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?
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.
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.