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Thread: Unplayable splitting strategies

  1. #40


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    Quote Originally Posted by k_c View Post
    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
    Isn't this optimal in the sense of maximizing EV among *all* possible strategies? That is, the "Z" in CDZ+ is out of place; "zero memory" is pretty consistently agreed to mean a strategy that depends only on the cards in the current hand (vs. any additional cards or "state" of the round). Maybe ICountNTrack can confirm/weigh in on this small shoe example for comparison (since I know he has implemented optimal EV-maximization).

    I'm also not sure what EVPN(CDZ-) means; CDZ- refers to a well-defined strategy (maximize CD strategy temporarily prohibiting splits, then apply that strategy in evaluating candidate splits), in which case it has a definite expected value, no matter how we choose to implement an algorithm to compute it. Maybe you meant something like E(CDPN), or E(something-like-CDPN-that-we-haven't-named-yet-but-is-demonstrably-executable-at-the-table-unlike-CDP-or-CDPN)?

    In that case, yes, there is certainly opportunity to define and evaluate playing strategies that depend on more information about the round than just the cards in the current hand (but constrained to information actually available at the time of each necessary playing decision). CDP1 is such a strategy. And one could certainly do better, with some yet-to-be-named-strategy S, so that E(CDZ-)<=E(CDP1)<=E(S)<=E(OPT, i.e. what you're calling CDZ+).

    E

  2. #41


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    Quote Originally Posted by ericfarmer View Post
    Isn't this optimal in the sense of maximizing EV among *all* possible strategies? That is, the "Z" in CDZ+ is out of place; "zero memory" is pretty consistently agreed to mean a strategy that depends only on the cards in the current hand (vs. any additional cards or "state" of the round). Maybe ICountNTrack can confirm/weigh in on this small shoe example for comparison (since I know he has implemented optimal EV-maximization).

    I'm also not sure what EVPN(CDZ-) means; CDZ- refers to a well-defined strategy (maximize CD strategy temporarily prohibiting splits, then apply that strategy in evaluating candidate splits), in which case it has a definite expected value, no matter how we choose to implement an algorithm to compute it. Maybe you meant something like E(CDPN), or E(something-like-CDPN-that-we-haven't-named-yet-but-is-demonstrably-executable-at-the-table-unlike-CDP-or-CDPN)?

    In that case, yes, there is certainly opportunity to define and evaluate playing strategies that depend on more information about the round than just the cards in the current hand (but constrained to information actually available at the time of each necessary playing decision). CDP1 is such a strategy. And one could certainly do better, with some yet-to-be-named-strategy S, so that E(CDZ-)<=E(CDP1)<=E(S)<=E(OPT, i.e. what you're calling CDZ+).

    E
    My intention was to hopefully compute optimal SPL1. I thought that CDZ- referred to optimal pre-split strategy so I thought CDZ+ would refer to optimal post-split strategy.

    I took it that your original problem was that not only was CDP unplayable at the table but also that the EV computed by CA couldn't be accounted for. I don't know why but was wondering if SPL1 had the same problem except hopefully more simple. As long as strategy is fixed we know there are no problems for any number of splits.

    k_c

  3. #42


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    Quote Originally Posted by k_c View Post
    My intention was to hopefully compute optimal SPL1. I thought that CDZ- referred to optimal pre-split strategy so I thought CDZ+ would refer to optimal post-split strategy.
    The earliest discussions of notation that I am familiar with were back in the bjmath.com days, where although the "Z" was consistently established to indicate "zero memory," meaning that strategy is only a function of the current hand, I never saw us settle on notation to refer to perfect EV-maximizing play. (There was discussion back in 2003 with Steve Jacobs proposing "CDB," where the "B" indicated knowledge of cards in all past "branches" of a round, which would effectively mean perfect play in the sense of maximizing EV for the round... but I neither observed nor used this notation myself at any time after that discussion. I've used "OPT" to refer to this in the past in my own internal scribblings; I suppose we haven't really settled on anything because nobody was able to actually compute it, other than ICountNTrack.)

