1. ## blackjack 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?

2. a blackjack is when you are dealt an ace and either a ten, jack, queen, or king.

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.

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.

“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.

6. yes, i subtract it from 1. but was my combinatorial reasoning correct in coming up with P(A and B)?

7. Originally Posted by kkoutsothodoros
was my combinatorial reasoning correct in coming up with P(A and B)?
Yes that is what I got.