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Thread: Unplayable splitting strategies

  1. #27


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    Quote Originally Posted by Norm View Post
    That isn't close to what I was saying.
    Well I understand that when you add up everything lots of unlikely deck compositions happen like there are a lot of unlikely streaks but when you are faced with some improbable deck composition (but not all improbable deck compositions) standard strategy indices are basically useless.

  2. #28


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    Quote Originally Posted by ericfarmer View Post
    No. I see now that I likely added to the confusion by phrasing as a question at all, since it was intended to be rhetorical. That is, to the question "should we stand or double down with 6-8 vs. 8, in the split situation we've described, if we want to achieve an overall EV of 1548/715 for the round, no more and no less, using CDP strategy as computed by *either* "Eric's" or "MGP's" method?", the answer is, "neither, because if we always stand in that situation, or if we always double down in that situation, we have no hope of achieving the computed EV of 1548/715."

    Note that I'm not "asking" anything here, I'm *claiming* that CDP-- *and CDPN*-- are broken as specifications of strategies, since they are unplayable, the situation described in this post being a concrete example. For some reason, the discussion keeps gravitating toward computational methods (what you label "MGP" and "Eric" in the tables of formulas you keep quoting)... it makes no difference which of these two computational approaches are used, the problem (with the *definition of the random variable*, which has a well-defined expected value no matter how we choose to compute it) remains.



    This is exactly where the problem can arise, and it's still a problem for CDPN as well as CDP, which is what I seem to have not clearly communicated yet. This is where the problem arises in the example described in this post, albeit in the PPxxxx case for SPL3 instead of Pxxx for SPL2. The expected value that results from "all 3 hands are played the same way and all 3 take into account that 2 paircards are removed" cannot (in general) actually be achieved.

    I say "in general" because finding examples of this sort of thing has turned out to be hard to find. That is, consider the example I've described so far ((0,0,0,0,0,11,0,5,0,0), splitting 6s vs. 8). I had to search quite a bit to find that example, first looking for "nice" shoes with the appropriate differences in EV as we move from CDZ- to CDP[n]... but even after that reasonably automatable search, I still had to "re-evaluate" those same EVs by brute-force enumeration and playout of shoe arrangements, looking for situations where the *indicated* strategy variations *didn't* yield the corresponding EV.

    This is a detail that I'm not sure I've actually made explicit here yet. That is, it *is* possible to realize 1548/715 as an overall CDP EV for the split in this example... but to do so, not only do we need a dealer willing to deal all of our split hands out to two cards each before asking for a "non-split" strategy decision, but even then we still have to execute a strategy that looks surprisingly complicated. Coming back to the rhetorical question situation above, having split and resplit 6s to a maximum of 4 hands, and observing the first of those hands as 6-8, what should we do?

    We've already said that we can't *always* stand, and we can't *always* double down. It turns out that we have to distinguish *two* of the other three split hands (any two will do, but they have to be *fixed*), and stand if those hands both get fleshed out to 6-6, otherwise (if we draw an 8 to either of them) double down.

    But it's worse than that-- we have to execute the same "conditional stand/double" strategy for *each* of the other three split hands as well: for each, distinguish and fix two of the three "other slots," and stand only if both of those two other hands are 6-6.

    E
    Why would you need a dealer to play to 2 cards for each hand? I haven't actually played blackjack in years. Let's get really basic. What is the card drawing order for Pxxx?

    Doesn't the very first card dealed have to be a paircard that is split?

    I think I understand what you're saying now. So for PPxxx it could actually be:

    PxPxx
    PxxPx
    xPPxx
    xPxPx

    Did I do that right? In this case it's very simple to calculate the strategy for each different type of x:
    x
    x-P
    x-PP

    So would using those theoretically accomplish what you're asking? Maybe we weren't fine grained enough.

  3. #29


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    Quote Originally Posted by MGP View Post
    Why would you need a dealer to play to 2 cards for each hand? I haven't actually played blackjack in years. Let's get really basic. What is the card drawing order for Pxxx?

    Doesn't the very first card dealed have to be a paircard that is split?

    I think I understand what you're saying now. So for PPxxx it could actually be:

    PxPxx
    PxxPx
    xPPxx
    xPxPx

    Did I do that right? In this case it's very simple to calculate the strategy for each different type of x:
    x
    x-P
    x-PP

    So would using those theoretically accomplish what you're asking? Maybe we weren't fine grained enough.
    Short answer: no. Longer answer: although I've focused on SPL3 in my explicit running example in this thread, you've jumped back and forth a bit between SPL2 and SPL3 in your comments-- I've been able to follow you up until this last comment, where now I'm confused, because your count of Ps and xs seems off: in what game rules does PPxxx live?

