# Probability of the 2,3,4 and 5 of Clubs dealt one each into 4 hands of 13 cards

Simple card game “Spades”, 13 cards to 4 players. Big score is from a bid and made zero tricks. Play begins with each player required to play lowest Club in hand. What is the probability of the 5 of Clubs taking that first trick (and the zero bid being defeated)?

So, your question is what is the probability that each of the $$2\clubsuit, 3\clubsuit, 4\clubsuit, 5\clubsuit$$ each were distributed to different people?
Well, someone gets the $$2$$. What is the probability that the $$3$$ went to someone else? This will be $$\frac{39}{51}$$ as there are $$51$$ remaining equally likely positions in players hands that the $$3$$ could occupy, $$39$$ of which are in a hand different than the hand holding the $$2$$.
Given that the two and the three are in different players' hands what is the probability that the $$4$$ went to someone else? This will be $$\frac{26}{50}$$
Finally, given that the two three and four all went to different players' hands, the probability that the $$5$$ went to the final remaining hand will be $$\frac{13}{49}$$
$$\frac{39\cdot 26\cdot 13}{51\cdot 50\cdot 49}$$