Is there an efficient way to solve De Mere's puzzle? This question comes from Bertsekas' Introduction to probability, 2nd ed.


A six-sided die is rolled three times independently. Which is more likely: a sum of 11 or a sum of 12?

I understand the solution 100%, but I am wondering if there is an efficient way to enumerate all the combinations; indeed the solution just lists them without going into how to obtain them.
I had to do it case by case, ie exhaust all possibilities of numbers in the 1st roll. e.g. Beginning with the first roll have a 1, I will enumerate all combinations of $1+x+y$ ($x$ being the 2nd roll and likewise for $y$) such that the total is 11...keep doing this all the way till the first roll is 6.
But this is a very inefficient way to do it, is there a better way of doing it?

 A: As I understand it, the question is not to calculate the probabilities, but simply to determine which is more likely, $11$ or $12$. Let $X$, $Y$, and $Z$ denote the result of the first, second, and third roll.
If $6\le X+Y\le10$, there is one way to make a total of $11$ with the third roll, and one way to make a total of $12$; even chances, as we are assuming a fair die.
If $X+Y\le4$ or $X+Y=12$, there is no way to make a total of $11$ or $12$.
If $X+Y=5$, there is one way to make $11$, no way to make $12$; if $X+Y=11$, there is  one way to make $12$, no way to make $11$. So the question boils down to: which is a more likely total with two dice rolls, $5$ or $11$? There are $4$ ways to make $5$ but only $2$ ways to make $11$ with two rolls, so $5$ is a more likely total than $11$ with two rolls, so $11$ is a more likely total than $12$ with three rolls.
A: One way is generating functions. They are more sophisticated but not always efficient for a human to compute: You would imagine something like expanding
$$
(x + x^2 + x^3 + x^4 + x^5 + x^6)^3
$$
The coefficient of $x^k$ is the number of ways that $k$ can be obtained as the sum $a + b + c$ where $1 \leq a,b,c \leq 6$.
A: Let $A(k,n)$ be the number of ways of getting a sum $k$ by throwing a dice $n$ times.
Then we have the recursion $$ A(k,n) = \sum_{j=1}^6 A(k-j,n-1) $$
with the initial conditions
$$A(k,0)= \begin{cases} 
 1 & \text{if } k=0\\
 0 & \text{elsewhere}
\end{cases}$$
This at least gives you a simple way to compute any value. Example in Python
Or with pencil and paper ... or a spreadsheet:


Further: If you are familiar with probability, you can answer the question without computing the values. A single die has a uniform (discrete) pmf with mean $(1+6)/2=7/2$. Then sum of thre dice, then, is the triple convolution of a uniform, which should be a bell shaped distribution, symmetric around the mean $3 \times 7/2 = 10.5$ - hence the maxima should be at $10$ and $11$. Hence we should have $p(11) > p(12)$
