To convince yourself of this, suppose that the dealer is initially dealt only one card, but when you hit, she first draws a card for herself and then hits you. This is clearly equivalent to the actual game situation. The probability of you busting does not depend directly on the card the dealer draws for himself, but in the card you receive. For any specific card in the remaining deck, the probability that you get it is the product of the probability that the dealer did not receive the card and the probability that you did receive the card: (48/49) · (1/48) = (1/49), just like the probability of getting any given card if the dealer were not dealt another card.
With this, we can solve the problem. You will bust if you get a card from 8 to 10 or a face card. There are 6 · 4 = 24 of these cards in a regular deck, but the game is showing 2 of them, so there are 22 remaining. Thus, the probability of you busting is 22/49 &asymp 0.45. The fact that the dealer was dealt a card whose value you don't know does not affect your knowledge of the remaining cards in the deck and you can consider probabilities as if it was still in the deck.
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