J. A. Nairn false claims about first to calculate split evs for finite decks

[DSchles]

Sorry. Didn't realize the expectations being quoted weren't for BS. So, the numbers you're giving are for perfect play, reckoning every card in the deck. Not too much call for that!

Don

[ericfarmer]

Guess I'm missing something here, Eric. Can you explain under what circumstances someone would be drawing to A,8,T? Or even how he would GET to A,8,T vs. A, given that he would stand with A,8, with A,T, or with 8, T?

Don

Sorry for the confusion, I made a silly clerical error in my code searching for these examples, and worse, I didn't even squint at the result to see if it made sense. Let's try this again.

To clarify, iCountNTrack, this *isn't* a situation with a depleted shoe. This is 1D, S17. Ignoring the previous jacked up example, consider the hand (3,3,3) vs. dealer 7. We encounter this situation with probability about 0.0000147751. But we encounter the superset hand {10,3,3,3} vs. dealer 7 with the *larger* probability about 0.0000197002.

E

[MGP]

Thank you. Cacarulo was instrumental too but can't remember if he posted on the forums for the discussion and he did all the numbers for Don's book. I appreciate the update.

I've updated my write-up to reference all of the participants in that discussion, at least those that I have archived. I tried finding the actual bjmath.com posts in the wayback machine, but unfortunately it only has limited snapshots, and even those only have the views of the thread trees, not the content of the actual posts. All that I have copies of are my posts, with the surrounding "in reply to" contexts.

[MGP]

What I mean is the EVs for each type of hand are multiplied by their probabilities to get the net EV. I.e. the prob of EV(N) + EV(N-N) <> prob of 2*EV(N-N). but the net probs*EVs of the 2 columns are equal. Based on your question I assume you've done this and that they are, but it is still interesting to me that the numbers worked exactly both ways.

[k_c]

Since everyone seems interested in old split code I decided to post my old split algorithm. It's a lot of work to try to reasonably present.

http://www.bjstrat.net/splitAlgorithm.html

k_c

[aceside]

Since everyone seems interested in old split code I decided to post my old split algorithm. It's a lot of work to try to reasonably present.

http://www.bjstrat.net/splitAlgorithm.html

k_c

I just read your own stories about blackjack things and was able to link your name k_c to your website bjstrat.net. Your work is very helpful and has a long history of time evolution. Thank you for your hard work.

[MGP]

Very cool k_c, that's way shorter than my code!

Eric, that's a good example hand. And I know that's not how optimality works, especially with most normal gradient descent type situations where the local minima can be hard to get out of. I do agree with you that we can't be positive, but I still find it highly unlikely that more than 2 cycles through will find any changes and that the changes found by going through twice are only local minima.

[MGP]

Actually there was one more thing that I failed to express clearly. What you are saying makes sense theoretically for a non-split hand. But what I was saying about decreasing probabilities of a split hand effecting the overall TD strategy still holds. All post-split hands by definition have a lower probability than the normal hand since the other pair card(s) must have been dealt. So right off the bat it's at least 4/13 if not 1/13 less weighted than any of the non-split hands so the finding of a local-minima based on the sum of all post-split hands is practically almost nil.

[ericfarmer]

Actually there was one more thing that I failed to express clearly. What you are saying makes sense theoretically for a non-split hand. But what I was saying about decreasing probabilities of a split hand effecting the overall TD strategy still holds. All post-split hands by definition have a lower probability than the normal hand since the other pair card(s) must have been dealt. So right off the bat it's at least 4/13 if not 1/13 less weighted than any of the non-split hands so the finding of a local-minima based on the sum of all post-split hands is practically almost nil.

I'm not sure what you're arguing here? You're right that the probability of encountering a particular hand (let's say ace-2 vs. dealer 5, for a concrete example) after a split is less than the probability of encountering that same hand in the initial deal. (Although it's not quite as simple as a 1/13-ish reduction, since as in this ace-2 vs. dealer 5 case, assuming SPL3 for example, we get multiple-- up to four-- additional opportunities to encounter ace-2 vs. dealer 5, albeit itself with a (conditional) distribution with support {2,3,4}, corresponding to all of the possibly four distinct split hands after splitting 2-2. And when the critical decision might be hit vs. double down, then we're also talking about an "extra factor of two" in the potential impact on the difference in *expected value* of the round, as opposed to just the probability.)

