New World of Darkness

successes and rerolls

Lots of people like to know the odds for the New World of Darkness system. Put shortly, you roll a pool of d10s and count all 8s and higher as a success. Additionally, you reroll and add for each 10 you get. In some cases you can reroll on a 9 or even on an 8.

To get the odds in AnyDice, you need to use a custom function to get the desired explosion effect, like the following program does.

function: nwod R:n again N:n {
 if N >= R { result: 1 + [nwod R again d10] }
 result: N >= 8
}

NWOD: [nwod 10 again d10]

loop N over {1..10} {
 output [lowest of 10 and NdNWOD] named "[N]d"
}

That program will roll a die at most ten times, which is more than enough. It then shows the results for pools with up to ten such dice in them. Finally, it also limits the number of successes to ten at most, to get rid of the long tails of highly unlikely outcomes.

Here are the at-least graphs for the three variants.

NWoD 10 again

NWoD 9 again

Note that the odds for at least one success are the same as the 10 again option. Only the odds for more than one success get a boost.

NWoD 8 again

• Mon, 05 Sep 2011 07:30:36 +0000 Nargosiprenk

  Hey, ¿how do you do to see the individual rolls average? For example, if I roll 1d20, I have the same probabilities of roll a 3 than a 16. But, if I roll 3d20, ¿how are the probabilities of rolling one 12 at least, one 15 at least, etc?

  • Mon, 05 Sep 2011 07:31:00 +0000 Nargosiprenk

    Sorry the Off Topic

    • Tue, 06 Sep 2011 20:57:03 +0000 Jasper

      No problem!

  • Tue, 06 Sep 2011 21:00:41 +0000 Jasper

    You can use the normal option to get odds per individual value and use the table to get more precise values.

    To count how many 12s you get with 3d20, you can use [count 12 in 3d20] or 3d(d20 = 12).

    If you want to know how many dice out of 3d20 rolled at least a 12, you can use 3d(d20 >= 12).

• Thu, 08 Sep 2011 12:30:46 +0000 nikink

  Hiya, how would I view the result of nwod dice rolls with 7 again, or 6 again etc? I tried subbing in "7" instead of "10" within "NWOD: [nwod 10 again d10]" but the system seemed to stall. Is it because the function is set to look for 8+ as a success, thus loops through to 7 then gets stuck never getting to 8?

  • Sat, 10 Sep 2011 13:25:06 +0000 Jasper

    It stalls because the computation becomes too heavy and times out. I wrote that NWOD program with clarity and not efficiency in mind. An easy way to get it to work would be too add

    set "maximum function depth" to 5

    at the top of the program, or use some lower number (the default is currently 10). This limits the recursion, sacrificing precision for speed.

    However, the program assumed that if you reroll you always have a success. This is correct from 10 again down to 8 again, but below that I'm not sure how you'd like to interpret that.

• Sun, 11 Sep 2011 04:49:29 +0000 nikink

  Thanks. :) I wasn't assuming 7 to be a success, just a cause to roll again and perhaps get 8+ as normal. I have been tempted to allow the nWoD rules to do this, but have never been sure how badly or not the success curve would then be thrown out. And continuing down the line to 2+ re-roll (which I expect would be really dumb, might as well give auto-successes for 5+ I think.)

  • Sun, 11 Sep 2011 10:41:26 +0000 Jasper

    Rerolling on 7 when it's not a success means that a result of 7 doesn't exist, thereby increasing the odds of success. One way to get this is by replacing all d10 with d{1..6, 8..10}. Or with d{1..5, 8..10} if you throw out both 6s and 7s, effectively rolling d8s.

• Wed, 14 Sep 2011 08:04:26 +0000 Anonymous

  Ah! Of course it does! You're good at this! :) Thanks :)