How can I calculate my chance of success given my ability + skill vs a difficulty? How is this complicated when the difficulty is rolled as well?

# Vampire: The Masquerade – How to Calculate Probability of Success

statisticsvampire-the-masqueradeworld-of-darkness

#### Related Solutions

## Average The Skills

If he has to use two skills, average the two skills together and then make one roll. In this case, that'd be a single roll to get 50 or below, since he has 50 in both skills (so the average is 50).

If he was better at one skill than another, it'd look slightly different. Say he has a 50 in Stonecarving and 25 in Artistry. That makes the average of them 37.5, so he'd have to get a 37 or below (or a 38, depending on how you want to round).

That basically treats it like he's using both skills and has to succeed on using them in combination, rather than having to succeed on separate rolls for both. It also keeps it to a single roll with similar odds, and is relatively simple to implement for players.

## Alternative - Geometric Mean

The downside to averages is that if you're really good at one skill (say 100 in Stonecarving) and really bad at the other (0 in Artistry), you still have a 50 in the combined skill. That might not be what you had in mind, as someone with no artistic talent doesn't suddenly gain it just because they are working with stone.

In this case, an alternative method is to take the Geometric Mean. For two skills, that is this formula:

$$\sqrt{skill_1 \cdot skill_2}$$

So, if you have 100 in Stonecarving and 0 in Artistry, you do \$100 \cdot 0\$, which is 0. The square root of that is 0. As a result, you now need to at least have 1 skill point in Artistry in order to attempt the combined result. If you did have Artistry 1, you'd get \$100 \cdot 1 = 100\$, the square root of which is 10. As you add points in Artistry, your chances will quickly increase.

For my previous example of 50 and 25, you'd get \$50 \cdot 25 = 1250\$, the square root of which is 35.3.

The main downside to this method is that in a tabletop game, it's extremely hard to calculate without a calculator. Even with one, it requires a more complicated understanding of math and is more time consuming. If you put this in a rule book, there will be people who won't understand what you want them to do. For something like a video game where it's calculated by the software, that isn't a problem.

(Thanks to Peteris and Vatine for the suggestion!)

## Alternative - Minimum/Maximum

A very simple method for combining skills is to use either the minimum skill in the two of them, or the maximum skill in the two of them. The maximum means you're just using the skill you're better at, while the minimum means you're using the skill you're worse at.

In the case of the minimum, it simulates the idea that you have to succeed on what you're weaker at in order to accomplish the goal. This lets you do it in a single roll, and is very easy to understand. It also has some issues, in that if you're extremely good at Stonecarving and so so at Artistry, your Stonecarving gets ignored in this system as you only roll on your lower one (Artistry).

Because of that, I don't think it really accomplishes what you intend very well, but it's ease of use is a significant upside over the other suggestions.

(Thanks to Neil Slater and Ellesedil for suggesting.)

**You want a table that models "stacked" probabilities.**

For example, if the chance of any individual attempt's success is 40%, your table might look like this:

\begin{array}{cc} n\text{ to success} & \text{d}100\text{ roll}\\ \hline 1 & 01-40\\ 2 & 41-64\\ 3 & 65-78\\ 4 & 79-87\\ 5 & 88-92\\ 6 & 93-95\\ 7 & 96-97\\ 8 & 98\\ 9 & 99\\ 10 & 100\\ \end{array}

But how do you generate this table?

Given the probability \$p\$ of succeeding on any given attempt, then the probability of succeeding

**the \$n^\text{th}\$ attempt is given by $$P(n)=(1-p)^{n-1}\cdot p$$**

*on*This is because in order to succeed on the \$n^\text{th}\$ attempt we must first *fail*, with a probability \$(1-p)\$, \$n-1\$ times, *then* succeed (with probability \$p\$).

Thus in the example above we see that \begin{align*} P(1)&=p=0.4\\ P(2)&=(1-p)\cdot p =0.24\\ P(3)&=(1-p)^2\cdot p \approx0.14\\ P(4)&=(1-p)^3\cdot p \approx 0.09\\ \vdots\quad& \hspace{2cm}\vdots \end{align*}

But in order to *stack* the probabilities we recognize that the "break points"--the highest number in each of the percentile ranges--are given by the *sum* of all probabilities up to the \$n^\text{th}\$.
Luckily, this is just a geometric series: $$\sum_{i=0}^{n-1}{(1-p)^i} = \frac{1-(1-p)^n}{p}$$

With this in hand it's easy to generate a table of "break points" for any given \$p\$:

\begin{array}{c|ccccccc} & \text{d}100\text{ roll}\\ n\text{ to success} & p=0.1 & p=0.2 & p=0.3 & p=0.4 & p=0.5 &p=0.6 &p=0.7\\ \hline 1 & 01-10 & 01-20 & 01-30 & 01-40 & 01-50 & 01-60 & 01-70\\ 2 & 11-19 & 21-36 & 31-51 & 41-64 & 51-75 & 61-84 & 71-91\\ 3 & 20-27 & 37-49 & 52-66 & 65-78 & 76-88 & 85-94 & 92-97\\ 4 & 28-34 & 50-59 & 67-76 & 79-87 & 89-94 & 95-97 & 98-99\\ \hline 5 & 35-41 & 60-67 & 77-83 & 88-92 & 95-97 & 98-99 & 100\\ 6 & 42-47 & 68-74 & 84-88 & 93-95 & 98 & 100\\ 7 & 48-52 & 75-79 & 89-92 & 97-97 & 99\\ 8 & 53-57 & 80-83 & 93-94 & 98 & 100\\\hline 9 & 58-61 & 84-87 & 95-96 & 99 & \\ 10 & 62-65 & 88-89 & 97 & 99 & \\ 11 & 66-69 & 90-91 & 98 & 100 & \\ 12 & 70-72 & 92-93 & 99 & & \\\hline 13 & 73-75 & 94-95 & 99 & & \\ 14 & 76-77 & 96 & 99 & & \\ 15 & 78-79 & 96 & 100 & & \\ 16 & 80-81 & 97 & & & \\ \vdots & \vdots & \vdots\\ \end{array}

**Notes:**

^{1. In a few places the same single number appears twice. (p=0.2, n=14,15, for instance.) In these cases multiple break-points round to the same percent-value. You could randomly choose from among n values if this is rolled, simply take the lower, or devise some other scheme.}

^{2. Obviously, there is a non-zero probability that success would take longer than I've indicated, ending my tables at the first appearance of a rounded-to-100 percent-value. However, by construction there is less than a 1/2% chance that any n greater than the last presented could occur. I felt fine leaving the long tail off the table.}

## Best Answer

I would recommend using anydice.com. The creator of anydice.com has written a program on that site to calculate success probabilities for NWoD, but the dice mechanics are very similar and shouldn't need much tweaking. Here is the post. It should give you a rough idea.

I tweeted the creater of Anydice.com and he suggested this:

Which gives this sort of graph: