blackjack proabability question

Jun 2007
63
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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?
 
Jun 2007
63
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a blackjack is when you are dealt an ace and either a ten, jack, queen, or king.
 
Jun 2007
63
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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.
 
Jun 2007
63
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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.
 

Plato

MHF Helper
Aug 2006
22,507
8,664
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.
 
Jun 2007
63
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yes, i subtract it from 1. but was my combinatorial reasoning correct in coming up with P(A and B)?