    At any rate, let's continue to focus on the concrete SPL3 example that we started with. The round starts with dealing a 6 to the player, then an 8 to the dealer, then another 6 to the player, then the dealer's hole card. The player splits the pair...

    and to try to answer your question, let's further focus on the PPxxxx situation. Having split the original pair, the dealer deals another 6 (that's the first P in "PPxxxx"), the player splits *that* new pair, the dealer deals another 6 (that's the second P in "PPxxxx"), and the player splits again. At this point, there are four single-card hands on the table.

    Now, the dealer deals an 8-- this is the first x in "PPxxxx"-- to (the first) one of those single-card hands, so that the player is presented with 6-8 vs. dealer 8, and is asked whether to stand, hit, or double down. At the table, we must make a strategy decision *now*, with only those three other solo 6s populating the other three "x"s.

    You can verify that your CA reports an overall EV for this split of 1548/715. (Forget about the "MGP or Eric" algorithm business. Go back and update your code to use *your* algorithm to compute CDP strategy. The EV is still 1548/715.) The problem is that no fixed strategy decision for that first 6-8 hand-- no matter what decisions we make after that to complete the round-- will yield that overall EV of 1548/715.

  4. #30


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    Quote Originally Posted by ericfarmer View Post
    Short answer: no. Longer answer: although I've focused on SPL3 in my explicit running example in this thread, you've jumped back and forth a bit between SPL2 and SPL3 in your comments-- I've been able to follow you up until this last comment, where now I'm confused, because your count of Ps and xs seems off: in what game rules does PPxxx live?

    At any rate, let's continue to focus on the concrete SPL3 example that we started with. The round starts with dealing a 6 to the player, then an 8 to the dealer, then another 6 to the player, then the dealer's hole card. The player splits the pair...

    and to try to answer your question, let's further focus on the PPxxxx situation. Having split the original pair, the dealer deals another 6 (that's the first P in "PPxxxx"), the player splits *that* new pair, the dealer deals another 6 (that's the second P in "PPxxxx"), and the player splits again. At this point, there are four single-card hands on the table.

    Now, the dealer deals an 8-- this is the first x in "PPxxxx"-- to (the first) one of those single-card hands, so that the player is presented with 6-8 vs. dealer 8, and is asked whether to stand, hit, or double down. At the table, we must make a strategy decision *now*, with only those three other solo 6s populating the other three "x"s.

    You can verify that your CA reports an overall EV for this split of 1548/715. (Forget about the "MGP or Eric" algorithm business. Go back and update your code to use *your* algorithm to compute CDP strategy. The EV is still 1548/715.) The problem is that no fixed strategy decision for that first 6-8 hand-- no matter what decisions we make after that to complete the round-- will yield that overall EV of 1548/715.
    What exactly do you mean by "The problem is that no fixed strategy decision for that first 6-8 hand-- no matter what decisions we make after that to complete the round-- will yield that overall EV of 1548/715."? In the line before you explicitly stated the fixed strategy is CD with the extra 2 Ps removed.

  5. #31


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    Quote Originally Posted by MGP View Post
    What exactly do you mean by "The problem is that no fixed strategy decision for that first 6-8 hand-- no matter what decisions we make after that to complete the round-- will yield that overall EV of 1548/715."? In the line before you explicitly stated the fixed strategy is CD with the extra 2 Ps removed.
    In the line before, I did not state that the "fixed strategy is *CD*" (my emphasis). CD without further qualification isn't even well-defined, which is sort of the whole point here. I referred specifically to *CDP* (not just "CD," whatever that means), which as has been the point of this whole thread, corresponds to a well-defined random variable, whose expected value we can all agree on (namely, 1548/715 in this particular case), but that no simulation of a strategy realizable at the table (i.e., that makes that first 6-8 hand decision with only the three single-card, not-yet-fleshed-out pair hands beside it) can actually yield that 1548/715 expected return.

    I'm not sure how else to re-phrase it (although I get the impression that k_c gets the point here, and can perhaps help describe the problem in a way that I'm so far failing to do)-- our CAs have "CDP" as an option, and we all agree on what the reported EV is in the above situation. Your CA reports 1548/715; so, if that EV is realizable, what should that "fixed strategy decision for that first 6-8 hand" be? (Again, I'm being rhetorical in asking this question-- the answer is that there isn't one.)

