suppose that you are playing blackjack against a dealer. in freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack?

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- August 14th 2007, 11:59 AMkkoutsothodorosblackjack proabability question
suppose that you are playing blackjack against a dealer. in freshly shuffled deck, what is the probability that neither you nor the dealer is dealt a blackjack?

- August 14th 2007, 12:01 PMkkoutsothodoros
a blackjack is when you are dealt an ace and either a ten, jack, queen, or king.

- August 14th 2007, 01:45 PMkkoutsothodoros
I think I have an answer although I always question my reasoning.

Let A be the event - player gets a blackjack

Let B be the event - dealer gets blackjack

P(A) = (16*4)/(52 choose 2)

By symmetry P(B) is the same.

We have P(A or B) = P(A) + P(B) - P(A and B)

So if we calculate P(A and B) we can subtract P(A or B) from 1 to get the answer.

This is where I am unsure of myself.

The total number of possible outcomes is (52 choose 2)*(50 choose 2).

We give the first player 2 cards from 52 and the second 2 cards from the remaining 50.

The number of outcomes where both the player and dealer have blackjack is 16*15*4*3.

So we have P(A and B) = (16*15*4*3)/((52 choose 2)*(50 choose 2))

I think this gives the right answer. - August 14th 2007, 01:48 PMkkoutsothodoros
by the way, the answer in the back of my book is .9052. what i just got is reasonably close to that. but i still question how i got the answer.

- August 14th 2007, 02:30 PMPlato
Think carefully about what the question asks!

“What is the probability that**neither**you nor the dealer is dealt a blackjack?”

That is neither event happens: .

You don’t get a blackjack and the dealer does not get a blackjack.

Now use your answer for to finish. - August 14th 2007, 02:34 PMkkoutsothodoros
yes, i subtract it from 1. but was my combinatorial reasoning correct in coming up with P(A and B)?

- August 14th 2007, 02:42 PMPlato