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Probabilities in the World of Darkness

In the World of Darkness roleplaying games a player resolves an action by rolling a number of ten-sided dice. All dice which roll greater than or equal to a given difficulty value are counted as successes, and these are individually cancelled out by dice which roll ones.

If the resulting total is positive, then the action succeeds. If zero or negative, then the action fails. A special case of failure occurs when all dice are lower than the difficulty value and at least one die is a 1; when this happens, the action is said to have been botched.

What are the probabilities of a success, a failure, and a botch?

Consider a single roll of an $s$-sided die when the difficulty value is $d$. We are concerned with three outcomes:

  1. the roll is 1
  2. the roll is greater than 1 but less than $d$
  3. the roll is at least $d$.
These occur with the following probabilities: \begin{align*} p_1 &= \frac{1}{s} \\ p_2 &= \frac{d-2}{s} \\ p_3 &= \frac{s-d+1}{s}. \end{align*}

For $n$ rolls, let $N_i$ denote the number of times outcome $i$ occurs. Then the probability that $N_1$, $N_2$ and $N_3$ will have particular values $n_1$, $n_2$ and $n_3$ has a multinomial distribution, \begin{equation} P(N_1=n_1, N_2=n_2, N_3=n_3) = \frac{n!}{n_1!n_2!n_3!}\; p_1^{n_1} p_2^{n_2} p_3^{n_3}.\qquad(1) \end{equation}

An action succeeds if $N_3>0$ and $N_1<N_3$.

Let $r$ be the number of successes resulting from a roll of $n$ dice. If $N_1$ ones were rolled, then $N_3=N_1+r$. Since $N_1+N_2+N_3=n$, it follows that $N_2=n-2N_1-r$, and $N_1$ can vary from 0 to $(n-r)/2$.

Thus the probability of getting exactly $r$ successes is \begin{equation*} P(N_3-N_1 = r) = \sum_{i=0}^{(n-r)/2} P(N_1=i, N_2=n-2i-r, N_3=i+r), \end{equation*} which can be found by way of Equation (1). The probability of getting at least $r$ successes is therefore \begin{align*} P(N_3-N_1 \geq r) &= \sum_{j=r}^n P(N_3-N_1 = j) \\ &= \sum_{j=r}^n \sum_{i=0}^{(n-r)/2} P(N_1=i, N_2=n-2i-j, N_3=i+j). \end{align*}

An action fails if $N_3=0$ and $N_1=0$, or if $N_3>0$ and $N_1\geq N_3$.

In the first case, if $N_1=N_3=0$, then $N_2=n$. This contributes $P(N_1=0,N_2=n,N_3=0)$ to the total probability.

In the second case, let $j$ be the difference between the number of ones and the number of potential successes. If $N_3$ dice rolled at least $d$, then $N_1=N_3+j$. Since $N_1+N_2+N_3=n$, it follows that $N_2=n-2N_3-j$, and, while $j$ varies from $0$ to $n-1$, $N_3$ can vary from $1$ to $(n-j)/2$.

The probability of failure is therefore \begin{align*} & P(N_1=0, N_2=n, N_3=0) \\ +& \sum_{j=0}^{n-1} \sum_{i=1}^{(n-j)/2} P(N_1=i+j, N_2=n-2i-j, N_3=i). \end{align*}

An action is botched if $N_3=0$ and $N_1>0$.

Here $N_2=n-N_1$, and $N_1$ can vary from $1$ to $n$. The probability of a botch is therefore \begin{equation*} \sum_{i=1}^n P(N_1=i, N_2=n-i, N_3=0). \end{equation*}

Click here for tabulated probabilities.