11
\$\begingroup\$

I am trying to figure out the probability for one separate d10 to have a matching face with a pool of d10 comprised by up to 5 other d10s.

So just to make myself clear, the player rolls 5 d10s and a separate "proc" d10. What are the chances that the "proc" die matches with any of the other d10s?

I would like to know this probability for all 5 combinations (1d10, 2d10, 3d10, 4d10 and 5d10).

Also, after calculating the probability, if the "proc" die gets rolled twice, does it simply double?

\$\endgroup\$
  • 8
    \$\begingroup\$ I'm voting to close this question as off-topic because it belongs in math.se \$\endgroup\$ – goodguy5 May 11 '18 at 21:15
  • 4
    \$\begingroup\$ To be honest, we have questions about the probability on dice mechanics in the context of roleplaying games regularly; I assume you're designing a system, so if you give context on why this result is meaningful for a roleplaying game, it may magically become rpg.se-on-topic again. \$\endgroup\$ – Carcer May 11 '18 at 21:44
  • 4
    \$\begingroup\$ @SeraphsWrath It doesn't need to be edited in, it just needs to reopened. We just don't have a consensus on what tool questions are okay and which aren't. Editing might help the voting process here, but it also would make the question ugly and muddled so I think it's better to hold off on that until it's clear the anti-tool-questions subcommunity is gonna make a fuss on this one. \$\endgroup\$ – Please stop being evil May 11 '18 at 22:12
  • 5
    \$\begingroup\$ @Axoren please don’t answer in comments. \$\endgroup\$ – Purple Monkey May 11 '18 at 22:56
  • 8
    \$\begingroup\$ @Azoren If a question is closed, that means we do not want answers posted. That is the entire point of closing the question so that no more answers can be posted—so no answers get posted. Subverting that, particularly by abusing comments, which should never be answers, is just doubly wrong. Please don’t. \$\endgroup\$ – KRyan May 11 '18 at 23:05
10
\$\begingroup\$

Turn the problem around: it’s equivalent to ask if I roll 1 die, what is my chance to match it when rolling \$n\$ dice?

The answer is:

$$p = 1 - \left(9\over 10\right)^n$$

Rolling 2 dice first doesn’t double the chance - the 2 dice may get the same number and even if they are different the odds of matching 2 numbers is different:

$$p = \left({1 - \left(9\over 10\right)^n}\right){1\over 10} + \left(1 - \left(8\over 10\right)^n\right) {9\over 10} $$

| improve this answer | |
\$\endgroup\$
6
\$\begingroup\$

This can be modeled as a Binomial Distribution with \$N = 5\$ and \$ p = \frac 1 {10}\$.

We are trying \$N\$ times to match a die from the dice pool with the proc die. The chance of doing this for each die in the pool is \$p\$.

We want to calculate the probability of at least one of those \$N\$ trials being a success (the proc die being matched by a number in the dice pool).

The probability function associated with Binomial Distributions gives a probability for a number of successes.

$$\text{P}(\text{# of successes}) = \% \text{ chance of getting that many successes}$$

When we're talking about the probability of up to \$x\$ successes, we need the cumulative distribution function. For this, we use the following format.

$$\text{P}(\text{# of successes} \le x) = \% \text{ chance of getting at most $x$ successes}$$ $$\text{P}(\text{# of successes} < x) = \% \text{ chance of getting at less than $x$ successes}$$

We can flip that around from "less than" to "at least" by taking the complementary probability.

$$\begin{align}\text{P}(\text{# of successes} \ge x) &= 1 - \text P(\text{# of successes} < x) \\ &= \%\text{ chance of getting at least that $x$ successes}\end{align}$$

The following is the cumulative probability function of the Binomial Distribution (including a version for strictly less than)

$$ \text{P}(\text{# of successes} \le x) = \sum_{i=0}^{x} p^i(1 - p)^{N-i} $$

$$ \text{P}(\text{# of successes} < x) = \sum_{i=0}^{x-1} p^i(1 - p)^{N-i} $$

So now, we can use this for the at least case:

$$ \text{P}(\text{# of successes} \ge x) = 1 - \sum_{i=0}^{x-1} p^i(1 - p)^{N-i} $$

Specifically for the case of a 5d10 pool and a 1d10 proc die, where you just need to succeed once?

$$ \text{P}(\text{# of successes} \ge 1) = 1 - \frac{1}{10}^0\left(\frac{9}{10}\right)^{5} = 1 - \left(\frac{9}{10}\right)^{5} $$

