0

In the original Dungeons & Dragons game (D&D), each character has six ability scores determined by rolling three six-sided dice and taking the sum (i.e., 3d6 taken 6 times). What is the most likely set of scores to be produced with this method?

A few breadcrumbs, I think:

  • We can take the probability of any individual 3d6 result from a page like this.
  • For any ordered collection of values, we can find the number of k-permutations via the formula for multinomial coefficients.

A few example probabilities:

  • A set of all 10's: $0.125^6 \times 6!/6! = 3.8 \times 10^{-6}$
  • Five 10's and one 11: $0.125^6 \times 6!/(5!1!) = 2.29 \times 10^{-5}$
  • Three 10's and three 11's: $0.125^6 \times 6!/(3!3!) = 7.63 \times 10^{-5}$
  • Three 10's, two 11's, and one 12: $0.125^5 \times 0.1157 \times 6!/(3!2!1!) = 0.212 \times 10^{-4}$

Note that each of these examples have sequentially increasing probabilities.

So, what set(s) of scores have the highest likelihood of occurring?

(Note that a few prior questions exist on SE Mathematics about probabilities of rolling particular ability scores for later editions of D&D, using fundamentally different mechanics, and are different queries: e.g., here and here.)

Daniel R. Collins
  • 8,380

2 Answers2

1

You can split the probability formula into two parts: the powers of each of the individual 3d6 probabilities, and the multinomial part. If there are any duplicates, the multinomial part will be no at most half of the maximum value of $6!$. So if a list $a_1,...,a_6$ of scores that have duplicates beat a list of distinct scores $b_1,...,b_6$, it must be that $P(a_1) \times ... \times P(a_6) > 2 \times P(b_1) \times ... \times P(b_6)$.

If we take $b_1,...,b_6 = 8,...,13$ then the computation of $ 2 \times P(b_1) \times ... \times P(b_6)$ turns out to be higher than $P(10)^6$, which is the higher than the highest thre LHS can be (as 10 is joint most likely). I won't do the computation here; it is trivial but messy. This means it is impossible for there to be duplicates in the most likely list. As such, the answer is just the $6$ most likely distinct values, $b_1,...,b_6 = 8,...,13$.

Zoe Allen
  • 4,380
0

  • The score with the highest probability in a throw of $3$ dice is theoretically at the expected value of $\frac{1+6}2\times 3 = 10.5$

  • That for six such throws is $6\times 10.5 = 63$

  • Now unless the six scores are distinct, the multinomial coefficient will have some score repeated, and just one repetition will introduce a reducing factor of $2!$ in the denominator, lowering the probability

  • The $6$ scores must thus be distinct, and centered around $10.5$

  • So the best scores are $8,9,10,11,12,13$ in any order

Added, Part chart of sum frequencies of three dice throws

$\begin{array}{|c|c|}\hline Sum & Frequency\\ \hline 6,15 & 10 \\ \hline 7,14 & 15\\ \hline \color{red}{ 8,13} & 21 \\ \hline \color{red}{9,12} & 25 \\ \hline\color{red}{ 10,11} & 27 \\ \hline \end{array}$

It should be clear from the table that no other $\color{red}{combo\; of\; 6\; results}\;$ can match it, and that multiple occurrences of the same sum will lower the overall probability by having a division factor of $\geq 2$ .

true blue anil
  • 41,295
  • I think I see room for more precision in the answer around the claims, "the highest probability in a throw of 3 dice is... 10.5" (it's discrete) and what "drastically lowering the probability" means. – Daniel R. Collins Feb 19 '23 at 21:00
  • @DanielR.Collins: (1) Random variables may be discrete, but the expected value need not. (2) OP's question was re the optimal scores needed, not about making a chart of probabilities for many possible results, OP was computing alternatives only for groping towards the optimum. By immediate arrival at the optimal overall score using the concept of expected value, the fact that there is only one set of scores without duplicates that give the optimal probability, and that repeated values reduce the optimal by $\approx 2$ times or more, I think the question OP asked has been addressed – true blue anil Feb 19 '23 at 22:10
  • The critical fact here is that the probability of an $8$ is more than half the probability of a $10$. @DanielR.Collins While you might not like the wording, I understood the first paragraph to say that the most likely sums were the ones closest to $10.5,$ and the probabilities drop off strictly monotonically and symmetrically as we get farther from the expected value. This is simply a well-known feature of the sums when multiple copies of a standard $n$-sided die are thrown. – David K Feb 19 '23 at 23:50
  • @DavidK: I am not sure whether the above comment was addressed to me or to Daniel Collins. I am also not sure of what the defect in my answer is ( someone has downmarked it) Anyway, I have added a chart to improve the answer – true blue anil Feb 21 '23 at 13:59
  • With 3d6, the most frequents are 10,11 and then 9 or 12. When rolling 6 times 3D6, the most frequent is certainly one of these 3 propositions (9,10,10, 11,11,12) (9,10,10,11,12,13) or (8,9,10,11,12,13) ; I don't consider (9,10,11,11,12,13) or other similar, as it is duplicates of (9,10,10,11,12,13). I think that the question is to find which of these 3 propositions is the most frequent. – Lourrran Feb 21 '23 at 15:36
  • @Lourrran: From OP's working, I believe the highest probability is being sought, and that is what my answer will give. Bacause that is the only option that will have an attached multinomial coefficient of $\Large\frac{6!}{1!1!1!1!1!1!}$ – true blue anil Feb 21 '23 at 15:47
  • I have no idea who downvoted. Nor why there are not more upvotes yet besides mine. – David K Feb 21 '23 at 17:50
  • Thanks, I wasn't suggesting that you had downvoted ! $;;$ :) – true blue anil Feb 21 '23 at 18:08
  • You are right. I should read the full discussion before posting. – Lourrran Feb 21 '23 at 18:38