Okay since George's friend has two queens there are only two chances of him getting a royal flush because the two queen's are already used and a royal flush consists of five consecutive cards from nine to king in the same suit. So George wrote out:
ROYAL FLUSH:
(2C1)(5C5)
(2C1) because there are only two suits left to make a royal flush since the two queens are used up.
(5C5) because there are five cards to choose from and the five cards have to be consecutive from nine to king.
STRAIGHT FLUSH:
(2C1)(6C1)(4C4)
(2C1) because you choose a suit.
(6C1) because there's six cards that can create a straight flush.
(4C4) because you could choose four of the cards around the card you chose from 6C1.
you choose a suit, then you choose one of the numbers from A to 6. Then you choose the 4 numbers around it. Example: for the suits, clubs and diamonds, both 7 AND queen are missing. So, the only straight flushes that can be made are the ones from A-6. For the 3rd suit, Clubs, ONLY the 7 is missing. So, you don't even have to choose a suit, for it's already been decided. So then, you have A-6 and 8-K. For the 4th suit, NO CARDS are missing, so you can pick one of 13 numbers and pick 4 of the 4.
FOUR OF A KIND:
(11C1)(4C4)(10C1)(4C1)
(11C1) because since the queens and sevens are taken out then that means that there are only 11 other possible face cards that can be four of a kind.
(4C4) you need all four cards of the same face card.
(10C1) because there are 10 cards left for the kicker to be, so choose a face value for the kicker.
(4C1) the suit that the kicker will be.
HIGHER FULL HOUSE:
(6C1)(4C3)(10C1)(4C2) + (6C1)(4C3)(1C1)(2C2)
(6C1) because there are only 6 other cards that can beat the seven's. So, choose 1 of the 6 face cards that are higher than the seven.
(4C3) the face card determines which other cards you need to get. So, you need three of the four face card.
(10C1) there are ten available cards to be the pair because the sevens are used up and queens and what you have that made the three of a kind.
(4C2) because of the four cards of that face you only need two of them to make it a pair.
(6C1)(4C3)(1C1)(2C2)
choose a number for 3 of a kind, then get 3 of those numbers, then choose a pair, which have to be queens, then make it's pair
CHANCES OF BEATING:
((2C1)(5C5) + (2C1)(6C1)(4C4) + (11C1)(4C4)(10C1)(4C1) + (6C1)(4C3)(10C1)(4C2) + (6C1)(4C3)(1C1)(2C2))/ (47C5)
(47C5) is divided because it is all the possible hands that he could make with the remaining cards.
ANSWER:
1928/ 1533939 = 0.0012569
