# Dice rolling mechanic where modifiers have a predictable and consistent effect on difficulty

I am looking for a dice rolling mechanic that makes it such that increasing or decreasing a modifier on the roll has a constant multiplier effect on the probability of the outcome.

Say you have to make a roll for STAT. Such a roll has a probability of success of 50%. Now say you roll with a mod of -1, this roll has a probability of success of 25%. -2 has a probability of 12.5%. -3 is 6.25% and so on, always halving. The other way around it should be the same but for the probability of failure, always being divided by the same factor.

It doesn't have to be a multiplier of 0.5, in fact I'd much rather it was a multiplier of 0.66-0.75, not such an extreme change. The default unmodded value doesn't need to be 50% chance of success either, it can be something else.

Is there any kind of dice rolling mechanic I can use to simulate something like this?

## Roll a number of extra dice along with the base roll

I will start with an example for 0.75 and then explain how this can be more generalized. First, you have your base STAT roll with no modifiers which can be whatever you want. For this example, let's assume it has a 60% chance of success (such as a d20 that needs to roll at most a 12, it doesn't really matter here).

Now if you want to roll STAT-1, you still do the base STAT roll, but you also roll an extra 1d4. In order for STAT-1 to succeed, the base STAT roll must succeed and the 1d4 can't have a result of 1. This gives STAT-1 a chance of success equal to 0.6*0.75.

For STAT-x, you simply roll and extra xd4 and if any of these d4 rolls a 1 it is a failure, giving you the progression you want.

For STAT+1, you roll an extra 1d4 but this time, it is a success if the base STAT roll succeeds, or if the 1d4 rolls a 1 (or if both things happen). This causes the chance of failure to go from 0.4 to 0.4*0.75. Similarly for STAT+x, you roll x additional 1d4 and you get a success if any of them is a 1.

Using an extra d4 like this gives you a progression of 0.75 because that's the probability of not rolling a 1 on a d4. You can adjust this to other numbers by using different dice and different ranges of numbers. For example using a 1d3 would allow a 2/3 (0.66...) progression.

• anydice.com/program/1f5f3 this works perfectly. I am wondering if there's any way to generalise it both ways, though something tells me no. Dec 16 '20 at 17:20
• @MiguelBartelsman What do you mean by generalizing both ways?
– Sdjz
Dec 16 '20 at 17:54
• As it is right now, the calculations someone has to do when a modifier is negative are different from the ones when a modifier is positive. This is just a very minor issue and it's probably not even possible to achieve it otherwise. Dec 16 '20 at 18:02
• @MiguelBartelsman Ah I understand, I'm not seeing such a way at first glance but it might be possible
– Sdjz
Dec 16 '20 at 18:27

# Roll more dice

You should roll additional XdN where X ist your modifier and $$\1\pm\frac{1}{N}\$$ will be your multiplier.

If it is a negative modifier, any 1 rolled on a dN will make the check fail, no matter the result of your STAT roll. This results in a probability of $$\text{STAT}\times\left(1-\frac{1}{N}\right)^X$$

If it is a positive modifier, you may reroll your STAT roll for every N rolled on a dN. This results in a propability of $$\text{STAT}\times\left(1+\frac{1}{N}\right)^X$$

To modify your propability just choose a die. On a d8 this produces a 12,5% in-/decrease, on a d6 you get a 16,67% in-/decrease.

You are asking for a (discrete version of the) Laplace distribution. I know of two reasonably efficient ways of approximating this:

## Opposed exploding dice

Each side rolls an exploding die (i.e. on a max roll, roll the die again and add it to the total) and adds their modifier. Higher total wins. Here's an example on AnyDice using exploding d10s.

Compared to an ideal Laplace distribution:

## Opposed step dice

In this type of system, instead of adding a modifier to their roll, the two sides use larger and larger dice as they advance. A geometric series of die sizes is equivalent to a Laplace distribution. Here's an example near-geometric series:

d3, d4, d5, d6, d8, d10, d12, d16, d20, d24, d30, d40, d50, d60...

### Proof

Suppose two players roll step dice against each other, with respective sizes $$\A\$$ and $$\B\$$, with ties being broken by coin flip. Without loss of generality, let the first player be the underdog, i.e. $$\A \leq B\$$.

We can visualize the chance of the underdog $$\A\$$ winning according to the area on a 2-D diagram, with possible rolls for $$\B\$$ on the horizontal axis and for $$\A\$$ on the vertical.

The total area is $$\AB\$$, out of which the area where $$\A\$$ wins is $$\\frac{1}{2} A^2\$$. So the probability of $$\A\$$ winning is

$$p_a = \frac{1}{2} \frac{A}{B}$$

We can transform this into an equivalent fixed-die system by taking logarithms:

$$a = h \log_2 A \\ b = h \log_2 B \\ x = b - a \\ p_a = \frac{1}{2} 2^{-x / h}, x \geq 0$$

This is a Laplace distribution where $$\h\$$ is a half-life of our choosing. The above step dice sequence has $$\h \approx 3\$$, meaning the underdog's chances are cut in half about every three die steps they are behind.