Request: Success Level / WFRP Dice

[TheJadeKnight]

Would it be possible to include code to get success levels à la WFRP?

This would be useful for a number of games: WFRP 2nd & 4th editions (and some implementations of 1st) WH40k RPG (all versions) Maybe Age of Sigmar RPG

Basically, you roll a d100, compare it to a target number, and in most of these, assign 1 success level per every 10 full points it's beaten by (i.e., the floor of the delta/10).

Might structure it something like this: If d100<=(N) then (S) = floor(((N)-d100)/10) If d100>(N) then (F) = floor((d100-(N))/10)

Where (S) = degrees of success and (F) = degress of failure.

One last note: For WFRP 4th edition, it works slightly differently—rather than subtracting the results directly, you only compare the 10's. So for 4th, it'd be like this: If d100<=(N) then (S) = floor(N/10)-floor(d100/10) If d100>(N) then (F) = floor(d100/10)-floor(N/10)

You can still have a success of +0 or failure of -0.

If you really want to go the extra mile, you could add the descriptors for WFRP 4th: +6 or more = Astounding Success +4 or +5 = Impressive Success +2 to +3 = Success +0 to +1 = Marginal Success -1 to -0 = Marginal Failure -2 to -3 =Failure -4 to -5 = Impressive Failure -6 or less = Astounding Failure

Also, for WFRP 4th, doubles = critical (success) or fumble (failure). Not important that the descriptors or critical be explicitly mentioned, but just throwing this out there.

[claycle]

I second this.

[mstankey]

Is there any progress on this?