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Thread: Unplayable splitting strategies

  1. #1


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    Unplayable splitting strategies

    I made a speculative, in-passing, I'm-pretty-sure-mostly-ignored comment buried in an earlier thread, that having more time to think about makes me a bit more confident that I'm not completely off in the weeds. I welcome your thoughts on the following:

    Blackjack CAs offer a variety of options for specifying how pair split hands may be played (CDZ-, CDP, etc.). My point in this post is to suggest that you probably *never* want to select some of those options-- including one supported in my CA-- because they represent a playing strategy that you can't actually execute at the table.

    Short version: specifically, CDP (supported by my CA as well as by MGP's) and CDPN (supported by MGP's) are effectively "unplayable." Instead, you almost always want either CDZ- (supported by mine, MGP's, and k_c's), CDP1 (supported by mine, as well as k_c's), or truly optimal, which is supported only by ICountNTrack's.

    Note that this isn't an issue of strategy "complexity." Granted, those CDP and CDPN options do represent relatively complex strategies, since they specify stand/hit/double actions that depend on more information than the "simpler" CDZ- and CDP1. But compare with ICountNTrack's truly optimal strategy calculation: it's even more complex-- much more so-- than CDP or CDPN, but it's still possible at least *in principle* to write it down, so to speak, in a way that a player could actually follow that optimal strategy at the table. This isn't the case for CDP (or by extension, for CDPN).

    Nor is this an issue of accuracy or exactness of the CA calculations involved: the CDP and CDPN options do indeed represent well-defined random variables whose expected returns can be computed exactly. But it's impossible to sit down and play a round with a strategy whose outcome corresponds to either of those random variables.

    It's an interesting problem to prove this-- more to follow.

    Eric

  2. #2


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    Quote Originally Posted by ericfarmer View Post
    I made a speculative, in-passing, I'm-pretty-sure-mostly-ignored comment buried in an earlier thread, that having more time to think about makes me a bit more confident that I'm not completely off in the weeds. I welcome your thoughts on the following:

    Blackjack CAs offer a variety of options for specifying how pair split hands may be played (CDZ-, CDP, etc.). My point in this post is to suggest that you probably *never* want to select some of those options-- including one supported in my CA-- because they represent a playing strategy that you can't actually execute at the table.

    Short version: specifically, CDP (supported by my CA as well as by MGP's) and CDPN (supported by MGP's) are effectively "unplayable." Instead, you almost always want either CDZ- (supported by mine, MGP's, and k_c's), CDP1 (supported by mine, as well as k_c's), or truly optimal, which is supported only by ICountNTrack's.

    Note that this isn't an issue of strategy "complexity." Granted, those CDP and CDPN options do represent relatively complex strategies, since they specify stand/hit/double actions that depend on more information than the "simpler" CDZ- and CDP1. But compare with ICountNTrack's truly optimal strategy calculation: it's even more complex-- much more so-- than CDP or CDPN, but it's still possible at least *in principle* to write it down, so to speak, in a way that a player could actually follow that optimal strategy at the table. This isn't the case for CDP (or by extension, for CDPN).

    Nor is this an issue of accuracy or exactness of the CA calculations involved: the CDP and CDPN options do indeed represent well-defined random variables whose expected returns can be computed exactly. But it's impossible to sit down and play a round with a strategy whose outcome corresponds to either of those random variables.

    It's an interesting problem to prove this-- more to follow.

    Eric

    In my estimation the most basic decision regarding a pair is whether or not to split in the first place. In order to answer this most basic question some sort of post split strategy needs to be considered. Just about any reasonable post split strategy will agree on the answer for a full shoe. The keyword here is reasonable, meaning a "good enough" estimate of optimal CDZ. If your post split strategy is to split at every opportunity and hit until you bust, don't split!

