Sorry for asking a "do the work for me" type question, but I'm looking at the AnyDice documentation and appear to be missing something. I'm hoping to not just get the question answered, but learn how to do it in the future so I won't need help with AnyDice for this system again in the future. There was one other question regarding success-based probabilities, but I didn't seem to glean any insight from it, it was pre-written programs that didn't appear to show the actual dice-rolling code.

I am playing a Resonance 7, Willpower 5, Registering 6 Technomancer with 11 boxes of Physical Condition. I'm trying to figure out how likely I am to succeed if I try to a register a Force 7 sprite, and how likely I am to take Physical damage in the attempt.

The math (so non-Shadowrun players can help, since I'm likely to use AnyDice for World of Darkness and Exalted as well) will work like so:

Success: I will roll 13 d6 against the sprite's 14 d6, and need to get more 5's and 6's (hits) than it does.

Damage: Each hit the sprite rolls in the above test adds 2 DV to the damage I will take. I reduce the DV by 1 for each hit on a 12 d6 roll before applying the rest to my health. With 11 Stun health, I would need to take 13 DV before it started rolling into Physical.

How do I model this in AnyDice?


1 Answer 1


In this particular case, Anydice's count function is the right choice. So to count the number of hits you'll get (on average), you get:

 output [count {5, 6} in 13d6]

And for the sprite:

 output [count {5, 6} in 14d6]

To compare how often you'll get more hits than it does use the formula:

 output  [count {5, 6} in 13d6] > [count {5, 6} in 14d6]

Basically you should be able to use these three formulas to figure out (on average) your final health.

To count the number of hits you'll take you could try:

 output  [count {5, 6} in 14d6] - [count {5, 6} in 13d6]

This will give you the average difference in the number of hits.

  • \$\begingroup\$ This is really helpful. So then, for the damage, it'll be like..... "output [count {5, 6} in 14d6] * 2 - [count {5, 6} in 12d6]". And the third formula works perfectly for that probability - I'll probably need another die or two before I break even on that exchange. I only seem to take Physical 4% of the time, though, so I'm good. \$\endgroup\$ Jul 15, 2014 at 21:32
  • \$\begingroup\$ @DavidL That seems right \$\endgroup\$
    – wax eagle
    Jul 15, 2014 at 23:01

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