VtM5 has many quirks that make their calculation of probabilities really hard. But ignoring hunger dice, how can I calculate the distribution of success of a given pool considering that each pair of 10's counts as two successes (a regular success is just d10>5).

So 10,1 is just one success, but 10,10 is four.


Turns out it is simpler than I thought, but I must use a function:

function: roll ROLL:s  {
 result: (ROLL > 5) + ((ROLL = 10)/2)*2

I just sum every success and for every pair of 10's I add 2 successes.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.