    I think what you may be talking about is merging pre and post split strategies into one and only one strategy that is the best overall basic strategy. For a single deck this may be a problem but as more decks are added this happens naturally. I abandoned CDP and CDPN because I found that they are not necessarily more optimal than another lesser information strategy.

    k_c

  3. #3


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    Quote Originally Posted by k_c View Post
    I think what you may be talking about is merging pre and post split strategies into one and only one strategy that is the best overall basic strategy. For a single deck this may be a problem but as more decks are added this happens naturally. I abandoned CDP and CDPN because I found that they are not necessarily more optimal than another lesser information strategy.
    I should probably additionally clarify: not only is this not about complexity, or about accuracy, but it's also not about optimality. That is, let's ignore for now the question of which playing/splitting strategies yield *better* or *worse* expected returns. The issue here is that the expected return being computed-- no matter whether it's large or small-- can't actually be *realized* at the table.

    Above is my response to your comment; below is the original follow-up that I was composing and intending to post:

    Following up on the OP, we'll prove that CDP strategy is not playable with an explicit counterexample. Imagine playing a round (S17, DAS, SPL3) from a depleted shoe with just sixteen cards remaining: eleven 6s and five 8s. This is a conveniently constructed example for a couple of reasons. First, the shoe is small enough that we can brute-force enumerate all possible arrangements of cards and corresponding outcomes of the round (there are only 4368 of them, and we'll need even fewer as we'll see shortly), allowing us to explicitly verify any fancier calculations that our CA might report.

    For example, what is the overall expected return from the round, assuming that we play CDZ- strategy for this shoe (the simplest strategy to implement, where we are forced to always take the same action for a given hand, whether it is from the initial deal or drawn to a split pair)? Our CA reports 0.49198717948717946 (as a fraction of initial wager), which we can confirm by brute force as 2149/4368. (Note that MGP's CA has a bug that manifests for small shoes like this, with the current version incorrectly indicating an overall expected return for CDZ- strategy of 0.0652472527472528. That's okay-- we'll be focusing on CDP instead shortly.)

    Now let's further focus our example to consider a particular initial deal of a pair of 6s against the dealer's 8, i.e., so that nine 6s and four 8s remain face-down in the shoe, of which only 715 possible arrangements remain.

    The CDZ- strategy is to split the pair, resplitting at every opportunity... and otherwise doubling down on each resulting split hand. This is the second reason this is a convenient example: playing strategy is so simple to specify, and thus to evaluate and verify by brute force. Here, the resulting expected return is 2.1202797202797203, which we can again confirm as exactly 1516/715.

    How does the CA compute this value? We can follow the recursive calculations described in this paper, where in this example p=9/13, s=13, n=4, and the "base case" expected returns are:

    Code:
    E[X;0,0] = 74/143
    E[X;1,0] = 34/55,
    E[X;2,0] = 118/165
    E[X;3,0] = 4/5
    E[X;4,0] = 6/7
    
    E[P;0,0] = 106/165
    E[P;1,0] = 134/165,
    E[P;2,0] = 1
    E[P;3,0] = 76/63
    So far, so good. But now instead of CDZ-, let's consider CDP strategy, where we are allowed to vary our playing strategy based not just on the current hand and dealer up card, but also on the number of additional pair cards removed from the shoe. Applied to the same 6-6 vs. 8 hand, the CA reports an improved expected return from the split of 2.165034965034965 (or 1548/715); peeking under the hood, we find that the only difference between CDZ- and CDP is that the value E[X;4,0] has changed to 58/63... where the CDP strategy with 4 additional pair cards removed is to *stand* on a drawn 6-8 instead of doubling down. (This makes some sense: with four additional 6s removed, we are sufficiently more likely to bust our 14 trying to double down.)

    But here we come to the crux of the problem: how do we play this split hand, in a way that will realize this CDP expected return of 1548/715?

