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I've thought of a dice mechanic that I'd like to test the numbers for, but it's complicated enough that I don't know how to model it in AnyDice. It's sort of a modified dice pool, where the successes are counted differently than most.

With a skill level of 3, the player rolls 3d6. 2, 3, 4, and 5 all count as one success, 6 counts as 2 successes, and 1 subtracts a success. So a roll of 2, 4, and 5 would be three successes, while a roll of 1, 3, and 6 would be two successes.

I know how to set up AnyDice for a standard dice pool with the counting function, but getting it to subtract ones is a bit beyond me. Is this possible to model in AnyDice? Thanks so much for your time!

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  • \$\begingroup\$ I really like that mechanic. It's a bit like a mix of Exalted 2e and oWoD - 10's (on d10) count twice, while 1's take away a success - but the the "difficulty" set to the minimum. \$\endgroup\$
    – gatherer818
    May 18, 2015 at 7:09
  • \$\begingroup\$ Just so you're aware, the average result will always be slightly lower than the number of dice. \$\endgroup\$
    – Mooing Duck
    May 18, 2015 at 19:25

2 Answers 2

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output 3d{-1,1,1,1,1,2} named "fiveSixths"

Your mechanic returns 1 success exactly 5/6ths of the time. But it's a lumpy bi-modal distribution on 1 and 3.5 total successes. Make sure to provide free aspirin to players, as calculating results will be entailed.

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    \$\begingroup\$ Custom dice are such an elegant solution to this kind of problem - and yet I always forget them. My half-written function will never become a full answer now, +1. \$\endgroup\$
    – Miniman
    May 18, 2015 at 7:07
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    \$\begingroup\$ Yeah. I had functions all typed up, and then I realised I could do it in a single line. So I did. \$\endgroup\$
    – Brian Ballsun-Stanton
    May 18, 2015 at 7:08
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    \$\begingroup\$ The problem with this distribution comes from the -1, which is is off by 2 from the next result, while the result for 6 is only off by one. Overall, I would not use this for this reason. If you adjust the value of 1 to just 0, it's symmetrical again (then you basically use a variation of Fudge dice). \$\endgroup\$
    – MrLemon
    May 18, 2015 at 8:15
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    \$\begingroup\$ If you want to do this for real-life dice, get some blank d6s and just draw on them. \$\endgroup\$
    – Gustav Bertram
    May 18, 2015 at 10:46
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    \$\begingroup\$ I'll probably test it both ways, but the point is for the average result to be common. It's based on the design philosophy of the Grimm RPG, where characters perform at their level of skill most of the time. Kind of a mashup of FATE and Grimm, actually. EDIT I see you deleted the comment that I was originally replying to. What I'll probably do is buy some regular dice with pips and just color the 1 side red and the 6 side blue. Easily recognizable and I can still use the dice for other things. \$\endgroup\$
    – Witch's Knight
    May 18, 2015 at 10:46
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Improving on Brian's answer; if you are going to compare multiple rolls, you can assign your "special die" to a variable and reuse it over and over again. Also, there are four 1 results in there so there's a shortcut for writing that as well.

W: {-1, 1:4, 2}

output 2dW named "Skill 2"
output 3dW named "Skill 3"
output 4dW named "Skill 4"

And that's one weird probability distribution! If you want to even it out, I'd recommend counting 2 and 3 as zero successes.

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