#dice math

Okay, so yes, it is very possible. Unfortunately, there is no "easy way" to do it. I don't know what formula you found, but I'm pretty sure the simplest would be.

1d6 + [(1d6 − 1] × 6) + [(1d6 − 1] × 36)

Which basically gives you three dice with the following values

Die 1: 1 2 3 4 5 6 Die 2: 0 6 12 18 24 30 Die 3: 0 36 72 108 144 180

Which provides a range of 1 [ 1 + 0 + 0 ] to 216 [ 6 + 30 + 180 ]

If you understand that you can stop reading now, the rest is just explanation of why that works.

To make it more clear what's going on we can also subtract 1 from the first d6, giving the following range

Die 1: 0 1 2 3 4 5 Die 2: 0 6 12 18 24 30 Die 3: 0 36 72 108 144 180

The range provided is now 0–215, and also essentially a d100 in base-6 read in base-10

if we look at the outcomes in order, D1 can return any number 0–5, then incrementing by 1 sets D1 back to 0 and D2 to 6. Then from there incrementing D1 gives you 6–11, at which point you once again set D1 back to 0 and increment D2 to 12. Similarly, when the D1 and D2 reach 35, incrementing by one sets them both back to 0 and increments D3 to 36. And if you're noticing that this sounds a lot like how binary works, yes, it is exactly the same except with 6 instead of 2—which makes it "senary." And then of course putting that 1 back onto the first die simply shifts the entire range by 1, from 0–215 to 1–216.

And basically, that is how to roll any value of X × Y, 1dX + [(1dY − 1] × X). and just for a little further example, let's look at a d18 rolled with 1d6 + [(1d3 − 1] × 6). Which gives the following values

Die 1: 1 2 3 4 5 6 Die 2: 0 6 12

Even though the dice are different, the math still works exactly the same.

what’s the benefit of a dice pool rpg system over just rolling a d100 or something?

Okay so, a single die (but also d% even though this is rolled as two dice) has a flat distribution. There is an equal chance of all results on the die. When you roll a d20 your odds of a 1 are the same as the odds of a 20. The d20 was intentionally used for attacks in D&D because of the swinginess of that probability.

When you use a dice pool, it creates a bell curve instead, broadening and stabilizing your results. This is why PbtA is built around the 7-9 mixed success result - this will happen with much more stable frequency than rolling 10+ on 1d20. Similarly, Exalted uses pool math that allows the player to estimate every 2 dice generating one success, around which all the base difficulties are structured. When you start fucking with things like ones taking away in classic World of Darkness, you destabilize the nice curve and get into bad swinginess again.

The smaller your single die, the tighter (andore difficult) the framework for where you can adjudicate failure. My butthole clenches when I see people using single d6 resolution and have failure on 3 or less.

I don't really think people understand the importance of mathematical stability underpinning your basic resolution, if you're going to do something with randomization at all.

When someone says "We'll just roll a d20" I ask, is it because you want that level of swinginess? Usually no. Usually the answer is because that's what that person is familiar with - which fuels some of my distaste for WotC's stranglehold on the ttrpg zeitgeist.

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Hello there. As you may know, I am a charming and attractive tabletop game designer, and I would like to share with you a little experience I recently went through, just as a look into the process of failed ideas and how they die sometimes.

I had an idea for an upcoming project (Dinosaur Erotica: The Role-Plying Game) Which involved each character having a die representing their level of investment in the developing relationship, and when that relationship was tested, both players would roll their die, with the ideal result being bother players rolling the same number.

Now my assumption here was that lowering the size of either die would raise the odds of rolling the same number—so a higher chance with a d10 and a d6 than with 2 d10s—which totally makes sense. However, then I crunched the numbers, and let me just show you what I found.

So one d6 and one d10, let's look at the results for that

1 | 1 2 3 4 5 6 7 8 9 10 2 | 1 2 3 4 5 6 7 8 9 10 3 | 1 2 3 4 5 6 7 8 9 10 4 | 1 2 3 4 5 6 7 8 9 10 5 | 1 2 3 4 5 6 7 8 9 10 6 | 1 2 3 4 5 6 7 8 9 10

For each result on the d6, there are 10 possible results on the d10, only one of which is the same as the d6. So out of 60 total results, there are 6 where both dice come up the same number, a 6/60 probability, or 1/10.

But now, let's see what happens with 2 d10.

1 | 1 2 3 4 5 6 7 8 9 10 2 | 1 2 3 4 5 6 7 8 9 10 3 | 1 2 3 4 5 6 7 8 9 10 4 | 1 2 3 4 5 6 7 8 9 10 5 | 1 2 3 4 5 6 7 8 9 10 6 | 1 2 3 4 5 6 7 8 9 10 7 | 1 2 3 4 5 6 7 8 9 10 8 | 1 2 3 4 5 6 7 8 9 10 9 | 1 2 3 4 5 6 7 8 9 10 10| 1 2 3 4 5 6 7 8 9 10

Person after person after person was just going, "Let's see, the probability of rolling a 6 is 1/6, so the odds of rolling two 6s is 1/36, and three 6's 1/216. So the odds of rolling three matching numbers on three dice is 1/216!"

Mind you, a few people did actually understand that a die has multiple numbers, but most people basically invented a sub-problem—the odds of rolling three of one specific number—solved that problem, and then just walked away, confident in their own intellectual superiority, without considering the original problem.

Actual explanation after the break.