What is the chance the product of four dice would equal 144?

Question

Question

Four fair six-sided dice are rolled.

Out of the 1296 possibilities, what would result in a product of 144?

I started out with listing all possible combinations that lead to this product.

6*6*4*1 $${4\choose2}{2\choose1}{1\choose1} = {4!\over2!1!1!} = 12$$

6*6*2*2 $${4\choose2}{2\choose2} = {4!\over2!2!} = 6$$

6*4*3*2 $${4\choose1}{3\choose1}{2\choose1}{1\choose1} = {4!\over1!1!1!1!} = 24$$

4*4*3*3 $${4\choose2}{2\choose2} = {4!\over2!2!} = 6$$

And then I add all those numbers together.

12+6+24+6 = 48

Obviously this method is inefficient and prone to error. I do not feel confident with my answer (as in, I think it's not even right) and want to know if there's a better way to do this.

As a side note I checked out What is the probability of the sum of four dice being 22? but was completely confused on how that worked, so I need some hand-holding here.

Answer 1

You don't always need a rigorous method, but since $144$ is a comfortable number, and there are only $4$ dice, I think it is possible to create a rigorous method without using a computer.

Notice that $5$ cannot be part of any solution - it is not divisible by $144$. We continue by separating this problem into cases: where one case has the first number be $1$, then another with $2$, $3$ and so on.

Case 1:
Since the first digit is $1$, then we have to factor $144$ in $3$ numbers. If the next digit is $2$ or $3$, then observe that the most extreme case (two $6$'s) only gets $36$, less than $72$ or $48$. However, when the next digit is $4$, we find the only other possibility $(1,4,6,6)$. Trying $6$ gives the same combination.

Case 2:
When the first digit is $2$, we need to factor $72$ in $3$ numbers. The next digit cannot be $1$, but when it is $2$ or $3$, we get the solutions $(2,2,6,6)$, and $(2,3,4,6)$. When we try $4$ or $6$, again we get the same combinations.

Case 3:
Now, with the first digit being $3$, we need to factor $48$ in $3$ numbers. $1$ doesn't work, $2$ gives the same solution $(2,3,4,6)$, but $3$ gives a new solution - $(3,3,4,4)$. We can check that $4$ and $6$ give the same solutions.

When the first digit is $4$ or $6$, there are no new solutions.

This shows that $(1,6,4,4)$; $(2,2,6,6)$; $(2,3,4,6)$; and $(3,3,4,4)$ are the only solutions, and we can use your method to calculate the total number of rearrangements.