I just looked up the Stanford Wong PBJ book. It shows that for the hand of (7,7)vs8, the HiLo index of hit/splitting is TC>=+5 for a 4-deck s-17 shoe but TC>=-1 for a 1-deck s-17 shoe. This seemingly supports my argument of playing different indices at different levels of dealing depths for the same 8-deck shoe with a fixed penetration of 7.5 decks.
Oh no. It does nothing of the sort. Of course indices are different according to the total number of decks -- but not with remaining decks. If you start with four decks and have one left, the one left is still based on four decks.
Example, what are the odds of getting a BJ? It's the odds of getting an ace then a ten plus the odds of getting a ten and an ace since either is a natural.
Single-deck (52 cards, 4 aces and 16 tens): (4/52 * 16/51) + (16/52 * 4/51) = 0.04826546003
Double-deck (104 cards, 8 aces and 32 tens): (8/104 * 32/103) + (32/104 * 8/103) = 0.04779686333
Simple combinations. That's true at the start, middle, or end of the pile of cards.
"I don't think outside the box; I think of what I can do with the box." - Henri Matisse
Card combination only matters on single deck in some cases. Multiple deck (2, 4, 6, 8D) indexes are almost identical because there are so many cards in the shoe so different card combination almost doesn't matter. For example, you only need to memorize the index 16 v 9 and don't need to memorize different indexes for AAT22 v 9 or 79 v 9. So Wong's books like all other BJ books just list two different index tables. One for single deck. And the other for multiple decks (applying to double decks, six decks and eight decks.) (Technically there are four kinds of tables because S17 and H17 indexes are different.)
For a 8-deck shoe, my practice in casinos has been like this: using different indices at different levels of dealing depths if the penetration was fixed at 7.5 decks. For example, when 7 decks of cards has been dealt, I would take insurance when the TC>=1.4 because I always treated the remaining decks as a new shoe.
Suppose you have a deck of cards that consists of two cards, a two and a three. Suppose you have one deck and draw two cards. What are the odds of drawing a total of 5? Obviously 100%. Now suppose you have two decks, two twos and two threes. What are the odds of drawing two cards for a total of five? The possibilities:
2 2 = 4
2 3 = 5
3 2 = 5
3 3 = 6
Your odds have dropped to 50%
Now, for each of the four draws above, what will be the second hand? Well, there will only the two cards left:
3 3 = 6
3 2 or 2 3 =5
3 2 or 2 3 =5
2 2 = 4
Still 50%.
What matters is total decks, not remaining decks.
"I don't think outside the box; I think of what I can do with the box." - Henri Matisse
Norm has given you an argument. I'll give you a different one. Suppose you're using your strategy and indices for a six-deck game and, suddenly, the shoe drops on the floor and five decks spill out, leaving only one deck remaining in the shoe. Are you now suddenly playing a single-deck game? Do you understand how utterly ridiculous such an argument would be? Or, in similar fashion, if you held the six decks in your hands and, instead of dealing from the top, you turned the pack over and began dealing from the bottom deck only, would you now be playing single-deck blackjack? Oh!
Nonetheless, there is an entirely different concept that is valid, and that is the basis for the floating advantage. If you are counting a shoe game, and you get to a point where only one deck is remaining, AND, you have a count of, say, zero, then your edge at that moment is, in fact, consistent with that of a single deck off the top. But this is a very different idea from the one that you have been espousing.
Don
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