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Thread: 3,3 VS 2 index DD NDAS

  1. #14
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    Quote Originally Posted by JohnGalt007 View Post
    Thanks for the update! How would you revise the calculation to account for 6,6?
    I don’t understand what Don is talking about, but here is what I would calculate. In this part, get ride of the 6,6 pair contribution to the hard 12 hand.
    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.”

    You only count the probability of getting a hard 12, excluding a 6,6 pair. Let me comment a little further. Don has threatened to ban me off from this website a few times!
    Last edited by aceside; 10-14-2024 at 12:52 PM.

  2. #15
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    Quote Originally Posted by DSchles View Post
    "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
    Thanks Don! I appreciate it. I had assumed my work was correct, but I'm always open to revisions if they're needed. I used combinations in my formulas above so I can more easily generalize the same process to other hands. Writing out the work in fraction form leaves fewer clues as to how to do that.

  3. #16
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    Quote Originally Posted by aceside View Post
    I don’t understand what Don is talking about, but here is what I would calculate. In this part, get ride of the 6,6 pair contribution to the hard 12 hand.
    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.”
    Hi Aceside, I think the issue here is that we only need to treat 6-6 separately from the other combinations that yield a hard 12 total for the purposes of deciding what to do with that hand (i.e. calculating the conditional EV of hitting/standing/doubling down/splitting with 6-6, versus calculating the conditional EV of hitting/standing/doubling down with 2-T, 3-9, 4-8, or 5-7). We don't need to do that for the purposes of calculating the probability of being dealt a hard 12 and how that changes the true count.

  4. #17
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    I’ll take a look at this part tonight.

  5. #18


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    We'll all rest easier once you do!

    Don

    P.S. I don't/can't ban anyone from this site. That's Norm's domain. But I have been known to make suggestions!

  6. #19


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    Quote Originally Posted by DSchles View Post
    P.S. I don't/can't ban anyone from this site. That's Norm's domain. But I have been known to make suggestions!
    I was thinking that, too.

    Come on Norm. This month is aceside's 6 year anniversary on this forum. Nothing has changed. His troll tactic of posting incorrect information hasn't changed in all that time. Please get rid of him permanently.
    Last edited by 21forme; 10-15-2024 at 08:52 AM. Reason: typo

  7. #20
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    Apparently someone disagrees with me. Still, let me post my calculation. A 2-card hard-12 includes only these four hands:

    T, 2;
    9, 3;
    8, 4;
    7, 5.

    The probability that you are dealt a hard-12 vs. 4 is 0.00375+0.00094+0.00082+0.00094 = 0.00645.

    Hence, the expected true count conditioned on the event that you are dealt a hard-12 vs. 4 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094) (2/(101/52)))/0.00645 = +0.726.

    This is slightly off from John’s result.





  8. #21
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    Quote Originally Posted by aceside View Post
    Apparently someone disagrees with me. Still, let me post my calculation. A 2-card hard-12 includes only these four hands:

    T, 2;
    9, 3;
    8, 4;
    7, 5.

    The probability that you are dealt a hard-12 vs. 4 is 0.00375+0.00094+0.00082+0.00094 = 0.00645.

    Hence, the expected true count conditioned on the event that you are dealt a hard-12 vs. 4 is ((0.00375)(1/(101/52))+(0.00094)(2/(101/52))+(0.00082)(2/(101/52))+(0.00094) (2/(101/52)))/0.00645 = +0.726.

    This is slightly off from John’s result.




    Hi aceside, I'm aware that this is the calculation assuming that we don't count 6-6 as a hard 12. My question is, why would that be a valid assumption? We "count 6-6 separately" from other totals of 12 solely for the purposes of evaluating the conditional expectation of the playing strategies available to the player, because we can split a pair of sixes whereas we cannot split a 2-T, 3-9, 4-8, or 5-7. None of that matters for the purposes of calculating the average true count conditioning on being dealt a hard 12. The discrepancy in understanding lies in the definition of "hard 12" for the purposes of calculating the average true count.

  9. #22


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    Quote Originally Posted by JohnGalt007 View Post
    Hi aceside, I'm aware that this is the calculation assuming that we don't count 6-6 as a hard 12. My question is, why would that be a valid assumption? We "count 6-6 separately" from other totals of 12 solely for the purposes of evaluating the conditional expectation of the playing strategies available to the player, because we can split a pair of sixes whereas we cannot split a 2-T, 3-9, 4-8, or 5-7. None of that matters for the purposes of calculating the average true count conditioning on being dealt a hard 12. The discrepancy in understanding lies in the definition of "hard 12" for the purposes of calculating the average true count.
    I have to agree. For whatever reason, I seem to have a plethora of 66 v2 hands. Though I avoid NDAS games, hitting vs splitting is not an unusual play. As stated above, “it is a conditional expectation of the playing strategies available”

    It comes out to - what is the index.

  10. #23


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    Quote Originally Posted by Freightman View Post
    It comes out to - what is the index.
    Well, then, to be fair, the index for 6,6, vs. 2 is very different from standing on hard 12 vs. 2. There's probably more to this discussion than meets the eye.

    Don

  11. #24


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    Hi, friends!

    Enjoyed this thread. Thank you, all. I couldn’t resist this time to add my contribution, too.

    roliin wrote:

    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.

    2 Decks, H17, NDAS, 3,3 v 2

    Hitting -0.142295496
    Splitting -0.202137284
    Frequency 0.000410022

    What the simulated index is telling us, basically, is that to overcome the difference (6%) in gains for action 1, we need to be patient and avoid splitting. Nothing strange, therefore, with the appearance of such a high index.


    John Galt wrote:

    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.


    2 Decks, H17, 12 v 4
    Hitting Standing
    12v4 -0.206889222 -0.195748581
    Hitting Standing Frequencies
    T,2 -0.204205449 -0.206124526 0.003748774
    9,3 -0.219573436 -0.195903587 0.000937193
    8,4 -0.214138826 -0.191941785 0.000820044
    7,5 -0.203561198 -0.178661647 0.000937193
    Total 0.006443205
    Hitting Splitting Frequency
    6,6 -0.202729595 -0.069138850 0.000410022
    Here we can see the 6,6 index playing in another league, and thus it must be kept separate. A mandatory split!


    With accurate EORs for 2 decks, h17, ndas.

    Generic: 12 v 4 Stand if TC >= -0.157476

    Composition-dependent:

    Hand EORs Indices Occurrence Product
    T,2 0.622041 0.003748774 0.002331891
    9,3 -0.557348 0.000937193 -0.000522343
    8,4 -0.448992 0.000820044 -0.000368193
    7,5 -0.615712 0.000937193 -0.000577041
    Totals 0.006443205 0.000864313
    Weighted Index 0.134143418

    The full-deck favorability, m2, for carrying out the action (hitting) equals -0.371119 (see BJA3, page 519). Nevertheless, here the CD index for T,2 vs. 4 behaves itself like the head of operations, and determines the final fate of the index. In other words: The majority of the time, 58%, your index to stand will be 0.7 by rounding up. On the other hand, for the rest of the two-card holdings, 42% of the time, your index to stand will be negative, around -0.5. What this sort of inner contradiction is telling us is that the use of composition-dependent indices, totally worthless when dealing with shoes, is not a bad idea at all for a DD enthusiast, or for others addicted to saving pennies.

    Enjoy!

    Sincerely,

    Zenfighter

  12. #25


    2 out of 2 members found this post helpful. Did you find this post helpful? Yes | No
    Readers: It isn't often that Zenfighter graces our pages with his presence. Always wonderful to see one of the giants in BJ research come around for a visit.

    Don

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