Probability that a 13 card hand has at least 1 card in each suit? I understand that you would use inclusion-exclusion but I'm completely lost as to how to do so. Any tips would be appreciated!
Would having all spades/hearts/clubs/diamonds as their own event be on the right track?
 A: There are $\binom{52}{13}$ equally likely hands. Now we want to count the hands that have at least one of each suit. 
It is perhaps more natural to attack the complement, and count the hands that are missing at least one suit. The argument is typical Principle of Inclusion/Exclusion (PIE). 
There are $\binom{39}{13}$ hands that have no $\spadesuit$. We get the same counts for hands missing the $\heartsuit$, for hands missing the $\diamondsuit$, and for hands missing the $\clubsuit$. 
But the sum $4\binom{39}{13}$ counts twice the hands that are missing, for example, both the $\spadesuit$ and the $\heartsuit$. There are $\binom{26}{13}$ of these. Consider all $\binom{4}{2}=6$ pairs of suits. We (sort of) want to subtract $6\binom{26}{13}$  from $4\binom{39}{13}$ to deal with the double-counting. 
But then we have subtracted too much, for we have subtracted once too many times the $4\binom{13}{13}$ single-suit hands. Thus the number of ways to have at least one suit missing is 
$4\binom{39}{13}-6\binom{26}{13}+4\binom{13}{13}$, It follows that the number of hands with at least one card from each suit is
$$\binom{52}{13}-4\binom{39}{13}+6\binom{26}{13}-4\binom{13}{13}.$$
