Question

This is a question for a stats course for science students... In one version of the game played by a dealer and a player, 2 cards of a standard 52-card bridge deck are dealt to the player and 2 cards to the dealer. For this exercise, assume that drawing an ace and a face card (Jack, Queen and King for each shape) is called blackjack. If the dealer does not draw a blackjack and the player does, the player wins. If both the dealer and player draw blackjack, a tie occurs.

(a) What is the probability that the dealer will draw a blackjack?

(b) What is the probability that the player will wins with a blackjack?

please explain the answer

Answer 1

4 cards in the pack are aces.
16 cards in the pack (4 each of king, queen, jack, 10) are worth 10.

a)
There are two ways to get 21: ace then 10, or 10 then ace.
p(ace then 10) = 4/52 * 16/51
p(10 then ace) = 16/52 * 4/51

=> p(blackjack) = 4/52 * 16/51 + 16/52 * 4/51 = 2 * 16 * 4 / (52*51) = 0.048 (approx)
=> p(dealer gets blackjack) = p(blackjack) = the number above.

BUT, if the player gets the first card then it goes (4/51)(16/49)(2) = 0.51 would be the dealers probability for getting a blackjack.

b)
For the player to win with blackjack, we must multiply p(blackjack) [above] by the odds that the dealer does not then get blackjack. If a player gets blackjack, there are then only 50 cards left for the dealer, including 3 aces and 15 tens (because the player has taken 1 ace and 1 ten).

=> p(dealer gets blackjack after player does) = 3/50 * 15/49 + 15/50 * 3/49 = 2 * 3 * 15 / (50*49)

=> p(dealer does not get blackjack after player does) = 1-p(dealer gets blackjack after player does)

=> p(player wins with blackjack) = p(player gets blackjack) * p(dealer does not get blackjack after player does) = 2 * 16 * 4 / (52*51) * (1 - 2 * 3 * 15 / (50*49)) = 0.046 (approx)