Black Jack Bust Probability Can you explain me how to calculate the probability that the dealer in black jack game will bust. (more than 21 points)
There are different opinions in the relevant literature, so i would like to know what you think.
Specification: 6 deck Black Jack (6x52 cards), continious shuffle (shuffle machine), dealer stands on soft 17.
The literature mentions 0,2816 or 0,2819 as the probability.
What do you think? How to calculate?
Thank you so much!
 A: There are, fortunately, not all that many hands!  Since suit doesn't matter, we only really care about the ten usable ranks - since 10s and face cards are identical, rules-wise, we can combine them all.  We can begin by generating sequences of draws; we will terminate each sequence when we reach 17 points.  This gives some $54\,433$ hands, from the pedestrian two 10s to the absurd six aces, a 6, four more aces, and a 5.  Then, each hand can be evaluated for probability and whether is a bust: since we know how many of each card are in the deck, we can evaluate the likelihood of drawing each particular card next.
We're kind of fortunate here: every final hand is prefix-free, that is to say there is no final hand that starts with another final hand, so we don't have to consider the possibility that a particular hand might be overshadowed by another one, so we can simply add up the probabilities.
With a short script, I have found the probability of a bust on a six-shoe deck with stand-on-soft-17 rules to be slightly over $0.2819$, or if we wish to be precise, $$\frac{16674395838503610450371}{59145560742900234888285}$$
