater difference in size leads to a wider flat line.)

this.
By adding together different distributions you can skew the result either way — skewing to the left is harder, but still possible.
Getting an exact match to the distribution you want may be so non-trivial as to be impracticable; but one close to it may be possible.

Knaight does a similar trick with his negative dice but also adds in a skewed binarisation.

So: what is the distribution you want ?
Or do the details not matter.

[Knaight]

Knaight does a similar trick with his negative dice but also adds in a skewed binarisation.

That's not actually a binary distribution at all anymore, as each die has 3 possible results.

[nedz]

That's not actually a binary distribution at all anymore, as each die has 3 possible results.

Oops, trianarisation — well it's a similar concept.

Actually reading your post again, more carefully this time, you could view your trick as relabelling. It's a bit like the Average Die trick: A D6 with the 1 relabelled as a 3, and the 6 as a 4.

[Knaight]

Oops, trianarisation — well it's a similar concept.

Actually reading your post again, more carefully this time, you could view your trick as relabelling. It's a bit like the Average Die trick: A D6 with the 1 relabelled as a 3, and the 6 as a 4.

It's definitely a mapping technique (which pretty much goes with the territory with dice pools), and it is a bit like the Average Die trick - except you're introducing a skew deliberately. It's more like a d6 with the 6 relabeled as 1, which has a 1-5 distribution that averages a bit less than 3. It also has the benefit of the mean being exactly 6 in the d6 case, which needs d12s to pull off with the binomial distribution.

On an unrelated note - 2d8, 8s relabeled as 1 has a mean of 6.25. It's not the most common result, but it is the average due to the introduced skew. So there's another option.

[nedz]

On an unrelated note - 2d8, 8s relabeled as 1 has a mean of 6.25. It's not the most common result, but it is the average due to the introduced skew. So there's another option.

The most common result is 2

I occasionally use a skewed D6:{1,2,3 -> 1; 4,5 -> 2; 6 -> 3}

There are any number of mappings so any pattern is possible.

[Knaight]

The most common result is 2

Not so much - it's actually 7. The 6 average* has to do with how 1-5 are so much more common than 9-13.

*To one sig fig.

[fusilier]

Math teacher here: If you want some not-easy-to-get-with-dice distribution, one other option is not to use dice. :)

For an in-person game, for example, you might use cards/pieces of paper drawn out of a hat. You could visually "make" your initial distribution by setting up your cards with say, one "1" card, two "2" cards, twenty-seven "3" cards, five "4" cards, ten "5" cards, or whatever until it looked right, then just put all the resulting cards in a bag and draw one out instead of rolling a die each time. This lets you have any mix of probabilities you want since you're not doing any adding together so you won't produce a bell curve unless that's the distribution you chose with the mix of cards in the first place. (Make sure to put the card you just drew back in each time unless you want an effect where numbers get less likely when they've happened already.)

You can also do this by re-visiting the old D&D Percentile-based Random Encounter tables model (I have no idea if these are still a thing - the last D&D edition I played regularly was 2nd edition AD&D). Assign a bigger "chunk" of the table to the results you want to have be more common and a smaller "chunk" to the results you want to have be less common. Each number is equally likely to be produced, but entire clusters of numbers map to the same outcome and you distribute the size of the clusters to make things more or less likely as appropriate for the distribution you actually want.

You could also make a custom spinner with regions of sizes that mapped to how often you wanted those outcomes, but I'm not sure I'd want to use spinners in an ongoing game since it can be hard to get ones that work very well.

Anyway, if using a system where you're adding dice together and then doing a bunch of fiddling doesn't get you the distribution you want or seems to have so much overhead that it's a major pain, you might try a completely different random generator instead.

I've played wargames that have used "chits" (small numbered pieces of cardboard drawn out of a cup or bag), and played both wargames and rpgs which used tables (roll dice look up result on table).

I actually have a fondness for tables -- don't play many games these days that use them though. :-/