View Full Version : Reverse Bellcurve Dicerolling

Morph Bark

2014-02-12, 09:28 AM

For stat generation in DnD, you always end up with a bell curve because you've got the result of three dice, with most results being centered around the middle somewhere.

Is there a dicerolling method where you are more likely to get the extremes rather than the average results? A reverse bellcurve, basically?

1337 b4k4

2014-02-12, 09:37 AM

For stat generation in DnD, you always end up with a bell curve because you've got the result of three dice, with most results being centered around the middle somewhere.

Is there a dicerolling method where you are more likely to get the extremes rather than the average results? A reverse bellcurve, basically?

Yes actually, there is, but it's a bit convoluted. I looked this up once for a discussion, and I have an anydice example (http://anydice.com/program/11c7). Basically in this example, we roll 4d6 + 11, if that result is greater than 24, we’re going to subtract 21 and use that for the result, otherwise we’ll keep the roll as the result, then we will add 11 to whatever that result was. It's not quite a perfect bell curve but it does work. Sorry, I don't quite remember enough of the details to give you a generalized formula, but if you play around with that on anydice you should be able to get something like what you want.

Note that it's not a perfect bell curve, but it's close enough for government work.

CarpeGuitarrem

2014-02-12, 09:58 AM

The only simple-ish way that comes to mind is "roll 2d6, flip a coin--heads makes the result positive, tails makes it negative". It violates a normal probability distribution, so you'd have to seriously warp the results of normal dice to do it.

Rhynn

2014-02-12, 10:05 AM

Well, for 3d6, you could just flip the results and produce an exact "mirror" curve by doing the following...

If the roll is 10 or below, add +8 to it. If the roll is 11 or above, subtract -8 from it. This way, the averages (10, 11) become the extremes (18, 3) and the extremes (18, 3) become the averages (10, 11).

Sith_Happens

2014-02-12, 10:09 AM

1 + (Highest possible roll result) – (Roll) will get you the roll's distribution curve flipped upside-down.

Or maybe not, need to think about it a bit more.

EDIT: Stuck it into Anydice, it just reflects the curve about its axis of symmetry, a.k.a. does nothing.:smallredface:

Morph Bark

2014-02-12, 10:32 AM

Considering this method won't be used at the table (I was considering it for a forum game I GM) the extra math would probably just make it more of a hassle for players who aren't good with it, so I'll likely not go with it.

Thanks for the swift resolution.

CarpeGuitarrem

2014-02-12, 10:41 AM

No problem! Your other option would be something like "roll 3d6 on a table" and then you'd have results on the table that started extreme in the middle and then tapered to average results on each side.

Rhynn

2014-02-12, 10:46 AM

No problem! Your other option would be something like "roll 3d6 on a table" and then you'd have results on the table that started extreme in the middle and then tapered to average results on each side.

That's how I laid them out to get the +/- 8 formula.

CarpeGuitarrem

2014-02-12, 11:00 AM

Somehow, I totally missed your post. Whoops. Yeah, that'd work fairly well.

Radar

2014-02-12, 11:27 AM

As I see from the give ideas, it can be generalised to:

for a given symmetric distribution with a finite minimum and maximum, if the result is below average, add (maximum-minimum)/2, if the the result is above average, subtract (maximum-minimum)/2.

This method basicaly cuts the distribution in half and puts the halves in the opposite order. This won't give you exactly a reverse Bell curve, but for any practical purpose it's close enough.

Jay R

2014-02-12, 11:37 AM

No, there is no way to do it by rolling dice, without translating the result. The Central Limit Theorem states that as you add more dice (or any other random number generators) together, the resulting distribution approaches a bell curve.

But there are several ways you can translate the roll to get what you want. Theoretically, you could make a d30 marked as follows:

1, 1, 1, 1, 1

2, 2, 2, 2

3, 3, 3

4, 4

5

6

7, 7

8, 8, 8

9, 9, 9, 9

10, 10, 10, 10, 10

But if you added several of those together, you would still start getting a bell curve.

Ask yourself what exactly you need, and what it's intended to simulate. Then either build your own table, or come back here for more specific advice.

Dimers

2014-02-12, 01:30 PM

There's an easy way to get an inverse bell curve for a single die -- don't recall where I saw it, but the example was for a d20. You roll 2d20 and use whichever result is farther from 10.5. Sadly, I also don't recall what mechanism they used to break ties, e.g. a roll of 14 vs a roll of 7.

You could write an app and use a random number generator with a mapping function.

Mastikator

2014-02-14, 03:46 AM

There's an easy way to get an inverse bell curve for a single die -- don't recall where I saw it, but the example was for a d20. You roll 2d20 and use whichever result is farther from 10.5. Sadly, I also don't recall what mechanism they used to break ties, e.g. a roll of 14 vs a roll of 7.

Just reroll on ties?

