Challenging questions again- . -"

(a)
full house including 3 of a kind (x) and a pair (y), the first 2 cards can be any cards (x) or (y), the last 3 cards can be (x,x,y) or (y,y,x) in any order (3 different orders)
therefore,

52/52 * 51/51 * 2 * (3/50 * 2/49 * 3/48) * 3

= 9/9800

(b)
f4 of a kind including 4 of a kind (x) and one any card (y), the first 2 cards can be any cards (x) or (y), the next 4 cards can be either (x,x,x) or (y,y,y)
therefore,

52/52 * 51/51 * 2 * (3/50 * 2/49 * 1/48)

= 1/9800


questions (c) and (d) need the ans of (e) and straight cant be start of J, Q, K, there are only 40 cards are qualified. as the card in hand can be in any orders, 5! is the multipler for the combination


(e)
40/52 * 1/51 * 1/50 * 1/49 * 1/48 * 5!

= 1/64974

(c)
40/52 * 4/51 * 4/50 * 4/49 * 4/48 * 5! - 1/64974
(any straight including straight flush) (ans (e))

= 5/1274

(d)
52/52 * 12/51 * 11/50 * 10/49 * 9/48 - 1/64974
( any flush including straight flush ) (ans (e))

=1277/649174

a)
13(1/13 * 1/12 * 1/11 * 48/49 )
= 14/539
b)
13(1/52*3/51*2/50*1/49)
=1/83300