View Full Version : [Math help] Expanding dungeon world dice range

Madeiner

2016-10-12, 05:20 PM

Hi there.

I am modyfing a cyberpunk hack for a PbtA system.

PbtA uses a 2d6 +- 3 system ,where 10+ is success and 6- is failure.

I want to reduce the weight of a single +1.

Ideally, i want it to behave as around the equivalent of +0.5.

I want to use this to have more gradual, incremental progression of bonuses for a longer running campaign, especially being able to use a "deeper", more vertical equipment list.

I also want to (roughly) keep the distribution range for PbtA, which is summarized here. The bell curve i feel is necessary for PbtA, and i don't want to alter the balance which has surely been tested.

http://apocalypse-world.com/forums/index.php?topic=2787.0

I' ve spent the last 2 hours on anydice and i can't do it.

It works for single values (example: using 2d12 and setting success at 19 and failure around 11) but i can't get the curve right.

Is it at all doable mathematically?

I can do this very easily using single dice: a +1 on a 1d10 system is worth exactly double a +1 on a 1d20 system.

meschlum

2016-10-13, 02:42 PM

A major part of your problem comes from the fact that a +1 isn't always worth the same amount when using more than one die. As an illustration, two +1 bonuses (equal to +2) don't change the odds the same way with 2d6 as the 'expected' shift from twice the effect of a +1 bonus.

6- fail is 5/12 chance of failure (41.67%), reduced to 5/18 (27.78%) by a +1 (13.89% drop). Add another +1, and you're down to 1/6 chance of failure (16.67%), a reduction of 11.11% from your chance with a +1. So does a +1 reduce your chance of failure by 13.89% or by 11.11%?

So 'half' the effect of a +1 is undefined as it depends on what you're comparing it to!

There are analytical approaches that can help you narrow down the sort of numbers you're looking for, though - with a pile of assumptions about what you're after.

1) We're using dice totals rather than lookup tables. Otherwise, just create a table with the cutoffs you want (or more than one table, for different bonuses), and roll a d100 or d1000 or so.

2) We're sticking to 2 dice. Adding more dice gives you more distributions to play with, but you've indicated you want to keep the current shape. Plus, it makes the math simpler (even if it does limit options).

3) You want your base odds of success / failure to be close to those used initially. Obvious, but needs to be said.

4) You want a +1 to change the odds of both success and failure by roughly half the amount of change you get with a +1 on 2d6. So with a +1 modifier, the odds of failure should be close to 25/72 (34.72%), halfway between 5/12 and 5/18.

5) You don't care too much what happens to the odds of success / failure for anything besides a +1. This includes a +2, which may no longer be equivalent to a +1 on 2d6, and a -1 (or -2), which may not shift the odds the same way.

If you're happy with all the above, we can work something out.

The initial probability of failure (6-) is 41.67% (5/12). With a +1 on 2d6, it becomes 27.78% (5/18), so we're aiming for 34.72% (25/72).

The initial probability of success (10+) is 16.67% (1/6). With a +1 on 2d6, it becomes 27.78% (5/18), so we're aiming for 22.22% (2/9).

Using two S sided dice, and a total of T, where T is less than or equal to S+1, the probability of getting T or less is:

T * (T + 1) / (2 * S^2)

Adding +1 to the die roll means we now want a total of T - 1 or less.

So we get when assessing the odds of failure:

T * (T + 1) / (2 * S^2) = 5 / 12

T * (T - 1) / (2 * S^2) = 25 / 72

Which can be simplified to:

T / S^2 = 5 / 72

T = 5 S^2 / 72

Substituting in the value of T gives us:

5 * (5 * S^2 / 72 + 1) / 144 = 5 / 12

5 * S^2 / 72 + 1 = 12

5 * S^2 / 72 = 11

S^2 = 158.4

So we get S in the 12-13 range for the best fit (and T around 11).

Performing similar calculations for success indicates that we'd get S in the 11-12 range and T around 7 to get the desired results in that case.

Then comes the practical evaluation, because you can't really roll a die with 12.59 sides (or 11.22 sides). Try results with S and T values close to what we've established, and pick the ones that suit you best.

12 sided dice already exist, which is a plus.

You get 11 or less on two 12 sided dice with probability 38.19%, and you get 10 or less with probability 31.25%. Fairly close, slightly lower chances of failure without a +1, slightly higher than we'd like with a +1, but not bad.

You get 18 or more on two 12 sided dice with probability 19.44%, and a +1 bumps you up to 25%. Again, better chances of success than what you'd get otherwise, but fairly reasonable.

The method is fairly susceptible to approximations, so you can get somewhat different results depending on how much margin of error you allow for. The overall range is correct, but it can be worth checking values of S up to 50% away (6 to 18 or so here) from the initial estimate. For instance, using d16s (not practical as a single die), you can get a nice fit using 14 or less for failure (41.01% chance, 35.55% with a +1) and 24 or more for success (17.58% chance, 21.48% with a +1).

The core formula for probability is going to be more useful if you want to experiment, of course.

Knaight

2016-10-13, 06:57 PM

If you want to keep roughly the same distribution, the easy way to do it is to use anydice and compare your new system to 2d6*2. 2d11+2 comes very close (although if you're using physical dice it's not ideal). 2d10+3 is also solid, albeit less so. Then you want to move the cutoff points so that you get similar behavior near the points of interest. 2d10+3 does that very well, using 13 and 19. For this curve fitting what you really care about is the range where successes are conditional on skill and how that changes things, with the distribution between 4-9 in the original system of particular interest, and 6-18 in the 2d10+3 system.

Of course, actually hauling around a +3 is pointless, so adjust the starting points. 2d10 using 10- and 16+ works out to be fairly similar. Here's a pseudotable:

Die Code, P(Failure), P(Success)

2d6, 41.67%, 16.67%

2d10, 45%, 15%

2d6+1, 27.78%, 27.78%

2d10+2, 28%, 28%

2d6+2, 16.67%, 41.67%

2d10+4, 15%, 45%

2d6+3, 8.33%, 58.33%

2d10+6, 6%, 64%

So, reasonably close. If you use a 0-5 scale for 2d10 (replacing a 0-3) it fits even better, but it does make each point worth a bit more.

Madeiner

2016-10-18, 05:56 PM

Thanks guys.

Math was hard to understand, but i think i found some good values, being:

2d12, using 19+ for a success and <12 failure.

For values of -2 to around +6, the points in the curve looks to be within 4% of the existing 2d6 curve, which is totally acceptable. Another plus is that around these values, the new pluses are worth just about half of the previous ones, which makes for easier converting.

If anyone wants to confirm the values are good, i'd appreciate it.

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