    Let's make the question more explicit: suppose that we have split our initial 6-6, and we draw another 6, resplitting... and *another* 6, resplitting again, so that we have the maximum of four incomplete split hands. Now we draw an 8 to the first of these hands. Should we double down on this 14 as we are usually supposed to, or should we stand to realize the improved E[X;4,0] value?

    The problem is that we don't have enough information at this point to make this decision, in a way that will yield the overall expected return of 1548/715. And we *won't* have that information until we have fleshed out at least *two* more of the *subsequent*, currently-incomplete split hands.

    In other words, and to summarize, CDP is not a playable splitting strategy, because of the way split hands are completed. Too many hands may be left *incomplete* (i.e., as single pair cards) to provide the information needed to execute the CDP strategy for the *current* split hand.

    If we imagine a dealer who instead allowed us to split, and resplit at every opportunity, fleshing out *all* resulting split hands to exactly *two* cards, before going back and playing out *any* of them to resolution; *then* it would indeed be possible to execute the CDP strategy to realize the expected return of 1548/715 reported by our CA.

    E

  4. #4


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    The real question here is for Don, and that is “how many hot dogs is this worth??”

  5. #5


    2 out of 2 members found this post helpful. Did you find this post helpful? Yes | No
    Quote Originally Posted by ericfarmer View Post
    I made a speculative, in-passing, I'm-pretty-sure-mostly-ignored comment buried in an earlier thread, that having more time to think about makes me a bit more confident that I'm not completely off in the weeds. I welcome your thoughts on the following:

    Blackjack CAs offer a variety of options for specifying how pair split hands may be played (CDZ-, CDP, etc.). My point in this post is to suggest that you probably *never* want to select some of those options-- including one supported in my CA-- because they represent a playing strategy that you can't actually execute at the table.

    Short version: specifically, CDP (supported by my CA as well as by MGP's) and CDPN (supported by MGP's) are effectively "unplayable." Instead, you almost always want either CDZ- (supported by mine, MGP's, and k_c's), CDP1 (supported by mine, as well as k_c's), or truly optimal, which is supported only by ICountNTrack's.

    Note that this isn't an issue of strategy "complexity." Granted, those CDP and CDPN options do represent relatively complex strategies, since they specify stand/hit/double actions that depend on more information than the "simpler" CDZ- and CDP1. But compare with ICountNTrack's truly optimal strategy calculation: it's even more complex-- much more so-- than CDP or CDPN, but it's still possible at least *in principle* to write it down, so to speak, in a way that a player could actually follow that optimal strategy at the table. This isn't the case for CDP (or by extension, for CDPN).

    Nor is this an issue of accuracy or exactness of the CA calculations involved: the CDP and CDPN options do indeed represent well-defined random variables whose expected returns can be computed exactly. But it's impossible to sit down and play a round with a strategy whose outcome corresponds to either of those random variables.

    It's an interesting problem to prove this-- more to follow.

    Eric

    I am reposting this from the previous thread which I also think was a bit overlooked. It would have been more interesting if the methodology used to compute splits made a big enough difference to alter the optimal playing strategy (like stand instead of splitting)

    my understanding of basic strategy is that its a EV maximizing strategy based solely on the player's hand, dealer's up card, number of decks in the shoe and game rules. Basic strategy has two flavors : composition dependent and total dependent, We would agree that total dependent basic strategy is by far the most used/known because of it's simplicity and effectiveness for shoe games ( i remember though seeing once at a 2D table a guy with a giant composition dependent handwritten cheat sheet ). Also probably note mentioning that total dependent basic strategy is not exactly pure total because it differentiates soft hands and pairs which are really composition dependent, but i digress...

    That being said, all methodologies for computing EV for a split hand offer a means to generate the "Basic Strategy" (which could have been generated using CVData through a sim). My split calculation algorithm where the playing strategy is recomputed at every split hand based on the cards on all the other split hands. While my methodology for computing the expectation value of a split hand offers the perfect play EV maximizing strategy (call it CDO), while EV(CDO) will always be greater or equal to (CDZ, CDP etc...), for current blackjack rules the difference is usually in the 4th decimal and will not result in a discrepancy of a playing decision.

