Standard playing card deck without jokers: Deal 13 cards. What is the probability at least 1 suit is not present? Similar questions:
What is the probability that 13 cards drawn from a standard deck has at least one card from each suit?
Deal 4 cards from a deck. What is the probability that we get one card from each suit?

What I tried:
P(at least 1 suit is not present) = 1 - P(all suits are present)
where
P(all suits are present) = 
$$\sum_{a}\sum_{b}\sum_{c}\sum_{d}\frac{\binom{13}{a}\binom{13}{b}\binom{13}{c}\binom{13}{d}}{\binom{52}{13}}$$
where $a+b+c+d=13; a,b,c,d \ge 1$
Is that right? If so, is there a better way to approach this? If not, why?
Following the second question:
$$\frac{\binom{13}{1}^{13}}{\binom{52}{13}}$$
Something else:
$$\frac{\binom{13}{1}^{4} \binom{48}{9}}{\binom{52}{13}}$$
which is ~75 (far greater than 1)
 A: Use inclusion/exclusion principle:


*

*Include the number of combinations containing at most $\color\red3$ suits: $\binom{4}{\color\red3}\cdot\binom{\color\red3\cdot13}{13}$

*Exclude the number of combinations containing at most $\color\red2$ suits: $\binom{4}{\color\red2}\cdot\binom{\color\red2\cdot13}{13}$

*Include the number of combinations containing at most $\color\red1$ suit: $\binom{4}{\color\red1}\cdot\binom{\color\red1\cdot13}{13}$


Finally, divide the result by the total number of combinations:
$$\frac{\binom43\cdot\binom{3\cdot13}{13}-\binom42\cdot\binom{2\cdot13}{13}+\binom41\cdot\binom{1\cdot13}{13}}{\binom44\cdot\binom{4\cdot13}{13}}\approx5.1\%$$
A: This is probably handled best as an inclusion-exclusion problem, counting the number of hands that omit at least one suit:
Let's take it from the bottom: The number of hands with exactly one suit is of course
$$ A_1 = \binom{4}{1} \binom{13}{13} = 4 $$
The number of hands with exactly two suits is
$$ A_2 = \binom42 \binom{26}{13} - \binom31 A_1 $$
where the second term corrects for the fact that $\binom42\binom{26}{13}$ counts each of the $4$ one-suit hands three times.
The number of hands with exactly three suits is then
$$ A_3 = \binom43 \binom{39}{13} - \binom21 A_2 - \binom32 A_1 $$
where the subtracted terms correct for counting hands that actually contained fewer than three suits.
Now your result would be
$$ 1 - \frac{A_1+A_2+A_3}{\binom{52}{13}} $$
