What is the probability of getting a “blackjack” hand (21) from drawing the first two cards from a fair deck?
Question by suezzle: What is the probability of getting a “blackjack” hand (21) from drawing the first two cards from a fair deck?
What is the probability of getting a “blackjack” hand (21) from drawing the first two cards from a fair deck?
*Assume no jokers (52 cards per deck).
*Assume randomly shuffled.
I usually throw in my own assumption of how to go about figuring the question out, and often put my own guess as to the answer, then ask you to check it…but I’m really not sure how to approach this, as the probabilities seem to “overlap” or however it’s worded…
Please don’t just give the answer. I want to see the rationale behind it.
Thanks!
I’m really not sure ho
Best answer:
Answer by roshambo74
I would imagine that since you need an ace and a 10,j,q or K,
You have a 4/52 chance of getting an ace, then a 16/51 chance of getting the 10 card you need.
(4/52)x(16/51)=0.0241
Multiply by 100 to get 2.41%
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Tags: Blackjack, cards, Deck, drawing, Fair, First, from, Getting, Hand, PROBABILITY
August 4th, 2010 at 9:37 pm
Your first two cards need to be an ace and any 10-value card in either order. There are 4 aces and 16 10-value cards in the deck.
Total number of possibilities for the first two cards is 52 * 51 = 2652. Of those:
Number of times you will have a A followed by 10-value is 4 * 16 = 64.
Number of times you will have 10-value followed by A is 16 * 4 = 64.
Total number of blackjacks = 64 + 64 = 128.
Probability of a blackjack on the first two cards = 128/2652 , or about 1 out of 21 times. (more precisely 1 out of 20.71875.)
Note to roshambo: They can be in either order, so you need add in the cases where a ten occurs as the first card and an ace is second (16/52)*(4/51).