    But suppose we have a game called Blackqueen with some weird rules that results in an optimum playing decision discrepancy between fixed split strategy and re-computed split strategy. A "Basic Strategy" player for Blackqueen will be at a disadvantage playing if he adopted the playing strategy from the fixed split strategy and not really using a "EV maximizing strategy" as defined in the first paragraph.

    Also please note that while during the process of generating the basic strategy, playing strategies are recomputed based on composition of the split hands, for the player using that generated Basic Strategy that's not the case since his decision is always only based on the split hand being played and dealer up card.
    Chance favors the prepared mind

  6. #6


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    This is a great question Eric and I totally agree with you if you use the final hand calculations the way you pointed out.

    What's interesting is that if you use the "effect of removal" hand calculations which also give the exact same EV, you can actually play out the CDPN strategy.

    Fun thought experiment!

  7. #7


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    Quote Originally Posted by MGP View Post
    This is a great question Eric and I totally agree with you if you use the final hand calculations the way you pointed out.

    What's interesting is that if you use the "effect of removal" hand calculations which also give the exact same EV, you can actually play out the CDPN strategy.

    Fun thought experiment!
    Can you explain this comment in more detail? I can't figure out how to interpret to make this make sense. This isn't an issue with the method of calculation of the expected value (it's with the *definition* of the random variable). Are you maybe contrasting CDPN with CDP (where the example described in the post uses CDP)? Because the same issue exists with CDPN; it assumes the ability to vary strategy decisions based on information-- indeed, *more* information than required for CDP-- which isn't generally available at the time the decision must be made.

  8. #8


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    Code:
            'Rounds
            'SPL1	    MGP					            Eric
            'xx	    2*EV(x)					    2*EV(x) 
            '
            'SPL2	    MGP					            Eric 
            'NN	    EV(N) + EV(N-N)			            2*EV(N-N) 
            'Pxxx	    3*EV(x-P)				            3*EV(x-P) 
            'NPxx	    EV(N) + 2*EV(x-PN)		                    EV(N-P) + 2*EV(x-PN) 
            '
            'SPL3	    MGP					            Eric 
            'NN	    EV(N) + EV(N-N)				    2*EV(N-N) 
            'PNNN	    EV(N-P) + EV(N-PN) + EV(N-PNN)		    3*EV(N-PNN) 
            'NPNN	    EV(N) + EV(N-PN) + EV(N-PNN)		    3*EV(N-PNN) 
            'PPxxxx	    4*EV(x-PP)				            4*EV(x-PP) 
            'PNPxxx	    EV(N-P) + 3*EV(x-PPN)			    EV(N-PP) + 3*EV(x-PPN) 
            'NPPxxx	    EV(N) + 3*EV(x-PPN)			            EV(N-PP) + 3*EV(x-PPN) 
            'PNNPxx	    EV(N-P) + EV(N-PN) + 2*EV(x-PPNN)	            2*EV(N-PPN) + 2*EV(x-PPNN) 
            'NPNPxx	    EV(N) + EV(N-PN) + 2*EV(x-PPNN)		    2*EV(N-PPN) + 2*EV(x-PPNN)

    I copied this from my CA code on github.

    The hands in the MGP column are the original ones I used for splits. They are based on the effects of removal. The Eric column are the hand types used that you figured out. They both give the exact EV off the top of the deck.

    If you take into account the removal of N and P cards the second column can't be played as you explained because they are dependent on the hand already having been played out.

    If you use the hands I used you can take into account the removal of any P(air) or N(on-pair) card and play the strategy best for that situation. I.e. CD-PN (CD taking into account P and N cards).

    In fact, in OBO (Australian rules) calcs the MGP hands give a closer number to the brute-force number which makes sense because the effects of removal actually have an effect on the EV.

