In table 26.2 of Don/Gron's book, the index is a 10. Does this mean you would not hit a 3,3 VS 2 @ +10? If that's what the sims say that's what they say. It just seems counter intuitive not to hit a hand you can't bust, especially against a two. Am I missing something?
Ordinarily, in DD NDAS games, the optimal strategy against a pair of 3s against a 2 is to hit. However, at and after a true count of +10, the optimal strategy with a pair of 3s against a 2 is to split your pair. This is because the expected value of splitting your 3s against becomes greater than the expected value of hitting a hard 6 composed of a pair of 3s given that the true count is at least +10.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.
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.
No problem! Happy to help.
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/(101/52). 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/(101/52). 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/(101/52). 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/(101/52). 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/(101/52). 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/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094)(2/(101/52))+(0.00041)(3/(101/52)))/0.00686 = 0.779.
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 up from 0.779), 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 to an integer 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.
Last edited by JohnGalt007; 08-13-2024 at 09:18 AM.
Thank you for the thorough answer. I'm still a little confused tho. "If you have 6,6 and the dealer has 4, the running count is 1+1+1=3. and the true count is 3". Why would the TC be 3 if the RC is 3 and nearly 2 full decks left out of 2? Wouldn't the TC be 1? Or are you just pointing out that this card combination (6,6 vs 4) with a TC of 3 happens 0.00041 percent of the time?
Thank you for your reply! I'll modify my post to reflect the fact that you should divide by the number of decks remaining, which is 101/52, rather than multiply by 101/104. This is what happens when you try to post accurate math after a 15-hour long coding tear!
For the hand 6,6 vs. 4, a pair of player 6’s cannot be treated as a hard 12. All these pair hands must be treated differently from hard total hands. This changes the probability numbers and thus means that the above calculation still needs to be revised.
"How would you revise the calculation to account for 6,6?"
You wouldn't, especially if aceside tells you it's wrong. It is, of course, perfectly correct. I don't care for your notation, though, which is clumsy and confusing. For 6,6 vs. 4, why not the much clearer and intuitive:
8/104 x 7/103 x 8/102 = 448/1092624 = 0.00041?
Don
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