SiuiS

2014-02-14, 04:05 AM

I believe that having two dice, one positive and one negative, can do this? It has been a long time though.

Let's see, if they're even, you get... A normal bell curve which extends into negative numbers and bellies out at 0.

If the negative die is larger, you get... Yeah that's not it. Without some reason to invert or not invert one die, this will go nowhere. Best I can figure is that you have one which does both functions; say, d20, any result 1-10 you add, any result of 11-20 you ignore the tens place by one value and subtract (20 = -10). That gives you a linear progression of -10 to +10. If you also, say, roll and add one d20, then you get anywhere from

Oh wow. Yeah this is why I have coffee before math. There's no way any method that has multiple middle generation processes but only one end generation process that will give you a reverse bell curve.

InQbait

2014-02-14, 04:18 AM

I have created a system where when you want to do something, you roll a "Luck" check. Which involves you rolling a total of 4d6. 2d6 of that are positive or "bonus" dice while the other 2d6 are "negative" and penalty dice. You total these together to get your final result. My players don't seem to mind it, however, I have found that it is the improper use of dice. I should just use d12s for this purpose from now on instead of d6s.

I wonder if any other playgrounders find this way of dice rolling is interesting or have come up with this kind of idea before I did. I really like this method, but I am not sure if it is the right way to go about things.

unbeliever536

2014-02-14, 07:19 AM

I have created a system where when you want to do something, you roll a "Luck" check. Which involves you rolling a total of 4d6. 2d6 of that are positive or "bonus" dice while the other 2d6 are "negative" and penalty dice. You total these together to get your final result. My players don't seem to mind it, however, I have found that it is the improper use of dice. I should just use d12s for this purpose from now on instead of d6s.

I wonder if any other playgrounders find this way of dice rolling is interesting or have come up with this kind of idea before I did. I really like this method, but I am not sure if it is the right way to go about things.

This isn't "incorrect" persay (nothing is; if you're making your own system, you decide how it works), it's just kind of kludgy. You'd get the same distribution if you just roll 4d6; all you're doing is subtracting the average from that. If you want to scale against zero, that's fine, just note that subtraction is slower for humans than addition. As for 2d6 vs 1d12, 2d6 gives you a rough bellcurve, while 1d12 gives you a uniform distribution. 4d6 vs 2d12 (which is what you're really comparing) gives you a fairly steep bellcurve vs a somewhat less steep one. It's your system and you have your own design goals, of course, so I can't really advise beyond telling you how the numbers work.

CarpeGuitarrem

2014-02-14, 12:42 PM

I have created a system where when you want to do something, you roll a "Luck" check. Which involves you rolling a total of 4d6. 2d6 of that are positive or "bonus" dice while the other 2d6 are "negative" and penalty dice. You total these together to get your final result. My players don't seem to mind it, however, I have found that it is the improper use of dice. I should just use d12s for this purpose from now on instead of d6s.

I wonder if any other playgrounders find this way of dice rolling is interesting or have come up with this kind of idea before I did. I really like this method, but I am not sure if it is the right way to go about things.

It's similar to how some Fate-based games use d6-d6 instead of rolling four dice labeled -/blank/+. Fiasco also has a variant of it (except that the value is always positive--you subtract the smaller color from the larger). It's pretty nifty and helpful, but with 2d6-2d6, that's a lot of math.

Knaight

2014-02-14, 01:05 PM

I believe that having two dice, one positive and one negative, can do this? It has been a long time though.

Just adding won't do this. What you want is some sort of way to get an absolute value that tends to be high, then zero center it via randomly determining whether it is positive or negative. For instance:

Roll 2d10, keeping only the highest die. Flip a coin - on tails it is negative this, on heads positive. If you want to include zero, treat the 0 mark as a 0 instead of 10. Using Anydice notation, you can get this via:

output (3-2*(1d2))*([highest 1 of 2d10]-1)

It looks more complex than it is, mostly because (3-2*(1d2)) is just notation to get the coin flip effect. As for the probabilities, there is a (1+.5|n|)% chance of getting any result from -9 to 9, including zero. This isn't a bell curve at all, but it does produce more pronounced extremes. If you do want it curved more, you can just add more dice, using the highest of 3d10 or 4d10 or whatever. This also weights it more towards the extremes, and drags the probability of zero way down.

CarpeGuitarrem

2014-02-14, 01:11 PM

Actually, Knaight, that gives me an idea for a quicker method of resolution. Roll two dice, one positive, one negative. Take the higher die. Actually, yeah! That would work. I'm just not sure what to do about ties. Flip a coin? (It'd only happen in 1/n circumstances, where n is the sides on the die...or maybe ties cancel to 0?)

Knaight

2014-02-14, 01:16 PM

Actually, Knaight, that gives me an idea for a quicker method of resolution. Roll two dice, one positive, one negative. Take the higher die. Actually, yeah! That would work. I'm just not sure what to do about ties. Flip a coin? (It'd only happen in 1/n circumstances, where n is the sides on the die...or maybe ties cancel to 0?)

