Unplayable splitting strategies

[k_c]

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:

*** 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

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

[MGP]

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:

*** 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

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

[k_c]

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 **

** 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

[ericfarmer]

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 **

** 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

I'm having trouble interpreting this. Before digging deeper, it's still unclear to me what the intended answer to my original question is? That is, having split and resplit 6s (specifically vs. dealer 8) from the given shoe, to four now-all-incomplete hands, and subsequently drawing an 8 to the first split hand (so that seven 6s and three 8s remain face-down in the shoe), how do we read the table of information above to determine what we should do at that point (stand or double down)? Which is it?

And what should the strategy be for the subsequent second, third, and fourth split hands as well, to complete the round? Once we know that, we can compute the overall EV for the original 6-6 vs. 8 split (since presumably we aren't trying to change the split-otherwise-double strategy for the other split situations) ... where the objective is to specify such a strategy that realizes the overall EV for the original 6-6 vs. 8 split of 1548/715=2.165, that CDP strategy suggests is achievable.

Of course, if the intent of the above response is to describe how to compute a strategy that eventually realizes a different overall EV for the split, i.e. any value other than 1548/715=2.165, then, well, okay... but then I'm not sure what the point is. We can describe any number of other strategies-- split-and-resplit-then-mimic-the-dealer has a particular EV that we could easily compute; truly optimal perfect play a la ICountNTrack has a particular EV that we (or rather ICountNTrack) could also compute, etc. But the whole point here is to demonstrate that the *particular* strategies, namely those referred to by CDP and CDPN, require making strategy-varying decisions that depend on information that the player doesn't have when he needs it.

E

[k_c]

I'm having trouble interpreting this. Before digging deeper, it's still unclear to me what the intended answer to my original question is? That is, having split and resplit 6s (specifically vs. dealer 8) from the given shoe, to four now-all-incomplete hands, and subsequently drawing an 8 to the first split hand (so that seven 6s and three 8s remain face-down in the shoe), how do we read the table of information above to determine what we should do at that point (stand or double down)? Which is it?

And what should the strategy be for the subsequent second, third, and fourth split hands as well, to complete the round? Once we know that, we can compute the overall EV for the original 6-6 vs. 8 split (since presumably we aren't trying to change the split-otherwise-double strategy for the other split situations) ... where the objective is to specify such a strategy that realizes the overall EV for the original 6-6 vs. 8 split of 1548/715=2.165, that CDP strategy suggests is achievable.

Of course, if the intent of the above response is to describe how to compute a strategy that eventually realizes a different overall EV for the split, i.e. any value other than 1548/715=2.165, then, well, okay... but then I'm not sure what the point is. We can describe any number of other strategies-- split-and-resplit-then-mimic-the-dealer has a particular EV that we could easily compute; truly optimal perfect play a la ICountNTrack has a particular EV that we (or rather ICountNTrack) could also compute, etc. But the whole point here is to demonstrate that the *particular* strategies, namely those referred to by CDP and CDPN, require making strategy-varying decisions that depend on information that the player doesn't have when he needs it.

E

I think I had some code to compute splits based upon whether the first card drawn to each hand is either a p or an n. I can't find it and I think I deleted it. I remember that I didn't like the method because it did not necessarily yield more optimal results than a lesser information computation.

The point was to try to get started in revisiting the problem by computing the first hand, which would be the first problem an actual player would face. Hands 2, 3, and 4 would be increasingly complex.

In my view this approach is an attempt at pseudo-optimality. Hand 1 is relatively simple. Hand 2 depends what is drawn to hand 1, hand 3 to what is drawn to hands 1 and 2, hand 4 to what is drawn to hands 1, 2, and 3. However, there is no guarantee that only 1 card is drawn to any given hand using best strategy so this may not really be more optimal and I guess that's what I didn't like about it.

I was just trying to outline something that might eventually answer your question if all computational complexities are solved.

k_c