K kkoutsothodoros Jun 2007 63 0 Aug 14, 2007 #1 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?

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?

K kkoutsothodoros Jun 2007 63 0 Aug 14, 2007 #2 a blackjack is when you are dealt an ace and either a ten, jack, queen, or king.

K kkoutsothodoros Jun 2007 63 0 Aug 14, 2007 #3 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.

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.

K kkoutsothodoros Jun 2007 63 0 Aug 14, 2007 #4 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.

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.

P Plato MHF Helper Aug 2006 22,507 8,664 Aug 14, 2007 #5 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: \(\displaystyle P\left( {\overline A \cap \overline B } \right) = P\left( {\overline {A \cup B} } \right) = 1 - P\left( {A \cup B} \right)\). You don’t get a blackjack and the dealer does not get a blackjack. Now use your answer for \(\displaystyle P\left( {A \cup B} \right)\) to finish.

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: \(\displaystyle P\left( {\overline A \cap \overline B } \right) = P\left( {\overline {A \cup B} } \right) = 1 - P\left( {A \cup B} \right)\). You don’t get a blackjack and the dealer does not get a blackjack. Now use your answer for \(\displaystyle P\left( {A \cup B} \right)\) to finish.

K kkoutsothodoros Jun 2007 63 0 Aug 14, 2007 #6 yes, i subtract it from 1. but was my combinatorial reasoning correct in coming up with P(A and B)?

P Plato MHF Helper Aug 2006 22,507 8,664 Aug 14, 2007 #7 kkoutsothodoros said: was my combinatorial reasoning correct in coming up with P(A and B)? Click to expand... Yes that is what I got. Reactions: kkoutsothodoros

kkoutsothodoros said: was my combinatorial reasoning correct in coming up with P(A and B)? Click to expand... Yes that is what I got.