in blackjack it is obvious that the dealer will break 8 times out of 13 when hitting his 16, does the same percentage apply when he is hitting his 12 thru 15?
No. search on blackjack statistics tables.
Thanks, but i also need to know why
A sentence like this triggers the “homework help” thought. But I will give you this much: Try looking at the Wizard of Odds. He’s good about explaining his math, but I don’t know if he’s got an explanation of the bust odds.
Besides, how many decks? What’s the count?
Let’s make this simple (i.e. forget the “count” and assume an infinite number of decks). Each rank has a 1/13 chance of being picked. Add the fact that an ace is 1 or 11 and there are four ranks that count as 10.
With a 16, the dealer doesn’t bust if he get an A-5 so he has a 1-5/13 or 8/13 chance of busting.
With a 15, the dealer gets a A and draws again (see above) or a 2-6 to stand. The probability of not busting is approximately (1/13 x 5/13) + (5/13) which is clearly bigger than 5/13 so the probablity of busting with a 15 is slightly less than with a 16.
A similar process shows that as the hand goes down (at least to 12), so does the probability of busting decreases. Knowing the count and number of cards known would give you the exact probabilities
Thank you. This was not hoework help. I have been to wizard of odds but i don’t think he discusses this precise question, at least I could not find it. Intuitively I arrived at this same conclusion but seeing your clear explanation puts my mind at ease. Thanks again.