Recently a friend commented that he thought the dice pool in the d10 Storytelling System actually gave mediocre results, and that by increasing the number of dice thrown you were actually decreasing your chances of success.

I'm actually certain of the opposite, that by having more dice, the probability of getting 8-9-10 in one of them grows. However, I've carefully forgotten my Statistics lessons from college, and would like to know more. Can anyone put together an explanation of the statistics of the system, and maybe a graph?

  • 4
    \$\begingroup\$ This question probably has what you want. \$\endgroup\$
    – Nigralbus
    Commented Mar 12, 2012 at 16:54

4 Answers 4


If I understand correctly, you need at least one d10 to get 8 or more to succeed, yes? And you get more dice if you have more ranks/skill level/whatever, right?

In that case, the other answers are correct, but I'd also like to have a go at explaining this:

Assuming a uniform distribution (meaning, all results are equally likely), reasoning with dice is pretty simple: To calculate the probability of an event, you just count the number of possible results that make the event true and divide it by the number of possible results. So, in our case the possible number of results is 10. The event of success has 3 results (8, 9, 10) and the event of failure has 7 results (1, 2, 3, 4, 5, 6, 7). So the probability of succeeding with one die is \$\frac{3}{10} = 0.3\$ and the probability of failing is \$\frac{7}{10} = 0.7\$.

To calculate the probability of independent events (like rolling normal dice which can't interact with each other), you just multiply the probability of the event occurring once with itself as many times as the event repeats. To extend the previous example, the probability of failing once is 0.7, the probability of failing twice is \$ 0.7 \times 0.7 = 0.7^2 \$, the probability of failing thrice is \$ 0.7 \times 0.7 \times 0.7 = 0.7^3 \$ and so on.

Now, if two events cover the whole spectrum of possibilities, then the sum of their probabilities is 1. For example, the events "failing three times with three rolls" and "getting at least one success with three rolls" are complementary. So, the probability of getting at least one success plus the probability of getting three failures is 1. Solving this trivial equation, you get that the probability of succeeding at least once with three rolls is \$1-0.7^3\$.

Here's a chart that does this same work for any number of dice from 1 to 10. 0.3 0.51 0.657 0.7599 0.8319 0.8824 0.9176 0.9423 0.9596 0.9718

To make this a bit more general, the probability of getting less than X with one Y-sided die is is \$\frac{X-1}{Y}\$. So, the probability of getting less than \$X\$ with \$N~Y\$-sided dice is \$\left(\frac{X-1}{Y}\right)^N\$. The probability of getting at least one \$X\$ or higher after rolling \$N~Y\$-sided dice is \$1 - \left(\frac{X-1}{Y}\right)^N\$. It should all make sense now, hopefully.


What your friend may be referencing is a quirk of the original World of Darkness rules, in which each one rolled cancelled a success, and TNs were the number needed for a die to count as a success. If the TN was very high (10 or greater) the more dice you rolled, the more likely you would be to roll more ones than successes, and therefore the higher the chance of failing the roll.

This is no longer the case with newer Chronicles of Darkness rules, in which the number needed for a die to be a success is constant.

As for a simple chart of the probabilities:


The probability of success definitely grows as you add more dice.

  • \$\begingroup\$ @AceCalhoon I'm not so sure about him confusing the two issues, he never mentioned how successes were cancelled in the conversation. \$\endgroup\$ Commented Mar 12, 2012 at 23:29
  • \$\begingroup\$ What I believe he assumed is that adding a die meant requiring one more success (NOT the same thing, of course). \$\endgroup\$
    – Nigralbus
    Commented Mar 13, 2012 at 11:08

I've already commented but here's a quick writeup of the probabilities in case you don't find the other answer to your taste.

Assuming a perfectly random die, yadda yadda...

The probability of rolling a 8 or more on a d10 is p(8,9,10 | 1-10) = 3/10

Adding dice, it actually becomes easier to determine by calculating the probabilities for 0 successes (which requires to calculate chances of 1-7 showing up on ALL dice, instead of trying to find the chances of having 8-10 on one OR two OR three... OR N dice which is just not practical).

So our formula becomes p(1-7 | 1-10) = 7/10 raised to the number of dice thrown.

Thus, with 2 dice you have 49/100 of no success => 51% chance of having at least one 8,9 or 10 on the table. With 3 dice it becomes 343/1000 chance of 0 successes => a 65.7% chance of at least one success, etc.

Without going all math-y and starting to speak of geometric sequences and other nightmare-inducing terms I've forgotten, let's just say that the more dice you add, the further down the probabilty of 0 success goes => higher are your chances of 1+ success.

Of course, some rolls will require more than one success. For this, I'll refer you to the aforementioned other question.


I had a similar question and chose to write a script that simulates the dicepools to make sure I was handling the 10-again rerolls properly. I also wanted to know the number of successes I could expect, so rather than simply compute the probability of a single success, I drew a chart that shows the distribution of how many successes you typically will get for a given dice pool.

As the image below shows, we have some common cases: Single die: Expect to fail, but there is ~25% chance of succeeding Three dice: Expect to get one success, and have an ~25% of getting two successes Five dice: Expect to get 2 successes. Only ~25% chance of failure.

Distribution of successes for different dice pool sizes

  • \$\begingroup\$ Yes, but I cap each individual die at 20 rerolls. \$\endgroup\$ Commented Jun 21, 2023 at 14:33

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