$52$ 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 $\frac{4\cdot13!\cdot39!}{52!}$.

  • $\begingroup$ 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 $12! \cdot 39! / 51! = 4 \cdot 13! \cdot 39! / 52!$. $\endgroup$ – antkam Mar 1 at 19:33

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