Probability of a hand is void in at least 2 suits

Question

I was calculating for the following question:

In a card game what is the probability of that a hand of 13 cards is void in at least 2 suits.

I think :

$$1- \frac{ \binom{52}{13} - 4 \times \binom{39}{13}} { \binom{52}{13}}$$

but my answer is wrong.

Answer 1

If a hand has 13 cards then the probability that a hand is void in at least 2 suits is, $$\frac{\binom{4}{2}\left(\binom{26}{13}-2\right)+4}{\binom{52}{13}}.$$ In fact a hand that is void in at least 2 suits will have two suits or one suit. In the first case we choose the suits in $\binom{4}{2}$ ways and the values in $\binom{26}{13}-2$ ways (we exclude the two cases where the cards are all of the same suit). In the second case we have to choose only one suit in $4$ ways.