That looks fairly straight forward. First put the four players in a specific order. There are 4! ways to do that. Now give the Ace, King, Queen, and Knave of one suit to each person. There are 4! ways to distribute the four suits among the four people. That leave 36 cards to be distributed at random, 9 cards to each person. There are \(\displaystyle _{36}C_9= \frac{36!}{9! 27!}\) ways to sort out one hand of 9 cards. Then there are \(\displaystyle _{27}C_9= \frac{27!}{9!18!}\) ways to sort out the second hand and \(\displaystyle _{18}\frac{18}{9!9!}\) ways to sort out the third 9 card hand, leaving the fourth hand.Q.The number of ways in which a pack of 52 cards of four different suits can be distributed equally among four players so that each player gets the Ace, King, Queen, and Knave of the same suit?
Mark, same as a deck of 52 cards with: 4 1's, 4 2's, ...., 4 13's
and each player gets 1 to 5 anywhere in his 13 cards...right?
Checked my coding...99.9% sure correct...
That makes the odds 0.0000000000000000000000000000003
Maybe that's reasonable?
Where did you get that answer?