There's the equality with Dale M's answer. However, this formula works for other cases, such as when you want the proc die to match at least 3 times with a dice pool of 10d10s and a 1d10 proc die.

$$ \text{P}(\text{# of successes} \ge 3) = 1 - \sum_{i=0}^{2} \frac{1}{10}^i\left(\frac 9{10}\right)^{10-i} $$


In case the notation isn't familiar to the reader:

$$ \sum_{i = 0}^n f(i) $$

The above notation is a shorthand for "add up \$f(i)\$ for each value of \$i\$ from \$0\$ to \$n\$, including \$0\$ and \$n\$.

| improve this answer | |
\$\endgroup\$
2
\$\begingroup\$

Let's see:

  • the probability to roll number X on a single d10 is \$0.1\$
  • the probability for rolling a specific number X at least once when rolling \$n\$ d10s can be calculated using the formula for a binomial distribution:
  • \$ P(k) = \binom{n}{k} \times p^k \times (1 - p)^{n-k} \$
  • In our case, \$n\$ is how many dice your secondary pool contains (let's use 5). \$p\$ is the probability for rolling the number X, and \$k\$ is how often X should appear. It should appear once or more, hence we have to calculate \$P(1)+P(2)+...+P(10)\$, or just \$1-P(0)\$, which is exactly the same value.
  • \$1 - P(0)\$
    \$= 1 - \frac{n!}{k!\times(n-k)!} \times p^k \times (1-p)^{n-k}\$
    \$= 1 - \frac{5!}{5!} \times 0.1^0 \times 0.9^5\$
    \$= 1 - 1 \times 1 \times 0.5905\$
    \$= 0.4095\$
  • The probability to roll the same number X on your single dice and in a pool of 5 dice is, therefore, \$0.1 \times 0.4095 = 0.04095\$. Since you don't just want the probability for a specific number X, but for any identical numbers, the probability for that case is just all probabilities for Xs added together. Or, in other words, just \$0.04095 \times 10 = 0.4095\$.
  • The probability for a different amount of dice, \$k\$, in your "dice pool" can be shortened to the following formula:
  • \$ P(k) = 1 - 0.9^k\$
  • This leads to the following probabilities for different numbers of dice in your dice pool:
    1. \$0.1\$
    2. \$0.19\$
    3. \$0.271\$
    4. \$0.3439\$
    5. \$0.40951\$
    6. \$0.468559\$
    7. \$0.5217031\$
    8. \$0.56953279\$
    9. \$0.612579511\$
    10. \$0.6513215599\$

If you want multiple (\$x\$) "proc" dice, use the same probabilities, but multiply them with \$10 \times probabilityToRollACertainNumberOnYourProcDice\$. Using the same formula as we previously did for the dice pool, we get:

\$1 - P(0)\$
\$= 1 - \frac{x!}{k!\times(x-k)!} \times p^k \times (1-p)^{x-k}\$
\$= 1 - \frac{x!}{0!\times(x-0)!} \times 0.1^0 \times (1-0.1)^{x-0}\$
\$= 1 - x!/x! \times 1 \times 0.9^x\$
\$= 1 - 0.9^x\$


Hence, to calculate the probability to roll a number X in one pool with \$n\$ "proc" dice as well as in another pool with \$s\$ dice, we can use the following formula:

\$P(X) = P_n(X) \times P_s(X) = (1-0.9^n) \times (1-0,9^s)\$

To get the probability that any number gets rolled in both dice pools, we therefore just have to multiply \$P(X)\$ with the number of possible values on the dice:

\$P = P(X) \times 10\$

Note that, if you use dice other than d10, you also have to change the formulas used for calculating \$P(X)\$ - just multiplying it with a different number does not yield the correct result.

| improve this answer | |
\$\endgroup\$
  • 3
    \$\begingroup\$ This answer would be improved by using MathJax instead of images and code tags, to improve legibility and accessibility. If it did, I would upvote it for being the most straightforward answer. rpg.meta.stackexchange.com/questions/5896/… \$\endgroup\$ – Bloodcinder May 12 '18 at 11:47
  • \$\begingroup\$ @Bloodcinder thanks, I wasn't aware that LaTeX styling was available here. Fixed it :) \$\endgroup\$ – PixelMaster May 12 '18 at 15:22
  • 1
    \$\begingroup\$ You can use a proper multiplication symbol in the mst by replacing each * with \times. \$\endgroup\$ – SevenSidedDie May 12 '18 at 17:17

Your Answer

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