If you use 2 key cards in an 8-deck shoe, after you see those key cards, you would bet more and play more spots as to receive the Ace, right? But how frequently does the Ace, due to the shuffle, end up nowhere near those 2 key cards? I assume it depends on the shuffle, but is there an answer "in general"?
Do you have to map the shuffle yourself (by simulation) and then figure out the probability that the Ace follows those 2 key cards by 2, 3, 4, 5, 6, ... cards?
I assume that the advantage is significant enough that you wouldn't even need to wong out too much or count cards even if this occurs once per shoe (or 3 times per hour).
Do you have to map the shuffle yourself (by simulation) and then figure out the probability that the Ace follows those 2 key cards by 2, 3, 4, 5, 6, ... cards?
I assume that the advantage is significant enough that you wouldn't even need to wong out too much or count cards even if this occurs once per shoe (or 3 times per hour).