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.

  • $\begingroup$ Actually, I think your method is very good and would be what I'd try to do as long as the question gives reasonable numbers $\endgroup$ – Shuri2060 Jul 10 '17 at 18:07
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    $\begingroup$ This is directly related to the prime factorization of the product you want to produce. If the dice had infinite sides, it wouldn't be all that complicated, given the prime factorization. Given the an upper limit on the value of each die, things are more difficult to enumerate. One case is simple at last: If even one of the nontrivial prime factors is greater than the number of sides on each die, the probability is 0. $\endgroup$ – Kajelad Jul 10 '17 at 18:12
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    $\begingroup$ If you are not sure if your answer is correct, well ... that's what computers are for. Here, with mathStatica/Mathematica: f = (1/6)^4; domain[f] = {{x,1,6}, {y,1,6}, {z,1,6}, {w,1,6}} && {Discrete}; Prob[x y z w == 144, f] returns $\frac{1}{27}$ - same as you obtained. $\endgroup$ – wolfies Jul 10 '17 at 18:32

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.


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