    Quote Originally Posted by k_c View Post
    I took it that your original problem was that not only was CDP unplayable at the table but also that the EV computed by CA couldn't be accounted for. I don't know why but was wondering if SPL1 had the same problem except hopefully more simple. As long as strategy is fixed we know there are no problems for any number of splits.

    k_c
    You are correct that CDP is unplayable at the table with realistic rules. However, if we hypothesize a dealer that fleshes out all split hands to two cards before accepting stand/hit/double decisions on any of them, then the expected return produced by the CDP calculation (in this example, 1548/715) *is* able to be realized/accounted for... it just takes an interestingly complicated playing strategy to realize it. I described this strategy in this earlier comment. (I don't want to give the impression that the strategy described there should be a priori intuitive or at all obvious-- it certainly wasn't to me, and it took quite a bit of experimentation with various strategies to yield an EV that agreed with that produced by CDP.)

    At any rate, you're right that "as long as the strategy is fixed we know there are no problems for any number of splits." That is, SPL1 *doesn't* have this same problem, essentially because CDP is the same as CDP1 in that case, and CDP1 does *not* have this problem in general (for SPL3 or SPL1). That is, when we try to describe CDP1 strategy to a player that wants to use it, our laminated card only needs to indicate *whether* the current hand is a result of splitting a particular pair card P. It doesn't need to indicate the additional dependent information like *numbers* of observed pair (or non-pair) cards that are not generally available to the player. (Note also that the fact that CDP1 is "problem-free" has nothing to do with the fact that it is *optimal* in any particular sense; it's problem-free simply because of the limited information on which the strategy depends. For example, it would make sense to define a strategy that plays CDZ- pre-split, but plays mimic-the-dealer post-split. Such a strategy is both computable *and* playable.)

    E

  4. #43


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    Quote Originally Posted by ericfarmer View Post
    The earliest discussions of notation that I am familiar with were back in the bjmath.com days, where although the "Z" was consistently established to indicate "zero memory," meaning that strategy is only a function of the current hand, I never saw us settle on notation to refer to perfect EV-maximizing play. (There was discussion back in 2003 with Steve Jacobs proposing "CDB," where the "B" indicated knowledge of cards in all past "branches" of a round, which would effectively mean perfect play in the sense of maximizing EV for the round... but I neither observed nor used this notation myself at any time after that discussion. I've used "OPT" to refer to this in the past in my own internal scribblings; I suppose we haven't really settled on anything because nobody was able to actually compute it, other than ICountNTrack.)



    You are correct that CDP is unplayable at the table with realistic rules. However, if we hypothesize a dealer that fleshes out all split hands to two cards before accepting stand/hit/double decisions on any of them, then the expected return produced by the CDP calculation (in this example, 1548/715) *is* able to be realized/accounted for... it just takes an interestingly complicated playing strategy to realize it. I described this strategy in this earlier comment. (I don't want to give the impression that the strategy described there should be a priori intuitive or at all obvious-- it certainly wasn't to me, and it took quite a bit of experimentation with various strategies to yield an EV that agreed with that produced by CDP.)

    At any rate, you're right that "as long as the strategy is fixed we know there are no problems for any number of splits." That is, SPL1 *doesn't* have this same problem, essentially because CDP is the same as CDP1 in that case, and CDP1 does *not* have this problem in general (for SPL3 or SPL1). That is, when we try to describe CDP1 strategy to a player that wants to use it, our laminated card only needs to indicate *whether* the current hand is a result of splitting a particular pair card P. It doesn't need to indicate the additional dependent information like *numbers* of observed pair (or non-pair) cards that are not generally available to the player. (Note also that the fact that CDP1 is "problem-free" has nothing to do with the fact that it is *optimal* in any particular sense; it's problem-free simply because of the limited information on which the strategy depends. For example, it would make sense to define a strategy that plays CDZ- pre-split, but plays mimic-the-dealer post-split. Such a strategy is both computable *and* playable.)

    E

    I believe that for SPL1 computing hand 2 allowing for a single p or n drawn to hand 1 cannot improve (hand 1 EV + hand 2 EV) to anything more than a fixed strategy for both hand 1 and hand 2.

