# $52$ cards are dealt among 4 players, determine the probability that a player gets all the spades

$$52$$ cards are dealt among 4 players, determine the probability that a player gets all the spades

Number of ways the cards can be dealt among $$4$$ players = $$52 \choose {13,13,13,13} = \frac{52!}{(13!)^4}$$

Number of ways player $$1$$ gets all the spades =$$13\choose 13 39\choose {13,13,13}$$ = $$\frac{39!}{(13!)^3}$$

So the required probability should be = $$\frac{4*\frac{39!}{(13!)^3}}{\frac{52!}{(13!)^4}}$$

Is this correct? If not, please tell where is the mistake so that I can learn

There's just one little typo in your working; it should be $$\frac{39!}{(13!)^3}$$ and not $$\frac{39!}{(13!)^4}$$. Otherwise it is correct.
Another, perhaps simpler way is to just consider 13 spades and 39 non-spades. There are then $$\binom{52}{13}$$ ways to deal the cards, 4 of which have one player with all the spades, for a probability of $$\frac{4\cdot13!\cdot39!}{52!}$$.
• Almost equivalently: One person must get the Ace of spades. There are ${51 \choose 12}$ ways to fill out his hand, only $1$ of which gives him all the other spades, for a probability of $12! \cdot 39! / 51! = 4 \cdot 13! \cdot 39! / 52!$. – antkam Mar 1 at 19:33