    I hope that helps.

  9. #9


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    Quote Originally Posted by MGP View Post
    Code:
            'Rounds
            'SPL1	    MGP					            Eric
            'xx	    2*EV(x)					    2*EV(x) 
            '
            'SPL2	    MGP					            Eric 
            'NN	    EV(N) + EV(N-N)			            2*EV(N-N) 
            'Pxxx	    3*EV(x-P)				            3*EV(x-P) 
            'NPxx	    EV(N) + 2*EV(x-PN)		                    EV(N-P) + 2*EV(x-PN) 
            '
            'SPL3	    MGP					            Eric 
            'NN	    EV(N) + EV(N-N)				    2*EV(N-N) 
            'PNNN	    EV(N-P) + EV(N-PN) + EV(N-PNN)		    3*EV(N-PNN) 
            'NPNN	    EV(N) + EV(N-PN) + EV(N-PNN)		    3*EV(N-PNN) 
            'PPxxxx	    4*EV(x-PP)				            4*EV(x-PP) 
            'PNPxxx	    EV(N-P) + 3*EV(x-PPN)			    EV(N-PP) + 3*EV(x-PPN) 
            'NPPxxx	    EV(N) + 3*EV(x-PPN)			            EV(N-PP) + 3*EV(x-PPN) 
            'PNNPxx	    EV(N-P) + EV(N-PN) + 2*EV(x-PPNN)	            2*EV(N-PPN) + 2*EV(x-PPNN) 
            'NPNPxx	    EV(N) + EV(N-PN) + 2*EV(x-PPNN)		    2*EV(N-PPN) + 2*EV(x-PPNN)

    I copied this from my CA code on github.

    The hands in the MGP column are the original ones I used for splits. They are based on the effects of removal. The Eric column are the hand types used that you figured out. They both give the exact EV off the top of the deck.

    If you take into account the removal of N and P cards the second column can't be played as you explained because they are dependent on the hand already having been played out.

    If you use the hands I used you can take into account the removal of any P(air) or N(on-pair) card and play the strategy best for that situation. I.e. CD-PN (CD taking into account P and N cards).

    In fact, in OBO (Australian rules) calcs the MGP hands give a closer number to the brute-force number which makes sense because the effects of removal actually have an effect on the EV.

    I hope that helps.
    No, it doesn't. As I said, the issue isn't with the method of calculation of expected value (i.e., the choice of column from your table). But at any rate, the problem scenario described in presented example occurs in the PPxxxx case... where there is no difference between the two columns/methods in your table. (This is admittedly a bit misleading, since what you call the "Eric" column doesn't really reflect what's happening in my CA, but no matter here.) So, I'll re-iterate my question: split 6s vs. dealer 8 from the given shoe, using CDP strategy. Use *your* CA-- i.e., using your "effects of removal" approach in the first column above-- to compute the resulting expected return. What is the resulting value? Now, what post-split playing strategy realizes that reported expected value? In particular, what does your computed strategy dictate when drawing two additional 6s (yielding four pair hands), followed by an 8?

  10. #10


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    Quote Originally Posted by ericfarmer View Post
    No, it doesn't. As I said, the issue isn't with the method of calculation of expected value (i.e., the choice of column from your table). But at any rate, the problem scenario described in presented example occurs in the PPxxxx case... where there is no difference between the two columns/methods in your table. (This is admittedly a bit misleading, since what you call the "Eric" column doesn't really reflect what's happening in my CA, but no matter here.) So, I'll re-iterate my question: split 6s vs. dealer 8 from the given shoe, using CDP strategy. Use *your* CA-- i.e., using your "effects of removal" approach in the first column above-- to compute the resulting expected return. What is the resulting value? Now, what post-split playing strategy realizes that reported expected value? In particular, what does your computed strategy dictate when drawing two additional 6s (yielding four pair hands), followed by an 8?
    Honestly I wish I could answer that but it's programmed with Visual Studio 2010. I don't have any way of recompiling it to change it over to those hands since I use the ones in the Eric column after you showed that was the actual EV.

