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I'm trying out different systems for a board game, and my (weak) probability skills are failing me.

The basic mechanic is as follows: you roll a number of six-sided dice (d6). Any die showing 5 or more is a success. A specific bonus can "upgrade" part of your pool to a larger die (let's say a 10-sided dice, or d10).

So for example:
regular roll: roll 7d6; any die showing 5 or more is one success.
same roll with bonus: roll 4d6 and 3d10; any die showing 5 or more is one success.

I'd like to learn how to calculate the chance of getting X successes with the mixed pool (either "X" successes or "X or more" successes).

this thread thaught me how to find the probability for the homogenous pool, but I have no idea how to change the formula to accomodate the differently-sized dice.

I would like to learn the general formula, to be adjusted for different size of dice (such as d8 and d12) and for a larger or smaller pool of mixed dice as well.

Thanks in advance, sorry for any English mistake I may have made.

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  • $\begingroup$ There is no "dices". The word "dice" is itself the plural of the word "die". $\endgroup$
    – user247327
    Commented Jun 23, 2017 at 12:18

2 Answers 2

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I jump directly to the second question:

same roll with bonus: roll $4d6$ and $3d10$; any die showing 5 or more is one success.

4d6: The probability that one die show a 5 or a 6 is $\frac26=\frac13$. Now you can use the binomial distribution to calculate that x dice out of $4$ have an success.

$$P(X=x)=\binom{4}{x} \left( \frac{1}{3} \right)^x\cdot \left( \frac{2}{3} \right)^{4-x}$$

Similar for $3d10$

$$P(Y=y)=\binom{3}{y} \left( \frac{6}{10} \right)^y\cdot \left( \frac{4}{10} \right)^{3-y}$$

Let´s denote $s$ as the number successes. We have to calculate the sum of the successes and fails to get the probability for $S=x+y$ successes and $3+4-x-y=7-s$ fails. At maximum we can have $7$ successes. Thus

$P(S\geq s)=P(S=s)+P(S=s+1)+\ldots+P(S=7)$

As an example I show how to calculate $P(S=5)$. For this purpose I made a table for the combination of the successes for $x$ and $y$

$\begin{array}{|c|c|}\hline x&y \\ \hline 2&3 \\ \hline 3&2 \\ \hline 4&1 \\ \hline\end{array}$

Note that $x=1$ is not possible since $y\leq 3$. This shows why there is no "simple" formula. The random variables $X$ and $Y$ are independent. Thus $P(X\cap Y)=P(X)\cdot P(Y)$. The single probabilities are

$$P(X=2,Y=3)=\binom{4}{2} \cdot \left( \frac{1}{3} \right)^2\cdot \left( \frac{2}{3} \right)^{4-2}\cdot \binom{3}{3} \cdot \left( \frac{6}{10} \right)^3\cdot \left( \frac{4}{10} \right)^{3-3}$$

$$P(X=3,Y=2)=\binom{4}{3} \cdot \left( \frac{1}{3} \right)^3\cdot \left( \frac{2}{3} \right)^{4-3}\cdot \binom{3}{2} \cdot \left( \frac{6}{10} \right)^2\cdot \left( \frac{4}{10} \right)^{3-2}$$

$$P(X=4,Y=1)=\binom{4}{4} \cdot \left( \frac{1}{3} \right)^4\cdot \left( \frac{2}{3} \right)^{4-4}\cdot \binom{3}{1} \cdot \left( \frac{6}{10} \right)^1\cdot \left( \frac{4}{10} \right)^{3-1}$$

Finally the single probabilities have to be summed up to get $P(S=5)$.

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"regular roll: roll 7d6; any die showing 5 or more is one success." With 6 sides, the probability of success, a single die showing "5 or more", i.e. 5 or 6, is 2/6= 1/3. The probability of no success is 2/3. The probability of exactly n dice showing 5 or more, the other 7- n not, is $\begin{pmatrix}7 \\ n\end{pmatrix}(1/3)^n(2/3)^{7- n}$ where $\begin{pmatrix}7 \\ n\end{pmatrix}$ is the "binomial coefficient", $\frac{7!}{n!(7- n)!}.

"same roll with bonus: roll 4d6 and 3d10; any die showing 5 or more is one success." The probability of "success" on the six sided die is, as before, 1/3. The probability of "success" on the 10 sided die is 5/10= 1/2. To find the probability of n successes, look at all the ways two non-negative integers can add to n: i= 1 to n, j= n- i. Calculate the probability of i successes in 4 rolls of a 6 sided die, as above. Calculate the probability of n- i successes in 3 rolls of the 10 sided die as above except that the probability of success is 5/10= 1/2 and the probability of non-success is 1/2. Multiply them together and then add for i going from 1 to n.

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  • $\begingroup$ "The probability of "success" on the $10$ sided die is $5/10= 1/2$" I would say it is $6/10$, because the favorable outcomes are $5,6,7,8,9,10$ $\endgroup$ Commented Jun 23, 2017 at 14:09

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