Test Difficulty Formula?

Is there an underlying formula for test difficulty?

Challenging is easy to remember — it’s one higher than the number of non-Artha dice. That means you need to spend Artha to pass a challenging test.

Is there a way to determine the highest routine or lowest difficult Obstacle without consulting the chart? I’ve tried to make sense of it, but I am vexed.

The simplest formulation I have is this:

The lowest difficult rating is:

<3 dice: #dice
<6 dice: #dice - 1
7+ dice: #dice - 2

No big deal if a simple formula doesn’t exist, but in that case it might be helpful to understand why. Does the given table aim to hit some kind of sweet spot for advancement? What would happen if I chucked it and used a simple algorithm instead, just to make it less intimidating to my player converts?

But… the advance table is on the char sheet, no need of a formula.

It’s not necessary, but it could be useful. Especially when automating certain parts of the process with computers, as is my wont.

But also, I’d really love to know why the table is the way it is. I trust that there is an interesting reason.

It’s essentially #D/3, rounded down, I believe. But you’re unlikely to ever have to deal with anything above a 7 at the start, except for maybe Steel.

EDIT: Totally wrong about the numbers, but in any case you should be able to internalize it easily enough in play - you’ll usually be playing with the same ranges of dice v obstacle.

Not on the char sheet I’m using…


Right, the issue is the rounding. If you were to graph it, I think you’d find that a straight line fits, so the equation is very simple: lowest Difficult Ob = k*#dice, for some k. But k is like five-sevenths or something. So now you have two options.

  1. Multiply the number of dice rolled by five sevenths (or whatever) and round off. Compare to Ob of test.
  2. Look at the chart.

As to why it is this way… When you’re at exponent 3, it’s pretty easy to get Ob 3 tests. Ob 2 is notably easier, and should be a Routine. But at exponent 7, it’s actually pretty hard to get tests at exactly Ob 7, and the reduction in difficulty from testing against Ob 6 is much smaller than was the case with B3 against Ob 3/Ob 2. So you need a slightly wider Difficult band, or else it’s a bitch to get tests at exactly the Ob you need. (I’m not explaining this very well, but… Just imagine if Difficult was always #dice, or always #dice-2. Think about the effects in each case for someone with a B3 skill, and for someone with a B7. You’ll see what I mean.)

Here’s a simplification that I quite like:

Challenging: Dice < Ob
Difficult: Dice <= Ob x 1.5
Routine otherwise

(1D/Ob 1 remains either Routine/Difficult at tester’s option)

This causes only 3 changes for Obs 1-6 and 1-9D:
[li]Ob 2 is now Difficult for 3D[/li][li]Ob 4 is now Difficult for 6D[/li][li]Ob 6 is now Difficult for 9D[/li][/ul]My guess is that the first of these is the biggest issue, simply because 3D/Ob 2 is so common. 3LP campaigns in particular, but every new skill will climb its way through B2/B3 areas where 3D is common.

At low ability exponents, it can be hard to get Routines, and now Ob 1 is the only Routine test for anything less than 4D. This affects small parties the most, as they have fewer helping dice available. For larger parties (who can hit 4D no problem), it might have the opposite effect, making Difficults really easy to come by.

The other two changes don’t strike me as consequential.

The only other changes are for when Ob is 7+ and when 10D+ are used. This happens in my games very infrequently, so I can’t really comment.

EDIT: The bit about BL perception tests was incorrect.

If what you’re looking for is a simple mathematical formula, something like this might work:

Diff. Ob = nearest integer(0.7*Dice + 0.5)

I’ve spent a fair amount of time looking for just such a formula to determine whether a test is routine or difficult.
I’ve noticed the following pattern:
1 Die: only 1 obstacle is classified as difficult (special case here)
2 Dice: only 1 difficult ob
3,4,5,6 Dice: 2 obstacles are difficult
7,8,9,10,11,12,13,14 Dice: 3 obstacles are difficult

It’s an expansion in powers of 2. The first two dice only have 1 difficult obstacle. The next four have 2 difficult obstacles. The next 8 have 3 difficult obstacles.
The pattern I see is that the next 16 should have 4 difficult obstacles.

It won’t be trivial to program into a computer, but there’s definitely a pattern that you can recognize.

If the expected value of the test is greater than or equal to the obstacle (round up) on Black shade dice, the test is routine.

If the test’s obstacle is greater than the number of dice, it is challenging.

Otherwise it is difficult.

Silverwizard, huh? Are you putting forth an alternative system? That’s clearly not how it works in the rules. It breaks down at 6D vs Ob 4 (routine in the rules, difficult in your system) and then again whenever you have a die pool greater than 7.

Dan2, I think you’ve got a mistake - for 3D only Ob3 is difficult. .

And Zabieru is right (I think) as to why the chart is the way it is. A 3 obstacle range gives you enough targets to shoot for so you don’t get frustrated trying to advance, while still making it painful in that you’ll have to be failing to advance. You don’t need to have 4 difficult obstacles for when you’re rolling 16 dice - if you’re rolling 16 dice you’re almost certainly looking at a routine test anyway. If the Obstacle is 13 and you want to get a difficult test, just refuse one of the helping dice or don’t use one of the ForKs or whatever is getting you to 16 dice. Roll 15 dice against Ob 13 and enjoy your difficult test. If you need that extra die (why? You’ll just fail anyway…) then spend a persona point.

Look, the simplest formula for remembering it at the table is the one listed in the first post of the thread. And as far as automation goes, that rule is really simple to program in any language.

Crap, you’re right. I never read the table closely enough apparently.

How about that! I don’t know what I got my idea from… Time to revise my idea. :smiley:

Thanks for the input, everyone.

I’m actually pretty satisfied with the approximation I put forth in the early posts, by the way. It obviates the need to check the chart, anyway. Still interested to know the “true” formula, but more importantly does anyone have any insight as to why it is done this way?

I feel there may be a reason, that has something to do with the ease of earning difficult tests as you go up.

Thanks for the responses!