After Reading Cruciel's Dice Riddle

[M. J. Young]

Cruciel's A Dice Mechanic Riddle got me thinking a bit, and although this lacks a few of the factors he wanted to include it struck me as an interesting mechanic that I can't remember having seen explored. I thought I'd toss it out for exploration.

I'm going to draw a few assumptions from his model. Some of these are not vital to the model so much as useful to the illustration.

• Attributes from 1 to 5 (variable A below)
• Skill ratings from 0 to 5 (variable S below)

Here's the technique.

Any task or challenge has a difficulty rating (variable D) not greater than twenty. For reasons which will become apparent, difficulty ratings below two are meaningless.

Faced with a task, the character uses A+2S against D. If A+2S=>D success is automatic, and no dice are rolled.

If D>A+2S, the player rolls d20 (or for a different feel, 2d10, but that would have greater handling time in this). If d20=>D, the character is successful. However, if d20=<A+2S, the player can reroll the dice, and may do so until d20>A+2S.

This effectively reduces the range of possible failure rolls as A+2S increases, proportionately increasing the percent chance of success. Handling time is up due to rerolls, but those groups who are willing to trade more math for fewer die rolls can easily substitute 8+d12, 10+d10, 12+d8, and 14+d6 when A+2S reaches 8, 10, 12, and 14, respectively, without changing the curve at all.

Is that clear? Anyone see it as useful?

--M. J. Young

[ffilz]

Interesting mechanic. Actually, a difficulty 2 is automatic success also since a 2+ on the d20 succeeds, and a 1 is re-rolled.

Basically what this mechanic does is fairly simply take any difficulty you don't automatically succeed at and evenly distribute the probability (for example, if A+2S=15, difficulty 16 is 100% success, 17 is 80%, 18 is 60%, 19 is 40%, and 20 is 20%, for A+2S=10, diffuculty 11 is 100%, 12 is 90%, 13 is 80%...20 is 10%). It also means no one ever has no chance of success with any difficulty (worst chance of success is 5.2% for A+2S=1 against a diffiuclty 20). It also provides accelerating benefits for better A+2S (for a 20 difficulty, percentages go 5.2%, 5.5%, 5.8%, 6.2%, 6.6%, 7.1%, 7.6%, 8.3%, 9%, 10%, 11.1%, 12.5%, 14.2%, 16.6%, 20%, for a 10 difficulty it goes 57.8%, 61.1%, 64.7%, 68.7%, 73.3%, 78.5%, 84.6%, 91.6%, 100%).

Fiddling around with the <= and >= changes things slightly, but doesn't change the basic flavor. I might be inclined to have the re-roll only on a roll < A+2S instead of <=.

Why not rate skills 0-10 though?

Frank

[ffilz]

Hmm, just did the computations for 2d10. Very different table. A 20 always remains hard, rising to at best a 6.6% chance of success with A+2S=15 (from a 1% chance of success for low A+2S). Difficulties >= 12 become universally harder. Difficulties <= 8 become universaly easier. Difficulties from 9-11 become eaiser or hader depending on A+2S:

A+2S 1: 9-11 are easier
A+2S 2: 9-10 are easier, 11 is the same
A+2S 3: 9-10 are easier, 11 is harder
A+2S 4: 9 is easier, 10-11 is harder
A+2S 5: 9 is the same, 10-11 is harder
A+2S 6+: 9-11 are harder

If desired, I could post the C code I used to make these analysises.

Frank

[M. J. Young]

I knew some of that already, but it's good to see it in print.

Why not rate skills 0-10 though?

Actually, because those were the numbers Cruciel was using in his riddle, and I had been thinking in those terms when I was toying with the idea. Changing it to 1-10 is something I considered, but since that's an easy enough change for anyone to make (and I did specify that the details were more there for purposes of illustration) I didn't bother with it.

Personally, yes, I would change it to 1-10.

--M. J. Young