    EV(p) = (EVPair_p + EVx_p) // draw a p card to hand 1
    EV(non_p) = (EVn + EVx_n) // draw a non_p card to hand 1
    EV(SPL1) = pP*EV(p) + (1-pP)*EV(non_p)

    EV(SPL1) = pP*(EVPair_p + EVx_p) + (1-pP)*(EVn + EVx_n)

    ** EVn = (EVx - pP*EVPair_p) / (1-pP) **
    ** EVx_n = (EVx - pP*Evx_p) / (1-pP) **

    EV(SPL1) = pP*EVPair_p + pP*EVx_p + (EVx - pP*EVx_p) + (EVx - pP*EVPair_p)
    EV(SPL1) = 2*EVx

    Compared to a fixed strategy where:
    EVx_x = EVx
    EV(SPL1) = EVx + EVx_x = 2*EVx

    I think that in order to begin to see an increase to the fixed strategy EV for SPL1 considering only p's and n's at least 2 cards need to be allocated to hand 1 (pp,pn,np,nn) and considered in hand 2. However, this is also where the hand begins to be unplayable.

    I can't prove that this is true but if it is couldn't the same principle apply to the x hands for multiple splits?

    k_c

  5. #44


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    May I ask a really stupid question please? What exactly is the dealer doing during a round for let's say SPL2?

    Is it?

    NN:
    P1P2 Split
    Deal N to P1
    P1N play P1, P2 (no second)
    Deal N to P2
    P2N play P2

    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N play P1, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2 Play P2
    Deal x to P3 Play P3

    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1 Play P1
    Deal x to P2 Play P2
    Deal x to P3 Play P3


    Or is it:
    NN:
    P1P2 Split
    Deal N to P1
    P1N, P2 (no second)
    Deal N to P2
    P1N P2N
    Play P1 and P2


    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2
    Deal x to P3
    Play P1, P2 and P3


    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1
    Deal x to P2
    Deal x to P3
    Play P1, P2 and P3

    Thanks

  6. #45


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    Quote Originally Posted by MGP View Post
    May I ask a really stupid question please? What exactly is the dealer doing during a round for let's say SPL2?

    Is it?

    NN:
    P1P2 Split
    Deal N to P1
    P1N play P1, P2 (no second)
    Deal N to P2
    P2N play P2

    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N play P1, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2 Play P2
    Deal x to P3 Play P3

    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1 Play P1
    Deal x to P2 Play P2
    Deal x to P3 Play P3


    Or is it:
    NN:
    P1P2 Split
    Deal N to P1
    P1N, P2 (no second)
    Deal N to P2
    P1N P2N
    Play P1 and P2


    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2
    Deal x to P3
    Play P1, P2 and P3


    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1
    Deal x to P2
    Deal x to P3
    Play P1, P2 and P3

    Thanks

    If I understand correctly you're asking if strategies are determined:
    Option 1: after each single card draw to each pair card
    Option 2: after the entirety of each of the possible drawing sequences of a single card to each pair card

    I think with either option there's a problem with x hands.

    I go back to my question about SPL1. Can SPL1 EV be optimized beyond a fixed strategy by considering the removal of a single card? I don't think so. For SPL1 initially there are 2 undrawn to pair cards.

    Let s = starting shoe composition where 2 pair cards and up card have already been removed
    Let EV(s) = optimal EV of drawing to first pair card from s

    Let x be first card drawn to first pair card where x can be any card in s
    EV(s_x) = optimal EV of drawing to second pair card with one x removed

    Weight and sum EV(s_x) for all possible x. I think this will compute equal to EV(s). If so this shows that knowledge of a single additional card removed cannot be used to optimize SPL1 EV beyond that of a fixed strategy.

    For multiple splits I would expect similar problems with x hands.

    k_c

  7. #46


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    Quote Originally Posted by MGP View Post
    May I ask a really stupid question please? What exactly is the dealer doing during a round for let's say SPL2?

    Is it?

    NN:
    P1P2 Split
    Deal N to P1
    P1N play P1, P2 (no second)
    Deal N to P2
    P2N play P2

    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N play P1, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2 Play P2
    Deal x to P3 Play P3

    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1 Play P1
    Deal x to P2 Play P2
    Deal x to P3 Play P3


    Or is it:
    NN:
    P1P2 Split
    Deal N to P1
    P1N, P2 (no second)
    Deal N to P2
    P1N P2N
    Play P1 and P2


    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2
    Deal x to P3
    Play P1, P2 and P3


    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1
    Deal x to P2
    Deal x to P3
    Play P1, P2 and P3

    Thanks
    Referring to these as "Option 1" and "Option 2" as k_c does, the short answer is that Option 1 is how it works at the table. I don't think this is a "stupid question," since this is at the heart of the problem I've tried to describe here. *Because* Option 1 is how it works at the table, I've claimed that it is impossible to split-and-resplit 6s vs. 8 in the example described in this thread, with *any* strategy, in such a way that would yield the EV reported by selecting the "CDP" option in any of our various CAs (which all agree on that value).

