Probability that an opponent have a Void in a trick-taking game In a trick taking game (like Whist or Spades) each player gets 13 cards out of a deck of 52 cards.  Assume you lead the first round with the A$\clubsuit$ which is not a trump.  The only option to loss this trick is if someone does not have $\clubsuit$s at all.
Given that your hand contains $n$ $\clubsuit$s:


*

*What is the probability that all 3 opponents have at least 1$\clubsuit$

*What is the probability that all  3 opponents have at least 2$\clubsuit$

*What is the probability that all 3 opponents have at least 3$\clubsuit$

 A: Here is a partial answer to get you started: Let’s say we want the probability that at least everyone has a club. Instead let’s work out the probability that there is someone (not you) who has no clubs at all. There are $13-n$ clubs of $39$ cards distributed amongst the other players (note you have the ace so $n\ge1$) so for one player, bob let’s say, to not have a club we imagine him drawing 13 cards from the cards not in your hand and get
$$\Bbb P(\text{Bob has no club}) = \frac{26+n}{39}\cdot\frac{25+n}{38}\cdots\frac{13+n}{26} = \frac{\binom{26+n}{13}}{\binom{39}{13}}$$
The last expression counts the number of ways to choose 13 cards from the $26+n$ non-clubs in the numerator and counts the ways to choose 13 cards from the 39 cards available in the denominator. Now by symmetry the probability for each individual other player not having a club is the same (but not independent: if you only have the ace then they can’t all have no clubs). Let’s now write down the probability that someone has no clubs:
$$\Bbb P(\text{someone has no clubs}) = \frac{\binom{26+n}{13}}{\binom{39}{13}} + \sum_{j=1}^{13-n}\frac{\binom{13-n}{j}\binom{26+n}{13-j}}{\binom{39}{13}}\left(\frac{\binom{13+n+j}{13}}{\binom{26}{13}}+\frac{\binom{26-n-j}{n+j}}{\binom{26}{13}}\right)$$
This corresponds to:


*

*Bob has no clubs. Or:

*Bob has $j$ clubs and:


*

*Alice has no clubs, or

*Alice has all the remaining clubs $13-n-j$ of them so the third player has none.



Finally you have $\Bbb P(\text{everyone has at least one club}) = 1 - \Bbb P(\text{someone has no clubs}).$ To deal with the other cases you can follow a similar pattern. 
