I've already seen You roll two dice. What is the probability of the event that they have no common factor greater than unity? but I'm looking into a similar but distinct problem, one I don't even know how to approach besides simply enumerating all possible results. Ideally I'd like to 'learn how to fish' instead of just be given fish.
As a concrete example, if the two 12-sided dice come up 4 and 12, they share ${1,2,4}$ as factors - so they would 'score' 3. The dice 7 and 7 would share ${1,7}$ (scoring 2) and two lots of 12 would share ${1,2,3,4,6,12}$ (scoring 6).
Is it easy to work out a probability distribution for this problem? Or is it not mathematically possible and I need to manually work out all 144 combinations?
With regards to the expected number of factors, we can calculate this easily without bothering to look at the probability distribution.
Letting $X$ be the random variable counting the total number of factors shared, and letting $X_i$ be the indicator random variable corresponding to whether $i$ is a factor of both the first and second die, we can see that $X=X_1+X_2+\dots+X_{12}$ and thanks to the linearity of expectation that $E[X]=E[X_1+X_2+\dots+X_{12}]=E[X_1]+E[X_2]+\dots+E[X_{12}]$, this despite any possible dependence between these.
Now, $E[X_i]=\Pr(X_i=1)$ is simply going to be equal to the probability that $i$ is a factor of each die result, and will be equal to $\frac{k}{12^2}$ where $k$ is the number of integer multiples of $i$ in the range $\{1,2,\dots,12\}$
We get then a calculation of:
$$E[X] = \dfrac{12^2+6^2+4^2+3^2+2^2+2^2+1^2+1^2+1^2+1^2+1^2+1^2}{12^2}$$
$$ = \dfrac{219}{144}= 1.5208\overline{3}$$
Actually finding the probability distribution for the number of factors however will be highly tedious brute force. For instance, to find those pairs of numbers in increasing order with two shared factors would be (2,2),(2,4),(2,6),(2,8),(2,10),(2,12),(3,3),(3,6),(3,9),(3,12),(4,6),(4,10),(5,5),(5,10),(6,8),(6,10),(7,7),(8,10),(9,12),... and take the count of all of these (and their reverses) and divide by $12^2$ to find the probability.
It can be done, but to reemphasize... if all we were wanting was to find the expected value then we don't have to do this! It is a waste of time and effort.