    However, *if* the rules of the game dictated that splits were instead played out according to Option 2, then it *is* possible to execute a strategy that will yield that CDP expected return value... but that strategy is more complicated than is suggested by our typical description of what CDP "means," namely, strategy depending on the "number of (additional) pair cards removed." But it *can* be done. (I've described this strategy in an earlier comment in this thread; let me know if I need to link to it again.)

    (This is where I might just misunderstand k_c's latest comment, but I disagree with the claim that "with either option there's a problem." As long as we stick to an unrealistic-but-hypothetical table that uses Option 2 rules, there *is* a strategy that yields the EV reported by the CDP strategy CA option. If k_c's point is simply that this EV isn't *optimal*, i.e. that we might be able to come up with some other strategy with a corresponding *higher* EV that is similarly efficiently computable with some new algorithm... okay, sure. My point throughout this thread has not been to *increase* EV by *improving/modifying* playing strategy, but merely to focus on CDP[N] in particular, noting that we have perhaps not understood the relationship between our *algorithms* for computing EVs and the (executability of) *strategies* required to realize those EVs. Or at least *I* didn't understand it as well as I thought.)

    E

  8. #47


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    Eric Farmer, k_c, and MGP, the three wisemen of combinatorial analysis.


    Sent from my iPhone using Tapatalk

  9. #48


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    Okay, I think I know what's going on here (maybe.)

    No one here is concerned over the overall return on EV for any strategy given. Everyone here who has a functioning CA can derive the optimal EV and strategy for some given strategy decision (the set of either CDZ-, CDP1, CDP, or CDPN) and can return the correct EV, that everyone agrees with.

    What I seem to be picking up (and both Eric and MGP have just now eluded too this) is that the overall EV is a red-herring. What I seem to be picking up is the fact that the *order in which the cards are drawn post-split have an influencing effect on the post-split strategy at hand* and that Eric wants to know what those post-split strategies are. If I may:

    Consider the pre-split hand [9 9]. Split it up into two hands [9 x, 9 x]. Give now a possible sequence of cards [9 6 5, 9 4 5]. That is, the first 9 draws a 6 and a 5 while the second 9 draws a 4 and a 5. Now, the question seems to be that is the strategy the same if we draw 6,5 for the first 9 and 4,5 for the second 9 the same strategy if we draw 4,5 for the first 9 and 6,5 for the second 9?

    The issue for Eric (if I am correct) is even for a small set of shoe subsets to enumerate, a person will experience combinatorial explosion trying to search for each possible, valid post-split strategy...due to the fact that the out-of-order execution of splitting is much deeper than in-order execution that Eric, MGP, k_c, and iCountnTrack use.

    If that is so, my next question would be: why does it matter? What is the overall goal of searching for some optimal post-split strategy if optimal EV can be derived from some CDP[N] strategy evaluation? As Eric pointed out, even trying to specify *one* strategy would be impossible nearly for a human player to use at the table.

  10. #49


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    Quote Originally Posted by dogman_1234 View Post
    Okay, I think I know what's going on here (maybe.)

    No one here is concerned over the overall return on EV for any strategy given. Everyone here who has a functioning CA can derive the optimal EV and strategy for some given strategy decision (the set of either CDZ-, CDP1, CDP, or CDPN) and can return the correct EV, that everyone agrees with.

    What I seem to be picking up (and both Eric and MGP have just now eluded too this) is that the overall EV is a red-herring. What I seem to be picking up is the fact that the *order in which the cards are drawn post-split have an influencing effect on the post-split strategy at hand* and that Eric wants to know what those post-split strategies are. If I may:

    Consider the pre-split hand [9 9]. Split it up into two hands [9 x, 9 x]. Give now a possible sequence of cards [9 6 5, 9 4 5]. That is, the first 9 draws a 6 and a 5 while the second 9 draws a 4 and a 5. Now, the question seems to be that is the strategy the same if we draw 6,5 for the first 9 and 4,5 for the second 9 the same strategy if we draw 4,5 for the first 9 and 6,5 for the second 9?

