What is the most likely set of scores produced by rolling 3d6 six times?

Question

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.)

Answer 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, $b_1,...,b_6 = 8,...,13$.

Answer 2

• 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 $\geq 2$ .