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by kakz » Fri Mar 02, 2012 9:25 am
A standard 52-card deck contains 13 cards (Ace, 2-10, Jack, Queen,King) in each of 4 suits (Spades, Hearts, Clubs. Diamonds). Each Ace can be worth either 1 or 11, at the discretion of the player; cards with numbers from 2-10 are worth the amount listed on the card; and Jacks,Queens and Kings are worth 10. A player is dealt two cards, without replacement. What is the probability that the sum of the two cards dealt to the player COULD be 18?
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by krusta80 » Fri Mar 02, 2012 9:52 am
kakz wrote:A standard 52-card deck contains 13 cards (Ace, 2-10, Jack, Queen,King) in each of 4 suits (Spades, Hearts, Clubs. Diamonds). Each Ace can be worth either 1 or 11, at the discretion of the player; cards with numbers from 2-10 are worth the amount listed on the card; and Jacks,Queens and Kings are worth 10. A player is dealt two cards, without replacement. What is the probability that the sum of the two cards dealt to the player COULD be 18?
I'm assuming the emphasis of COULD is to remind us of an ace and seven.

As a blackjack player, I personally wasn't about to forget a soft 18. :)

Ways to get eighteen:

9,9 -> 4C2 = 6 ways
A,7 -> 4*4 = 16 ways
8,T/J/Q/K -> 4*16 = 64 ways

Total ways to get eighteen is 86.

Total two-card hands for a full deck is 52*51 = 2652

Probability of being dealt an eighteen is 86/2652 = 43/1326
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