If 13 players are each dealt four cards froma 52 card deck w

If 13 players are each dealt four cards froma 52 card deck, what ist he probability that each player gets one card of each suit?

Solution

let\'s count how many total WAYS the cards can be dealt such that each player has one card of each suit. let\'s also say that order matters. in other words, john being dealt an ace of spades followed by an ace of hearts is a different WAY than john being dealt an ace of hearts followed by an ace of spades. the problem can also be solved using an \"order doesn\'t matter\" approach, and it will give the same answer. it doesn\'t matter which approach you use as long as you are consistent throughout. so once again, let\'s say order does matter. let there be 52 emtpy slots in a row in which the cards can be placed. the first 4 slots correspond to the cards of player one, the next 4 slots correspond to the cards of player 2, etc. so how many ways can the first 4 slots be filled with 1 card of each suit? well there are 13 choices for which spade will be selected * 13 choices for which heart will be selected * 13 choices for which club will be selected * 13 choices for which diamond will be selected = 13^4. now we have selected the 4 cards for the first 4 slots, and we must determine how many orders those 4 selected cards could be used to fill the first 4 slots. well that\'s just 4!. now let\'s fill the next 4 slots. well there are 12 remaining cards of each suit, so there are 12^4 ways to select the 4 cards for the next 4 slots and again there are 4! ways to order those 4 cards in those 4 slots. so far there are (13^4)*(4!)*(12^4)*(4!) ways to fill the first 8 slots so that the first 2 players get 1 of each suit. continue this pattern to see that there are

(13^4)*(4!)*(12^4)*(4!)*(11^4)*(4!)*.....

= ((4!)^13)*(13*12*11*....*2*1)^4 = ((4!)^13)*((13!)^4)

ways to deal the cards so that each player gets 1 card of each suit. then how many total ways are there to fill the 52 slots when there are no restrictions on who gets what suits? that\'s just 52!.

so the probability is [ ((4!)^13)*((13!)^4) ] / [ 52! ] .