    E

  6. #32


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    Quote Originally Posted by ericfarmer View Post
    In the line before, I did not state that the "fixed strategy is *CD*" (my emphasis). CD without further qualification isn't even well-defined, which is sort of the whole point here. I referred specifically to *CDP* (not just "CD," whatever that means), which as has been the point of this whole thread, corresponds to a well-defined random variable, whose expected value we can all agree on (namely, 1548/715 in this particular case), but that no simulation of a strategy realizable at the table (i.e., that makes that first 6-8 hand decision with only the three single-card, not-yet-fleshed-out pair hands beside it) can actually yield that 1548/715 expected return.

    I'm not sure how else to re-phrase it (although I get the impression that k_c gets the point here, and can perhaps help describe the problem in a way that I'm so far failing to do)-- our CAs have "CDP" as an option, and we all agree on what the reported EV is in the above situation. Your CA reports 1548/715; so, if that EV is realizable, what should that "fixed strategy decision for that first 6-8 hand" be? (Again, I'm being rhetorical in asking this question-- the answer is that there isn't one.)

    E
    Isn't the idea of CDP to define post split strategies of CDZ-(pRemoved)? In that case if 6-6 is split to 4 hands with no splits remaining
    from a shoe composition of 11 6s and 5 8s then the strategy you would be stuck with is CDZ-(2):
    6-6
    ----- versus 6 double
    ----- versus 8 double
    6-8
    ----- versus 6 hit
    ----- versus 8 double

    It seems the problem or solution, if one exists, would be in the other possible drawing sequences of splitting 6-6 from the above shoe comp,
    all which involve drawing 1 or more non-pair cards and where a hand of 6-6 cannot be played until zero splits remain.

    k_c

  7. #33


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    Quote Originally Posted by k_c View Post
    Isn't the idea of CDP to define post split strategies of CDZ-(pRemoved)? In that case if 6-6 is split to 4 hands with no splits remaining
    from a shoe composition of 11 6s and 5 8s then the strategy you would be stuck with is CDZ-(2):
    6-6
    ----- versus 6 double
    ----- versus 8 double
    6-8
    ----- versus 6 hit
    ----- versus 8 double
    We might be just differently indexing the number of pair cards removed, but as described in the post at the start of this thread (actually my 2nd post), the place where CDP strategy differs from CDZ- is with 4 pair cards observed to have been removed, not 2. This is sort of the point-- there *can eventually* be 4 pair cards observed, but not at the point that we have to make a strategy decision for that first 6-8 hand.

    Quote Originally Posted by k_c View Post
    It seems the problem or solution, if one exists, would be in the other possible drawing sequences of splitting 6-6 from the above shoe comp,
    all which involve drawing 1 or more non-pair cards and where a hand of 6-6 cannot be played until zero splits remain.

    k_c
    I might misunderstand you here, but note that CDP strategy, and the "component" expected values, agree with CDZ- in all of those other possible drawing sequences of splitting/resplitting; the only place where things differ is in the PPxxxx outcome (and the "problem" situation occurs when trying to play out that first "x" in that outcome).

    If there were anything we might call a "solution" here, IMO it's nothing really earth-shaking-- just don't use CDP or CDPN as a measure of realizable expected return. CDZ- works... and CDP1 does as well, since it doesn't suffer from this problem of asking the player to *change* strategy decisions as a function of information that the player doesn't yet have. (In the case of CDP1, the only additional information that can affect a player's decision is *whether* he has split the original hand or not.)

    E

  8. #34


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    Quote Originally Posted by ericfarmer View Post
    We might be just differently indexing the number of pair cards removed, but as described in the post at the start of this thread (actually my 2nd post), the place where CDP strategy differs from CDZ- is with 4 pair cards observed to have been removed, not 2. This is sort of the point-- there *can eventually* be 4 pair cards observed, but not at the point that we have to make a strategy decision for that first 6-8 hand.



    I might misunderstand you here, but note that CDP strategy, and the "component" expected values, agree with CDZ- in all of those other possible drawing sequences of splitting/resplitting; the only place where things differ is in the PPxxxx outcome (and the "problem" situation occurs when trying to play out that first "x" in that outcome).

    If there were anything we might call a "solution" here, IMO it's nothing really earth-shaking-- just don't use CDP or CDPN as a measure of realizable expected return. CDZ- works... and CDP1 does as well, since it doesn't suffer from this problem of asking the player to *change* strategy decisions as a function of information that the player doesn't yet have. (In the case of CDP1, the only additional information that can affect a player's decision is *whether* he has split the original hand or not.)

    E

    I'm trying to mimic CDZ- from a full shoe. Don't you determine CDZ- from a full shoe by simply recording the strategies for each hand as dealt from the top of a full shoe? That's how I do it and I save the results in order to use this as a CD basic strategy option to apply to less than full shoe. I would think CDZ- given 2 cards have been removed would be done the same except that these same 2 additional cards have been removed. For the example full shoe is {0,0,0,0,0,11,0,5,0,0}.

