When playing a game of blackjack, what are the odds that you will be dealt BlackJack on the first deal? Assume you are playing with one opponent and your opponent is the dealer.

Note: 10, jack, queen and king all are worth 10. Aces are worth 11. You have blackjack when the values of both your cards add up to 21. There is only one deck of cards being dealt out.

(In reply to Decks? by levik)

Isn't the clue in the last line of the question?.... "There is only one deck of cards being dealt out"

To my way of thinking the number of opponents should be irrelevant because what is important is that the two cards, out of a possible 52, that I receive make a blackjack. So whether it's the 1st & 3rd cards (with 1 opponent who is also the dealer) or the 5th and 10th cards (with 4 opponents and I'm the dealer) shouldn't make any difference. If this assumption isn't true then can somebody explain why, rather than just say I'm talking nonsense.

Anyway, if my assumption is true then to receive a blackjack either my first card is worth 10 and the second is an ace or vice versa, which makes the probability....

(16/52)*(4/51) + (4/52)*(16/51) = 128/2652

...which is approx 4.83%

Posted by fwaff on 2003-02-12 21:07:37