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Is there a formula to calculate the probability of busting in the Black Jack game after you are dealt two cards? I want the game to be just between the dealer and me. I do not want to know what the probability of the dealer busting is. So I am dealt two cards, I want to know what will be my probability of me busting after I decide to hit, or stand.

I am new to probability and I have had some hard time trying to figure this out.

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  • $\begingroup$ There are many factors involved. Are you using one deck or many? Can you see the dealer's up card or other players cards? $\endgroup$ – turkeyhundt Dec 7 '14 at 22:37
  • $\begingroup$ Can you spell out what you mean? You get two cards initially. You can't bust with those. Then you get to look at them, and you'll see what you have. Then the probability you'll bust depends on what you have to start. (And also what other cards you can see, and the total deck size.) Do you want two more cards? Without looking in between? $\endgroup$ – aes Dec 7 '14 at 22:38
  • $\begingroup$ I want the game to be just between the dealer and me. I do not want to know what the probability of the dealer bursting is. So I am dealt two cards, I want to know what will be my probability of me bursting after If decide to hit, or stand. $\endgroup$ – danilojara123 Dec 7 '14 at 22:44
  • $\begingroup$ I am using one deck $\endgroup$ – danilojara123 Dec 7 '14 at 22:57
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If this is a new deck or you have no knowledge of the cards that have gone before, then there are 3 known cards (your 2 and the dealer's 1 face-up).

If you have an Ace or your total is 11 or less then there is no chance of you busting.

If you have 20 then you will bust on anything but an Ace. If the dealer is not showing an Ace then the chance of busting is $\frac{49-3}{49}$. If he is showing an Ace then the chance is $\frac{49-2}{49}$.

The result is similar for 19 (Ace or 2), 18 (Ace, 2 or 3), 17 (Ace, 2, 3 or 4) & 16 (Ace, 2, 3, 4 or 5).

For 12 to 15 you also need to account for the fact that you may be holding some of the "safe" cards (e.g. 9 + 6 eliminates one of the "safe" 6s).

This can be extended to any number of decks.

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