If you mean that you are not dealt a blackjack, and keep hitting until you reach 21 or bust, you have a 13.0% chance of getting 21, and an 87.0% chance of busting. If you mean you have two cards that are not blackjack, and hit, what your chances are of getting 21 with one card: If your total is under 10 or over 20, zero.
Blackjack odds of Busting While Taking a Hit. This chart shows the probabilities of going bust ...
The probability of the dealer getting exactly a 9-card 21 under those rules is 1 in 32,178,035. Here is the probability for various numbers of decks and whether dealer hits or stands on soft 17. Probability of Dealer 9-Card 21
Is it possible to determine the probability and odds for getting a total of exactly 21 on your first 3 cards? This would be on any three cards totaling 21. If the probabilities for getting a blackjack mean that you’ll get one every 21 hands, how many hands will it take to get exactly 21 on the first three cards?
Your probability of getting an ace and then a 10 is 1/7 X 16/27, or 16/189. Again, you could get a blackjack by getting an ace and a ten or by getting a ten and then an ace, so you add the two probabilities together. 16/189 + 16/189 = 32/189. Your chance of getting a blackjack is now 16.9%.
If we want to calculate the probability of achieving 19, 20 or 21 points, all we must do is total the three probabilities just calculated. We obtain P = 9/47 = 19.14893%.
The odds to get a blackjack (natural) as arrangement: 128 / 2652 = .0483 = 4.83%. 4.83% is equivalent to about 1 in 21 blackjack hands. (No wonder the game is called Twenty-one!) Calculations for the Number of Cards Left in the Deck, Number of Decks
Another way of looking at it is to turn it upside-down to find the odds: 663 ÷ 32 is (aptly) about 21 – so if you played 21 games you'd only expect to get one blackjack. Something to think about How many games would you have to play so that the probability of getting at least one blackjack was greater than 0.5?
Blackjack being $2$ cards adding to $21$ i.e. $Ace + 10,J,Q,or K$ (or vice versa as order does not matter). The farthest I've really come is that the odds of the first player getting dealt a blackjack is $128\over 2652$. First case: Odds of getting an Ace are $4\over52$, odds of the next being 10,J,Q,or K are $16\over51$.