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)

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:
$$\frac{\binom43\cdot\binom{3\cdot13}{13}-\binom42\cdot\binom{2\cdot13}{13}+\binom41\cdot\binom{1\cdot13}{13}}{\binom44\cdot\binom{4\cdot13}{13}}\approx5.1\%$$
