TITLE: Calculating probability neither you nor dealer gets blackjack-please clarify symmetry comment
QUESTION [0 upvotes]: This question was asked and answered in this link:
Probability that Both Player and Dealer are not Dealt Blackjack
However, I still don't understand why due to symmetry the probability for getting black jack is the same for either you are the dealer. When the cards are dealt out, a total of 4 cards are dealt out. How come this doesn't come into play? Say one person is dealt first. Now you have 50 cards to deal the other player from. Wouldn't you calculate the probability of the second player getting blackjack based on the cards dealt to the first player?
Also, in a real blackjack game, the player gets one card and then the house. Doesn't that once again change the probability of getting blackjack for the two players?

REPLY [1 votes]: Imagine dealing four cards face down on the table. What is the probability that the first two cards consist of an ace and a 10/face? What is the probability that the last two cards consist of an ace and 10/face? The probabilities are the same; either way, you are looking at two random cards. But this is exactly the situation as the blackjack deal, as far as probabilities are concerned.
The same is true when the cards are dealt in a different order. Either way, both the player and the dealer have two random cards.
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