    Anyway my viewpoint is that for drawing sequence ppxxxx CDP strategy is clear. When an n is involved it seems more complicated.
    For example pnnn and npnn yield the same 3 hands (pn, pn, pn) but strategy of the first n in each case would need to be determined with a differing number of pRemoved when played in sequence in a blackjack game.

    I guess it's just a different approach to figure out that a problem exists.

    k_c

  9. #35


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    Quote Originally Posted by k_c View Post
    I'm trying to mimic CDZ- from a full shoe. Don't you determine CDZ- from a full shoe by simply recording the strategies for each hand as dealt from the top of a full shoe? That's how I do it and I save the results in order to use this as a CD basic strategy option to apply to less than full shoe. I would think CDZ- given 2 cards have been removed would be done the same except that these same 2 additional cards have been removed. For the example full shoe is {0,0,0,0,0,11,0,5,0,0}.
    Yep, this is correct... sort of. For that full shoe with 11 6s and 5 8s, the strategy for 6-8 vs. 8 is indeed to double down. And if we remove two 6s and recompute, the strategy is still to double down, and so on... but when we remove *four* additional 6s from the shoe, and recompute strategy, that we *stand* on 6-8 vs. 8 instead of double down, and that's the strategy departure that matters here. Now, there are-- or will be later in the resolution of a PPxxxx hand-- possible situations where we may observe that many 6s. But as you can verify, when we actually collect the individual conditional expected returns into one overall CDP EV (or CDPN, for that matter), the result is not (generally) achievable.

    You don't have to take my word for it-- I went searching for examples of this in shoes where things are easy to compute and verify "by hand." For example, in this split case of 6-6 vs. 8, there are only two possible useful strategies to consider, standing or doubling (hitting will never be optimal, since if it's desirable, then doubling is more so).

    Quote Originally Posted by k_c View Post
    Anyway my viewpoint is that for drawing sequence ppxxxx CDP strategy is clear.
    How is it clear? If it were clear, then someone would be able to answer my rhetorical/didactic question that I've been asking, namely, what's the "clear" strategy if that first "x" in PPxxxx is an 8, i.e., 6-8 vs. 8, with those three other 6s waiting to be fleshed out? Where remember, the goal is to realize the CDP value that each of our CAs reports of 1548/715 (or at least, yours *could* if you also implemented it).

    E

  10. #36


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    Quote Originally Posted by ericfarmer View Post
    Yep, this is correct... sort of. For that full shoe with 11 6s and 5 8s, the strategy for 6-8 vs. 8 is indeed to double down. And if we remove two 6s and recompute, the strategy is still to double down, and so on... but when we remove *four* additional 6s from the shoe, and recompute strategy, that we *stand* on 6-8 vs. 8 instead of double down, and that's the strategy departure that matters here. Now, there are-- or will be later in the resolution of a PPxxxx hand-- possible situations where we may observe that many 6s. But as you can verify, when we actually collect the individual conditional expected returns into one overall CDP EV (or CDPN, for that matter), the result is not (generally) achievable.

    You don't have to take my word for it-- I went searching for examples of this in shoes where things are easy to compute and verify "by hand." For example, in this split case of 6-6 vs. 8, there are only two possible useful strategies to consider, standing or doubling (hitting will never be optimal, since if it's desirable, then doubling is more so).



    How is it clear? If it were clear, then someone would be able to answer my rhetorical/didactic question that I've been asking, namely, what's the "clear" strategy if that first "x" in PPxxxx is an 8, i.e., 6-8 vs. 8, with those three other 6s waiting to be fleshed out? Where remember, the goal is to realize the CDP value that each of our CAs reports of 1548/715 (or at least, yours *could* if you also implemented it).

    E

    I'm not sure if I'm right or wrong but what I think is there are 3 separate CDZ- strategies that apply to CDP for 3 allowed splits:
    1. Strategy for full shoe (starting composition) - applies to a decision where no additional pair cards have been removed
    2. Strategy with 1 pair card removed from starting composition - applies to a decision where 1 additional pair card has been removed
    3. Strategy with 2 pair cards removed from starting composition - applies to a decision where 2 additional pair cards have been removed

    For drawing sequence ppxxxx there are always 2 additional (pertinent) pair cards removed for each decision so wouldn't strategy always be defined? I guess what I'm saying is that there is a strategy for each px that is fixed. What I'm not saying is what it computes to. I believe you when you say there is no workable fixed strategy that computes to the expected CDP value assuming there are no other CDZ- variations in the other drawing sequences.

    k_c

  11. #37


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    Simplify the probem?

