This question asks about the probability of a success in a Role Playing Game. However, the question, and its answers do not cover some of the complexities of the dice mechanic. In particular, it does not cover botches (one possible outcome) at all.

A player has a dice pool, based on some mechanic in the game irrelevant to this question. A dice pool is a variable number of dice a player may roll. There are rules about how many dice the player gets to roll, but that is irrelevant to this question. It can be any number of dice from 1 (a single die) to about 15. I am calling this P.

The dice have 10 sides labeled 1 through 10 inclusive (Called a 'd10' in our domain terminology)

When rolling dice, there is a target number, or difficulty number. How this number is generated is outside the scope of this question, but the number can be between 3 and 9 inclusive. The rules around this are explained below. I am calling this T.

When all the dice are rolled, there are some rules to determine the outcome:

  • Any die equal to or greater than T is counted as a success
  • Any die equal to 1 subtracts from successes

Such that...

  • If, after subtraction (if applicable), there are no die left greater than or equal to T, then the result is a failure.
  • If, after subtraction (if applicable), there is at least one die left greater than or equal to T, then the result is a success.
  • If no die rolled are greater than or equal to T, and at least one die is 1, then it is a botch

For a given P pool and T target, how do you calculate the probability of a success, failure, or botch in this system?

  • $\begingroup$ Please clarify. Does the failure condition only apply if the roll is not also a botch? Or is it possible for a result to be both a botch and a failure? (I'm trying to see if all three probabilities must sum to 1, or only P(success) + P(failure) = 1 with a botch as a "side effect".) $\endgroup$
    – wberry
    Jul 25, 2016 at 22:32
  • $\begingroup$ A botch is a type of a failure, so the set of all botches is a subset of the set of all failures. Does that help? $\endgroup$
    – Tritium21
    Jul 26, 2016 at 3:33
  • $\begingroup$ I think so. So it seems like the probabilities of successes and failures sum to 1, and the probabilities of botches and non-botches sum to 1. $\endgroup$
    – wberry
    Jul 26, 2016 at 14:11

1 Answer 1


I'll have to tackle this in stages as time permits. I expect someone will give a complete (and probably simpler) approach before I finish.

First, let's look at botches.

I'm going to ignore some of your notation and call the number of dice $n$.

If no die rolled are greater than or equal to T, and at least one die is 1, then it is a botch

First consider $P(\text{no dice }\geq T) = (\frac{T-1}{10})^n$

Now consider $P(\text{no } 1| \text{no dice }\geq T) = (\frac{T-2}{T-1})^n$

So $P(\text{botch}) = [1- (\frac{T-2}{T-1})^n]\cdot (\frac{T-1}{10})^n$

$\hspace{2cm} = \frac{(T-1)^n-(T-2)^n}{10^n}$

(assuming I didn't make any errors)

Second, the distribution of the number of individual-die successes after subtraction can be tackled by the method at this post. However, you seem to be after $P(\text{at least one success in total})$ (i.e. the overall roll succeeds) which I think may be amenable to relatively more simple-minded approaches (though they may well involve more work in the end). I'll look at that next edit.

  • 2
    $\begingroup$ Take your time, I have been needing this for ten years, a few days are... statistically insignificant. $\endgroup$
    – Tritium21
    Sep 26, 2015 at 6:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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