3
$\begingroup$

$52$ 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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.