    I understand what you're saying about the PPxxxx case. All the x's have the same EV but may have a different strategy based on the effect of removal. What you're describing is a full CD strategy. Once the P's are drawn, the EV and strategy for the hands would be the same because no other information about the cards is known and thus the "x" designation. There is obviously no strategy for either pair card other than to split.

    Now, if you wanted to DECIDE to split or not after the first pair card, that's a whole other ball game and I wouldn't even know how to name that strategy.

    But I guess I misunderstood that's what you were going for. I am just explaining that if you use the effects of removal hands you'd be closer to a fully CD strategy based on every card and it would be the correct strategy based on removing N and P cards in a way that can actually be played in real time.
    Last edited by MGP; 09-04-2021 at 02:46 PM.

  11. #11


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    So, I'll re-iterate my question: split 6s vs. dealer 8 from the given shoe, using CDP strategy. Use *your* CA-- i.e., using your "effects of removal" approach in the first column above-- to compute the resulting expected return. What is the resulting value? Now, what post-split playing strategy realizes that reported expected value? In particular, what does your computed strategy dictate when drawing two additional 6s (yielding four pair hands), followed by an 8?
    Edit: I think what I posted is relative to a starting shoe composition of {0,0,0,0,0,13,0,5,0,0} rather than {0,0,0,0,0,11,0,5,0,0} but the principle remains the same.

    This is how it could possibly be done. The EV values are from my gui CA, the purpose of which is to show a lot of rounded off values rather than a few very accurate ones. I manipulate the input so the desired condition is output and I hope there are no errors.

    Split 6-6 from shoe comp {0,0,0,0,0,11,0,5,0,0} (rank 1-10), 3 allowed splits
    Draw 2 sixes so there are 4 split hands with zero splits remaining with (shoe comp {0,0,0,0,0,9,0,5,0,0} (rank 1-10))

    ** prob p = 9/14, prob n = 5/14 **
    ** prob of first card 6 = 1 **

    ** if a p is drawn:
    EVPair_ppp - Stand: -.6970, Double: -.01515, Hit: -.007576 (best strat = hit)

    ** if an n is drawn:
    Code:
    *** Preliminary data needed to compute EVn_pp given a forced strategy ***
    Hand           Prob         Stand EV   Double EV    Hit EV       Best EV
    6-6 v 6        9/14*8/13    -.6970     -.01515      -.007576     -.007576
    6-6 v 8        9/14*5/13    .1515      .6424        .3212        .6424
    6-8 v 6        5/14*9/13    -.8182     -.01212      -.006061     -.006061
    6-8 v 8        5/14*4/13    -.09091    .2364        .1182        .2364
    EVx_pp(strat)  1            -.45057    .17582       .08791       .1803
    *** EVn_pp(strat) = (EVx_pp(strat) - 9/14*EVPair_ppp(strat)) / (1 - 9/14) ***

    EVn_pp(stand) = (-.45057 - 9/14*(-.6970)) / (1 - 9/14) = -.006996
    EVn_pp(double) = (.17582 - 9/14*(-.01515)) / (1 - 9/14) = 0.519566
    EVn_pp(hit) = (.08791 - 9/14*(-.007576)) / (1 - 9/14) = 0.2597848
    EVn_pp(best strat) = (.1803 - 9/14*(-.007576)) / (1 - 9/14) = 0.5184768

    Code:
    Hand 1     EV          strategy
    6-6        -.007576    hit
    6-8        0.519566    double
    ** Hands 2,3,4
    Hand 2 possible additional removals: pp, pn*2, nn
    Hand 3 possible additional removals: ppp, ppn*3, pnn*3, nnn
    Hand 4 possible additional removals: pppp, pppn*4, ppnn*6, pnnn*4, nnnn

    I have shown how hand 1 may be computed. Hands 2, 3, and 4 might be computed using similar logic.

