How do you find $f(x_1, x_3)$? $X_i$ is the number of times (out of 100) that a die's face has $i$ dots.
I know that $X_i\sim \text{binomial}(100, 1/6)$, so $f(x_i)={100 \choose x_i}(1/6)^{x_i}(5/6)^{100-x_i}$.
How do you find the joint probability mass function for $X_1$ and $X_3$ (the number of times you roll $1$s and $3$s, respectively? I'm unfamiliar with joint pmfs so I'm not sure how to even begin figuring this out.
 A: One way is to figure it out from the conditional PMF instead. First, a priori the distribution of $X_1$ is binomial($100$,$1/6$), as you've said. Now suppose you know the value of $X_1$ is $y$. Then you have $100-y$ remaining chances to roll a $3$. The only dependence between these two groups of rolls is that you can no longer roll a $1$ in the second group. Hence the conditional PMF of $X_3$ given $X_1$ is the PMF for binomial($100-X_1$,$1/5$).
Then we have the formula
$$p_{X_1,X_3}(x,y) = p_{X_3|X_1}(x,y) p_{X_1}(y).$$
Looking at the right side, the first term is, as a function of $x$ for each fixed $y$, the PMF of binomial($100-y$,$1/5$). The second term is the PMF of binomial($100$,$1/6$). Can you put it together from here?
A: Alternatively, from first principles.
The favoured space is permutations of $x_1$ probability $1/6$ type-1 successes, $x_2$ probability $1/6$ type-2 successes, and $n-x_1-x_2$ probability $4/6$ failures; all of which form disjoint partitions of a single trial (they are mutually exclusive and exhaustive).   
Just as the binomial distribution's probability mass function is constructed by counting permutations and multiplying by the probability of obtaining so many failures and successes in a certain order, we do the same for this multinomial distribution to obtain the joint probability mass function:
$$f_{X_1,X_3}(x_1,x_3) = \frac{n!}{\Box!\,\Box!\,(\Box)!}
\left(\frac{\Box}{\Box}\right)^{\Box}
\left(\frac{\Box}{\Box}\right)^{\Box}
\left(\frac{\Box}{\Box}\right)^{\Box}
$$
