Black Jack probability questions.?

Question

What is the probability of getting 2 blackjacks given you have 2 split aces?

and

After you insured your cards, what is the probability of winning insurance assuming a fresh deck and no face cards are showing?

Thank you for all your help!!

Answer 1

The answer to the first part is zero, because even when your ace is hit with a face card, it is not considered "blackjack" (which pays 3 to 2).
Assuming a 52 card deck, you have used 2 cards so there is a 50 card deck remaining. There are 16 cards that are tens. So there is a 16/50 chance of getting the first 21 and now there are 15 tens left out of 49 cards for a 15/49 chance. So the answer is 16x15/50x49 =0.098

Insurance means that the dealer has an ace and so do you.
You are betting even money that he does not have a ten underneath. Since 3 known cards are used (2 aces and one ten) there are 49 unknown cards remaining. 15 of them are tens. The dealers chance of blackjack is 15/49. (not such good odds for an even bet)

Answer 2

There are 16 ten-point cards in a 52-card deck. Most deals use a 6-deck stack, totaling 312 cards. The probability depends on how many cards you can see. The simplest assumption is that you are playing alone against the dealer, and see only 1 other card. That leaves 309 cards. If dealer's card is not 10 points,
P(2 blackjacks) = (96/309)(95/308) = 0.095827
If dealer's card is 10 points,
P(2 blackjacks) = (95/309)(94/308) = 0.093830
The probabilities get better if there are other players with no 10-point cards showing, and worse if there are other 10-point cards showing.

and

P(win) = 96/309 = 0.310680

Answer 3

1. it depends on how many decks and the dealer's upcard.
with one deck and the dealer having a non 10-count card, there are 49 cards you haven't seen and 16 of them produce 21. 16/49 *15/48 = 5/49

2. it depends on how many non-ten count cards you hold. Assume 0. Then there are 49 cards left, and 16 of them make dealer BJ. So prob. of BJ = 16/49

Answer 4

what is blackjack?