Bridge is a game where 4 people equally divide 52 cards among themselves. What is the probability that one of them will get 7 spades, 3 hearts, 1 diamond and 2 clubs?
I guess the first thing to do is to find the probability of first person getting this certain set of cards. Knowing that he is not returning drafted cards how can I do it?
And then I guess I also have to count probability of other people getting this set without first person returning his cards.
Can someone help me solve it?
On
The probability that the first person gets the specified distribution is $$p= \frac { \binom{13}{7}\binom{13}{3}\binom{13}{1}\binom{13}{2} } { \binom{52}{13} } $$ By symmetry, the probability that the second person gets the specified distribution is also equal to $p$.
Likewise for the third person and the fourth person.
Two players can't both get the specified distribution, else together they would have $14$ spades.
Thus, the $4$ events are mutually exclusive, so we can add the probabilities.
Therefore, the probability that some player gets the specified distribution is $4p$, where $p$ is as given above.
Your set up wasn't very clear...
I suppose the cards are just being randomly distributed so the total number of ways to distribute the card is $${52\choose13}{39\choose13}{26\choose13}{13\choose13}\cdot4!$$
Then the number of ways to get (at least) one hand with 7 spades, 3 hearts, 1 diamond and 2 clubs is $$\left({13\choose7}{13\choose3}{13\choose1}{13\choose2}\right){39\choose13}{26\choose13}{13\choose13}\cdot4!$$
Hence, the probability is $$\frac{\left({13\choose7}{13\choose3}{13\choose1}{13\choose2}\right){39\choose13}{26\choose13}{13\choose13}\cdot4!}{{52\choose13}{39\choose13}{26\choose13}{13\choose13}\cdot4!}=\frac{497646864}{635013559600}\approx0.0007837$$
In fact, you will notice that we actually only need to care about one hand and ignore the other three since these are randomly picked in the same ways with 39 cards left.