    When a single n is removed EV can be computed in terms of p cards removed right away. When more than one n card is removed I think it is possible to eventually express EV in terms of EVx_pRemoved and EVPair_pRemoved through a series of calculations. MGP has methods of dealing with multiple n cards removed. The split algorithm I developed continuously eliminates EVn_pRemoved by updating multipliers for EVx_pRemoved and EVPair_pRemoved for varying numbers of pRemoved for varying number of splits allowed so that no more than one n is ever considered at a time. However, here there are zero splits allowed so the algorithm is not immediately applicable. I think that your split algorithm somehow computes n hands to eliminate the possibility of any "wayward" n hands so that any EVn_pRemoved is equal to any other EVn with the same pRemoved. A "wayward" n hand occurs when an n is removed and the effect of this is immediately computed outside of any other context.

    That's about the best analysis I have to offer. Hopefully I'm at least not way out of line.

    k_c
    Last edited by k_c; 09-09-2021 at 03:28 AM.

  12. #12


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    Quote Originally Posted by k_c View Post
    Edit: I think what I posted is relative to a starting shoe composition of {0,0,0,0,0,13,0,5,0,0} rather than {0,0,0,0,0,11,0,5,0,0} but the principle remains the same.

    This is how it could possibly be done. The EV values are from my gui CA, the purpose of which is to show a lot of rounded off values rather than a few very accurate ones. I manipulate the input so the desired condition is output and I hope there are no errors.

    Split 6-6 from shoe comp {0,0,0,0,0,11,0,5,0,0} (rank 1-10), 3 allowed splits
    Draw 2 sixes so there are 4 split hands with zero splits remaining with (shoe comp {0,0,0,0,0,9,0,5,0,0} (rank 1-10))

    ** prob p = 9/14, prob n = 5/14 **
    ** prob of first card 6 = 1 **

    ** if a p is drawn:
    EVPair_ppp - Stand: -.6970, Double: -.01515, Hit: -.007576 (best strat = hit)

    ** if an n is drawn:
    Code:
    *** Preliminary data needed to compute EVn_pp given a forced strategy ***
    Hand           Prob         Stand EV   Double EV    Hit EV       Best EV
    6-6 v 6        9/14*8/13    -.6970     -.01515      -.007576     -.007576
    6-6 v 8        9/14*5/13    .1515      .6424        .3212        .6424
    6-8 v 6        5/14*9/13    -.8182     -.01212      -.006061     -.006061
    6-8 v 8        5/14*4/13    -.09091    .2364        .1182        .2364
    EVx_pp(strat)  1            -.45057    .17582       .08791       .1803
    *** EVn_pp(strat) = (EVx_pp(strat) - 9/14*EVPair_ppp(strat)) / (1 - 9/14) ***

    EVn_pp(stand) = (-.45057 - 9/14*(-.6970)) / (1 - 9/14) = -.006996
    EVn_pp(double) = (.17582 - 9/14*(-.01515)) / (1 - 9/14) = 0.519566
    EVn_pp(hit) = (.08791 - 9/14*(-.007576)) / (1 - 9/14) = 0.2597848
    EVn_pp(best strat) = (.1803 - 9/14*(-.007576)) / (1 - 9/14) = 0.5184768

    Code:
    Hand 1     EV          strategy
    6-6        -.007576    hit
    6-8        0.519566    double
    ** Hands 2,3,4
    Hand 2 possible additional removals: pp, pn*2, nn
    Hand 3 possible additional removals: ppp, ppn*3, pnn*3, nnn
    Hand 4 possible additional removals: pppp, pppn*4, ppnn*6, pnnn*4, nnnn

    I have shown how hand 1 may be computed. Hands 2, 3, and 4 might be computed using similar logic.

