Chances of going bust on the second hit? (blackjack) In a game of blackjack/twenty-one between 2 players, the dealer and another player, the dealer shuffles a regular 52 card deck and deals himself 6H,6D and the player 8S,8C. What is the probability that the player does not go bust on the second hit (providing he didn't go bust on the first)? -Aces have a value of 1, face cards have a value of 10. Dealer has 6h6d and player has 8s8c, so player's score before hitting is 16. To not bust on the first go, he needs a card lower than a 6. There are 2+4+2+4+16 = 28 cards that would make him go bust on the first go, out of 48 cards so probability of not busting first go = 20/48 = 5/12. Chances for the second hit: All 28 cards that would have made him go bust on the first hit are still in the deck of now 47 cards. I have tried to go from here but all my methods end up with probabilities that cannot be correct as they do not have a sum total of 1. Any help would be appreciated, thanks.

The player doesn't go bust on the second hit if she draws a pair of cards that sum to $5$ or less. All cards up to $5$ are still in the deck. The admissible pairs are $(A,A)$, $(A,2)$, $(A,3)$, $(A,4)$, $(2,2)$, and $(2,3)$, which is $2$ pairs of equal ranks of which there are $\binom42=6$ each and $4$ pairs of different ranks of which there are $4^2=16$ each, for a total of $2\cdot6+4\cdot16=76$ pairs out of $\binom{48}2=1128$, so the probability is $\frac{76}{1128}=\frac{19}{282}\approx6.7\%$.