Ties cancelling to zero gets weird mathematically. It basically produces a probability spike at zero. I'd almost prefer to create a more wave like oddity (odd ties go positive, even ties negative). That said I'm not nearly good enough with AnyDice notation to try and get the data on the probability of these*.

*Also I'm too lazy to do it by hand or with excel, as it's not exactly hard.

SiuiS

2014-02-14, 01:22 PM

Just adding won't do this. What you want is some sort of way to get an absolute value that tends to be high, then zero center it via randomly determining whether it is positive or negative. For instance:

Roll 2d10, keeping only the highest die. Flip a coin - on tails it is negative this, on heads positive. If you want to include zero, treat the 0 mark as a 0 instead of 10. Using Anydice notation, you can get this via:

output (3-2*(1d2))*([highest 1 of 2d10]-1)

It looks more complex than it is, mostly because (3-2*(1d2)) is just notation to get the coin flip effect. As for the probabilities, there is a (1+.5|n|)% chance of getting any result from -9 to 9, including zero. This isn't a bell curve at all, but it does produce more pronounced extremes. If you do want it curved more, you can just add more dice, using the highest of 3d10 or 4d10 or whatever. This also weights it more towards the extremes, and drags the probability of zero way down.

Yeah. I knew I was missing a basic step, really. The concept was there but as tired as I was it wasn't gonna happen.

Actually, Knaight, that gives me an idea for a quicker method of resolution. Roll two dice, one positive, one negative. Take the higher die. Actually, yeah! That would work. I'm just not sure what to do about ties. Flip a coin? (It'd only happen in 1/n circumstances, where n is the sides on the die...or maybe ties cancel to 0?)

There it is. Feels painful though. How are ties decided?

CarpeGuitarrem

2014-02-14, 01:25 PM

Ties cancelling to zero gets weird mathematically. It basically produces a probability spike at zero. I'd almost prefer to create a more wave like oddity (odd ties go positive, even ties negative). That said I'm not nearly good enough with AnyDice notation to try and get the data on the probability of these*.

*Also I'm too lazy to do it by hand or with excel, as it's not exactly hard.

I like this method for handling ties. Mind you, odd positive/even negative would bias the results slightly towards negative (because on a die, every even number is an odd number + 1), and vice-versa.

But it's closer. And it's fairly elegant, relatively-speaking. Having two dice of each (and picking the highest die out of all of them) would make the curve better (more extreme) and render ties less likely.

Rhynn

2014-02-14, 01:34 PM

-1d10, +1d10, take higher, odd ties +, even ties - gets you an average of -0.05 (the other way around would, indeed, be +0.05)... and has some quirks, like there never being a result of -1, more -10s than +10s, etc. ...

Knaight

2014-02-14, 01:43 PM

I like this method for handling ties. Mind you, odd positive/even negative would bias the results slightly towards negative (because on a die, every even number is an odd number + 1), and vice-versa.

It depends, really. If you use 1d10 as a 0-9 generator it's a slight positive bias instead, which was actually what I was aiming for.

-1d10, +1d10, take higher, odd ties +, even ties - gets you an average of -0.05 (the other way around would, indeed, be +0.05)... and has some quirks, like there never being a result of -1, more -10s than +10s, etc. ...

Yep. The question is whether these quirks are better or worse than the quirks of ties going to zero, or iffy enough that the longer method involving random sign is better. There are also somewhat fewer quirks using a -9 to +9 generator by treating the 0 on a d10 as a zero, in that it can at least produce all the numbers in the range.

CarpeGuitarrem

2014-02-14, 02:10 PM

I really need to reread the game, but this is definitely reminding me of Don't Rest Your Head. There, however, you're looking for the highest die from three colored pools...and instead of some pools being positive or negative, every pool is positive--but means something significant. (The pools were Exhaustion, Madness, and Discipline. Plus Pain dice, rolled by the GM.)

Segev

2014-02-14, 04:43 PM

Roll 3 dice: positive, negative, and "decider."

Take the higher absolute value as the winning die, but keep the die's sign for the final result.

On a tied absolute value, look at the decider die. If it's odd, result is positive; if it's even, result is negative.

Sith_Happens

2014-02-15, 04:28 PM

You're all making this too complicated. Just reroll ties.

NichG

2014-02-15, 04:49 PM

This should generalize the first algorithm posted: roll XdY. If the result is less than X*Y/2, add X*Y to the result.

This generates a number between X*Y/2 and 3*X*Y/2, weighted towards the extremes. You will want at least 2 dice to have any sort of weighting.

So for example 2d20 would generate results between 20 and 59, where 20 and 59 are the most common outcomes (~1/20 chance), and 41 and 40 are the least common outcomes (1/400 chance)

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