    Can the problem be simplified by looking first at the case of 1 allowed split?

    Code:
    Shoe composition: {0,0,0,0,0,11,0,5,0,0}
    Split EV for 6-6:
    Up card     CDZ- (Farmer console program)      CDZ+ (my console program - computes optimal SPL1)     CDP1 (from my gui program rounded off)
       6        -1.864801865%                      0.0217560217560218                                    -1.399%
       8        103.496503497%                     1.04055944055944                                      103.5%
    Is it in general possible to get a partially optimal split EV where EV(CDZ-) <= EVPN(CDZ-) <= EV(CDZ+) for a single split with simplified removals of just p's and n's?

    k_c
    Last edited by k_c; 11-17-2021 at 01:47 PM.

  12. #38


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    Quote Originally Posted by k_c View Post
    Can the problem be simplified by looking first at the case of 1 allowed split?

    Code:
    Shoe composition: {0,0,0,0,0,11,0,5,0,0}
    Split EV for 6-6:
    Up card     CDZ- (Farmer console program)      CDZ+ (my console program - computes optimal SPL1)     CDP1 (from my gui program rounded off)
       6        -1.864801865%                      0.0217560217560218                                    -1.399%
       8        103.496503497%                     1.04055944055944                                      103.5%
    Is it in general possible to get a partially optimal split EV where EV(CDZ-) <= EVPN(CDZ-) <= EV(CDZ+) for a single split with simplified removals of just p's and n's?

    k_c
    This is interesting; can you describe what is meant by "CDZ+", i.e. the optimal SPL1 is subject to what constraints? (I can't find reference to CDZ+ anywhere, other than earlier in this thread where it isn't clearly defined.) What is the hand(s) where strategy differs from CDZ- to yield that increase in EV from 740/715 to 744/715 (where 715 is the number of possible shuffled arrangements of the remainder of the shoe)?

    Thanks,
    E

  13. #39


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    Quote Originally Posted by ericfarmer View Post
    This is interesting; can you describe what is meant by "CDZ+", i.e. the optimal SPL1 is subject to what constraints? (I can't find reference to CDZ+ anywhere, other than earlier in this thread where it isn't clearly defined.) What is the hand(s) where strategy differs from CDZ- to yield that increase in EV from 740/715 to 744/715 (where 715 is the number of possible shuffled arrangements of the remainder of the shoe)?

    Thanks,
    E
    CDZ+ means making each post split decision considering all removals at the time of the decision.

    For SPL1 there are 2 hands, each with its own EV. hand1EV is simply the optimal EV of drawing to the first pair card allowing for one pair
    card removed. This is what is done to compute CDP1. For CDP1, EVx = optimal EV of first split hand and EV(CDP1) for SPL1 = 2*EVx where
    strategy for hand2 is the same as strategy for hand1. For CDZ+, hand2EV and strategy for one allowed split is variable depending upon what
    has been drawn to hand1 as well as what is currently drawn to hand2.

    *Here are some things I do to compute hand2:
    1. In the course of computing hand1 I use busted as well as unbusted hands. Busted hand1 hands have no effect on hand1EV and strategy but do
    effect hand2.
    2. For each hand1 composition I compute a parameter I call h2EVX. h2EVx is the optimal hand2EV
    for that particular hand1 comp. To compute h2EVx only unbusted hands need to be considered because for SPL1
    no more split hands follow. I do not save this set of hand2 hands, only the h2EVx parameter for this hand1 comp.
    3. I have parameters I call h2EVx_stand, h2EVx_double, and h2EVx_hit which depend on hand1 strategy. h2EVx_stand = h2EVx because no cards are
    drawn for the hand1 comp being referenced. hand2EV is computed while going through all hand1 hands using above values. If by some quirk of
    fate strategy of all hand1 hands is stand then hand2EV would equal hand1EV.

    I'm sure there's room for improvement.

    - This is what I get for the sample shoe comp -

    Shoe composition: {0,0,0,0,0,11,0,5,0,0}
    Split EV for 6-6 (SPL1), S17, DAS:

    ** versus 6 **
    hand1EV = -0.0069930069930069488
    hand2EV = 0.028749028749028734
    hand1EV + hand2EV = 0.0217560217560217852 = CDZ+(single split)

    ** versus 8 **
    hand1EV = 0.51748251748251750
    hand2EV = 0.52307692307692299
    hand1EV + hand2EV = 1.04055944055944049 = CDZ+(single split)

    k_c
    Last edited by k_c; 11-20-2021 at 07:49 PM.

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