    When a single n is removed EV can be computed in terms of p cards removed right away. When more than one n card is removed I think it is possible to eventually express EV in terms of EVx_pRemoved and EVPair_pRemoved through a series of calculations. MGP has methods of dealing with multiple n cards removed. The split algorithm I developed continuously eliminates EVn_pRemoved by updating multipliers for EVx_pRemoved and EVPair_pRemoved for varying numbers of pRemoved for varying number of splits allowed so that no more than one n is ever considered at a time. However, here there are zero splits allowed so the algorithm is not immediately applicable. I think that your split algorithm somehow computes n hands to eliminate the possibility of any "wayward" n hands so that any EVn_pRemoved is equal to any other EVn with the same pRemoved. A "wayward" n hand occurs when an n is removed and the effect of this is immediately computed outside of any other context.

    That's about the best analysis I have to offer. Hopefully I'm at least not way out of line.

    k_c

    Hi,

    Yes, you can use my burn card calculation to handle multiple N cards removed at a time. Interesting analysis k_c

  13. #13


    Did you find this post helpful? Yes | No
    I made another attempt at this because I didn't vary some of the data by up card as it should have been. I also tried to address a starting shoe comp of {0,0,0,0,0,11,0,5,0,0} where 4 hands of split 6s are ready to be played as I originally intended. I hope this is an improvement.

    ** prob 6 = 7/12, prob 8 = 5/12 **
    ** prob of first card of 6 = 1 **

    Code:
    ** hand 1 **
    EVPair_ppp versus 6: Stand EV = -.5556, Double EV = .1667, Hit EV = .0833
    EVPair_ppp versus 8: Stand EV = .3333, Double EV = 1.000, Hit EV = .500
    
    Compute EVx_pp(strat)
    ----------------------------------Up card = 6--------------------------------
    Hand     Prob          Stand          Double         Hit            Best
    6-6      7/12*6/11     -.5556         .1667          .08333         .1667
    6-8      5/12*7/11     -.7333         -.2667         -.1333         -.1333
    EVx_pp                 -.6364         -.0303         -.015155       .0303
    
    ----------------------------------Up card = 8--------------------------------
    Hand     Prob          Stand          Double         Hit            Best
    6-6      7/12*5/11     .3333          1.000          .5000          1.000
    6-8      5/12*4/11     .0667          .2167          .1083          .2167
    EVx_pp                 .23634         .7152          .35756         .7152
    
    
    Compute EVn_pp(strat) = (EVx_pp(strat) - 7/12*EVPair_ppp(strat)) / (1 - 7/12)
    ----------------------------------Up card = 6--------------------------------
    EVn_pp(stand) = (-.6364 - 7/12*(-.5556)) / (1 - 7/12) = -.74952
    EVn_pp(double) = (-.0303 - 7/12*(.1667)) / (1 - 7/12) = -.3061
    EVn_pp(hit) = (-.015155 - 7/12*(.0833)) / (1 - 7/12) = -.151992 (best)
    
    ----------------------------------Up card = 8--------------------------------
    EVn_pp(stand) = (.23634 - 7/12*(.3333)) / (1 - 7/12) = .100596
    EVn_pp(double) = (.7152 - 7/12*(1.000)) / (1 - 7/12) = .31648 (best)
    EVn_pp(hit) = (.35756 - 7/12*(.5000) / (1 - 7/12) = .158144
    
    Hand 1    Up Card     EV         Strategy
    6-6       6           .1667      Double
              8           1.000      Double
    
    6-8       6           -.151992   Hit
              8           .31648     Double
    
    ** Hands 2,3,4 **
    Hand 2 possible additional removals: pp, pn*2, nn
    Hand 3 possible additional removals: ppp, ppn*3, pnn*3, nnn
    Hand 4 possible additional removals: pppp, pppn*4, ppnn*6, pnnn*4, nnnn
    k_c

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