    The issue for Eric (if I am correct) is even for a small set of shoe subsets to enumerate, a person will experience combinatorial explosion trying to search for each possible, valid post-split strategy...due to the fact that the out-of-order execution of splitting is much deeper than in-order execution that Eric, MGP, k_c, and iCountnTrack use.

    If that is so, my next question would be: why does it matter? What is the overall goal of searching for some optimal post-split strategy if optimal EV can be derived from some CDP[N] strategy evaluation? As Eric pointed out, even trying to specify *one* strategy would be impossible nearly for a human player to use at the table.

    CDZ- and CDP1 are fixed strategies. They define a strategy no matter what is drawn.

    CDP and CDPN attempt to mix the removals of P's and/or N's with fixed strategies. This results in problems. Something may be learned by looking at these problems. However, in the end the problems outweigh any benefits in my opinion.

    k_c

  11. #50


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    Ok, now we're getting somewhere. I hope I can now more easily explain what I've been saying all along:

    Here are the possible hands I've been referring to:

    Code:
    Rounds         
    SPL2        MGP                    Eric 
    NN          EV(N) + EV(N-N)        2*EV(N-N) 
    NPxx        EV(N) + 2*EV(x-PN)     EV(N-P) + 2*EV(x-PN)
    Pxxx        3*EV(x-P)              3*EV(x-P)
    Here is how it happens at the table:

    Code:
    NN:
    P1P2 Split
    Deal N to P1
    P1N play P1, P2 (no second)
    Deal N to P2
    P2N play P2
    
    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N play P1, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2 Play P2
    Deal x to P3 Play P3
    
    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1 Play P1
    Deal x to P2 Play P2
    Deal x to P3 Play P3

    So, what I am saying is that in the MGP hands, the EVs are calculated based on the order of play. They are fixed play given the effects of removal of the relevant number of P and N cards removed, and thus a CDPN strategy. Note that the hand strategy for a post split hand always includes the first extra pair card removed)

    Code:
    NN:
    P1P2 Split
    Deal N to P1
    P1N play P1, P2 (no second)   ***Use Strat/EV/Prob(CD-N) for hand which has 1 N and 1 P removed. It is not using later hand information ***
    Deal N to P2
    P2N play P2   ***Use Strat/EV/Prob(CD-NN) for hand which has 2 Ns and 1 P removed ***
    
    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N play P1, P2 (no second)   ***Use Strat/EV/Prob(CD-N) for hand which has 1 N and 1 P removed. It is not using later hand information ***
    Deal P3 to P2 Split
    Deal x to P2 Play P2   ***Use Strat/EV/Prob(CD-NPx) for hand which has 1 N and 2 Ps removed. It is not using later hand information ***
    Deal x to P3 Play P3   ***Use Strat/EV/Prob(CD-NPx) for hand which has 1 N and 2 Ps removed. This is the same strategy as the previous hand ***
    
    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1 Play P1   ***Use Strat/EV/Prob(CD-Pxxx) for hand which has 2 Ps removed.  It is not using later hand information ***
    Deal x to P2 Play P2   ***Use Strat/EV/Prob(CD-Pxxx) for hand which has 2 Ps removed.  This is the same strategy as the previous hand ***
    Deal x to P3 Play P3***Use Strat/EV/Prob(CD-Pxxx) for hand which has 2 Ps removed.  This is the same strategy as the previous hand ***
    
    So as you can see, an MGP Hands CDPN strategy determines the best CD strategy taking into account the effects of removal. A CDPN strategy would treat the post-split hands differently than pre-split hands but take the removals into account.

    I thought this is what you were asking about Eric but I'm so lost on everyone's terminology and viewpoints I don't know. I just hope I explained what I've been trying to say from the start.

