Compute the probability that a bridge hand is void in at least one suit.
First I will select the suit I want to exclude from the $13$ cards, which is done in $4\choose 1 $ $=4$ ways, then I will choose the $13$ cards from the remaining $39$ cards.
So the probability should be equal to
$4\cdot$$\frac{39 \choose 13 }{52\choose 13} $
Is this correct?
Best Answer
Let $E_1, \;E_2,\; E_3, \;E_4$ denote the events that your hand is void in $\clubsuit$, $\diamondsuit$, $\spadesuit$, $\heartsuit$ respectively.
We are looking for the probability P $= P(\cup_{i=1}^{4} E_i)$
By the inclusion-exclusion identity, we have
P = $p_1-p_2+p_3-p_4$ where $p_i$ denotes the probability that your hand is void in $i$ number of suits and contains $4\choose i$ terms each with the probability
$$\frac{\binom{52-13\cdot i} {13}}{\binom{52} {13}}$$ except for the case for $p_4$ since $p_4=0$ is evident as the hand cannot be void of all the $4$ suits.
The required probability P will then be $$4\cdot\frac{\binom{39}{13}}{\binom{52}{13}} - 6\cdot\frac{\binom{26}{13}}{\binom{52}{13}} + 4\cdot\frac{\binom{13}{13}}{\binom{52}{13}}$$