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Originally Posted by

**roliin**
Not sure why I was so hung up on that last night. Trying to memorize indices for too many games I guess and forgetting about not splitting 3,3 in a NDAS game. I don't ordinarily play DD or NDAS. Thanks for clearing that up.

No problem! Happy to help.

Originally Posted by

**roliin**
Another question. DD, H17 the index for 12 vs 4 is 1. So anything below 1 you would hit, correct? If that's the case than how come BS isn't to hit 12 vs 4? Surely the count is below 1 the majority of the time.

Basic strategy is calculated on the premise that the deck from which all other possible hands that could be dealt is comprised of the double-deck minus the three cards that are on the felt, namely the two cards comprising your hard 12 and the dealer's upcard of 4. The EV of standing, hitting, and/or doubling down on 12 against a 4 with 2 decks and H17 is thus calculated using the 52*2 - 3 = 101-card shoe that's left to be played. In this case, assuming no knowledge of the remaining cards, the EV of standing is -17.87%, whereas the EV of hitting is -20.36%. Since the EV of standing is slightly larger, basic strategy recommends a "stand" in this situation. Importantly, removing the two cards that comprise your 12 as well as the dealer's upcard of 4 from this shoe changes the true count accordingly:

• If you have T,2 and the dealer has 4, the running count is -1+1+1 = 1, and the true count is 1*(104/101). This happens with probability (C(32,1)C(8,1)C(8,1))/(C(104,2)C(102,1)) = 0.00375.

• If you have 9,3 and the dealer has 4, the running count is 0+1+1 = 2. and the true count is 2*(104/101). This happens with probability (C(8,1)C(8,1)C(8,1))/(C(104,2)C(102,1)) = 0.00094.

• If you have 8,4 and the dealer has 4, the running count is 0+1+1 = 2. and the true count is 2*(104/101). This happens with probability (C(8,1)C(8,1)C(7,1))/(C(104,2)C(102,1)) = 0.00082.

• If you have 7,5 and the dealer has 4, the running count is 0+1+1 = 2. and the true count is 2*(104/101). This happens with probability (C(8,1)C(8,1)C(8,1))/(C(104,2)C(102,1)) = 0.00094.

• If you have 6,6 and the dealer has 4, the running count is 1+1+1 = 3. and the true count is 3*(104/101). This happens with probability (C(8,2)C(8,1))/(C(104,2)C(102,1)) = 0.00041.

Furthermore, the probability that you are dealt a 12 is 0.00375+0.00094+0.00082+0.00094+0.00041 = 0.00686

Hence, the expected true count conditioned on the event that you are dealt a hard 12 is ((0.00375)(1*(104/101))+(0.00094)(2*(104/101))+(0.00082)(2*(104/101))+(0.00094)(2*(104/101))+(0.00041)(3*(104/101)))/0.00686 = 1.558.

Note that I calculated the running count with the tags from the Hi-Lo system; a more accurate calculation would require the effects of removal for each card, not the approximations afforded by the tags.

The same is true when using the true count to adjust your playing decisions. When the true count drops below the expected true count of 1 (rounded down from 1.558), this is an indicator that the composition of the shoe has changed sufficiently to allow for the EV of hitting to surpass the EV of standing. This indicator is not 100% accurate, as the actual true count will be a real number with many decimal places and not a convenient integer. The errors of estimating the EV of the game are well-documented, e.g. Griffin's "Theory of Blackjack" and Schlesinger's "Blackjack Attack"; for the purposes of this discussion, suffice it to say the indexes are usually rounded down to reduce either the variance in one's bankroll when we choose one action over another, or in the penalty of making a technically wrong playing decision in those rare instances when the count (known to us during a game) indicates one action but the actual EV of the remaining shoe (unknown to us during a game) dictates another action.

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