  12. #51


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    Quote Originally Posted by MGP View Post
    Ok, now we're getting somewhere. I hope I can now more easily explain what I've been saying all along:

    Here are the possible hands I've been referring to:

    Code:
    Rounds         
    SPL2        MGP                    Eric 
    NN          EV(N) + EV(N-N)        2*EV(N-N) 
    NPxx        EV(N) + 2*EV(x-PN)     EV(N-P) + 2*EV(x-PN)
    Pxxx        3*EV(x-P)              3*EV(x-P)
    Here is how it happens at the table:

    Code:
    NN:
    P1P2 Split
    Deal N to P1
    P1N play P1, P2 (no second)
    Deal N to P2
    P2N play P2
    
    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N play P1, P2 (no second)
    Deal P3 to P2 Split
    Deal x to P2 Play P2
    Deal x to P3 Play P3
    
    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1 Play P1
    Deal x to P2 Play P2
    Deal x to P3 Play P3

    So, what I am saying is that in the MGP hands, the EVs are calculated based on the order of play. They are fixed play given the effects of removal of the relevant number of P and N cards removed, and thus a CDPN strategy. Note that the hand strategy for a post split hand always includes the first extra pair card removed)

    Code:
    NN:
    P1P2 Split
    Deal N to P1
    P1N play P1, P2 (no second)   ***Use Strat/EV/Prob(CD-N) for hand which has 1 N and 1 P removed. It is not using later hand information ***
    Deal N to P2
    P2N play P2   ***Use Strat/EV/Prob(CD-NN) for hand which has 2 Ns and 1 P removed ***
    
    NPxx:
    P1P2 Split
    Deal N1 to P1
    P1N play P1, P2 (no second)   ***Use Strat/EV/Prob(CD-N) for hand which has 1 N and 1 P removed. It is not using later hand information ***
    Deal P3 to P2 Split
    Deal x to P2 Play P2   ***Use Strat/EV/Prob(CD-NPx) for hand which has 1 N and 2 Ps removed. It is not using later hand information ***
    Deal x to P3 Play P3   ***Use Strat/EV/Prob(CD-NPx) for hand which has 1 N and 2 Ps removed. This is the same strategy as the previous hand ***
    
    Pxxx:
    P1P2 Split
    Deal P3 Split
    Deal x to P1 Play P1   ***Use Strat/EV/Prob(CD-Pxxx) for hand which has 2 Ps removed.  It is not using later hand information ***
    Deal x to P2 Play P2   ***Use Strat/EV/Prob(CD-Pxxx) for hand which has 2 Ps removed.  This is the same strategy as the previous hand ***
    Deal x to P3 Play P3***Use Strat/EV/Prob(CD-Pxxx) for hand which has 2 Ps removed.  This is the same strategy as the previous hand ***
    
    So as you can see, an MGP Hands CDPN strategy determines the best CD strategy taking into account the effects of removal. A CDPN strategy would treat the post-split hands differently than pre-split hands but take the removals into account.

    I thought this is what you were asking about Eric but I'm so lost on everyone's terminology and viewpoints I don't know. I just hope I explained what I've been trying to say from the start.

    I should mention that you could in fact do a full weighted average of all reached hands including post-split hands to determine a CDPN+ strategy (+ meaning in my use that pre and post split hands are taken into account). If you do that I am fairly certain the example you gave Eric would in fact end up with the correct strategy.

  13. #52


    Did you find this post helpful? Yes | No
    Quote Originally Posted by MGP View Post
    So, what I am saying is that in the MGP hands, the EVs are calculated based on the order of play. They are fixed play given the effects of removal of the relevant number of P and N cards removed, and thus a CDPN strategy. Note that the hand strategy for a post split hand always includes the first extra pair card removed)

    ...

    So as you can see, an MGP Hands CDPN strategy determines the best CD strategy taking into account the effects of removal. A CDPN strategy would treat the post-split hands differently than pre-split hands but take the removals into account.

    I thought this is what you were asking about Eric but I'm so lost on everyone's terminology and viewpoints I don't know. I just hope I explained what I've been trying to say from the start.
    No. I'm not sure how else to explain. You keep referring to "the *best* CD strategy," when this isn't about the *optimality* of any particular expected return, it's about the *achievability at the table*. You also keep referring to "MGP hands," presumably in contrast to the "Eric hands" that you include in your tables, as if it makes a difference... when they both yield the same expected value, a value which is not actually achievable at the table.

    You also keep focusing on SPL2, when the example scenario throughout this thread has been SPL3. You also keep focusing on CDPN, when the example scenario throughout this thread has been CDP. Perhaps you're thinking that CDPN is immune to the problem, and that it's confined to CDP? It's not... indeed, in the place where the problem manifests (PPxxxx), the calculation isn't different for the two. I only focused on CDP throughout because that's the first situation where I could find an example of the problem. As I've said repeatedly now, these example scenarios are not easy to find.

    E

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