But those two *separate* probabilities (the larger pre-split, and the smaller post-split) aren't really what matter anyway, when we're arguing about the (sub)optimality of a *zero-memory* strategy, i.e., when we're forced to take the same player action on ace-2 vs. dealer 5 no matter whether we encounter it after a split or not. And this isn't a purely academic discussion: consider a 6-deck shoe, with an ace, four, eight, and two tens removed. Now consider this same ace-2 vs. dealer 5 hand. What should our zero-memory strategy be if we encounter this hand during a round dealt from this modestly-depleted shoe? Should we hit or double down? (Since the problems I'm emphasizing here are really about the zero-memory-ness, not the total-dependent-ness, let's make the point *stronger* by setting the total-dependent-ness aside, and allow composition-dependent-- but still zero-memory-- strategy.)

This is a very narrow decision, but it can be verified that although a normal CDZ- strategy would be to hit, it's better overall to always double down instead.

If you're simply saying that this "probably/practically" doesn't happen *very often*, and/or that the effect of getting this decision wrong is small or "practically nil," then I suppose I would agree with this opinion. But if you're saying this *doesn't* happen, ever, and won't, ever, so our/your existing strategy optimization algorithms are already truly optimal, then I would point to the above as an explicit counterexample. (Granted, I think this is an interesting problem, because this is the *only* counterexample that I know of!)

E

[k_c]

You'll have to elaborate on what you mean by "their prob[abilities] differ too"? I'll re-iterate my question: consider being dealt an initial pair (of, say, 10s vs. dealer 6), that we split and repeatedly re-split at every opportunity, up to a maximum of *four* hands (SPL3 in our now-reasonably-established notation). Repeat this experiment in Monte Carlo fashion, ignoring all outcomes *except* those where we end up with exactly *two* split hands (i.e., when we draw a non-10 card to each half of the split). Let the random variable X1 be the final outcome of the first half of this split, and X2 be the outcome of the second half. Then is it true that E[X1]=E[X2]?

Although I'm not able to give a formal write up to this question, my answer would be that EVn = EV(n-n) eventually when all pair cards have been depleted.

This is what happens in an infinite shoe. Splits occur until an infinite number of pair cards have been removed and are eliminated from being drawn, leaving a constant ev for each split hand and no need to compute ev of simply drawing to a pair rather than splitting.

Other than that it might just be better to say that EVn-EV(n-n) approaches 0 as number of pair cards removed increases. Evidence of this can be seen in my split multiplier algorithm. In this case there would be values for 15 or 16 pair cards removed except they have been completely depleted.

Pair cards present in shoe: 14  Non-pair cards present in shoe: 35
Number of allowed splits for which to display data (1 - 9): 9
Allowed splits: 9       Expected number of hands: 3.15966

Pair cards removed      EVx Multiplier          EVPair_p Multiplier

0                       2.0000000000            0.0000000000
1                       1.1428571429            -0.5714285714
2                       0.6190476190            -0.3095238095
3                       0.3161094225            -0.1580547112
4                       0.1511827673            -0.0755913836
5                       0.0671923410            -0.0335961705
6                       0.0274877759            -0.0137438879
7                       0.0102280096            -0.0051140048
8                       0.0034093365            -0.0017046683
9                       -0.0317364536           0.0158682268
10                      0.0166166178            -0.0083083089
11                      -0.0033707576           0.0016853788
12                      0.0003088154            -0.0001544077
13                      -0.0000120762           0.0000060381
14                      0.0000001499            -0.0000000750
15                      0.0000000000            0.0000000000
16                      0.0000000000            0.0000000000

Press x to exit, i to reinput p & np, any other key to reinput num splits

k_c

[k_c]

You'll have to elaborate on what you mean by "their prob[abilities] differ too"? I'll re-iterate my question: consider being dealt an initial pair (of, say, 10s vs. dealer 6), that we split and repeatedly re-split at every opportunity, up to a maximum of *four* hands (SPL3 in our now-reasonably-established notation). Repeat this experiment in Monte Carlo fashion, ignoring all outcomes *except* those where we end up with exactly *two* split hands (i.e., when we draw a non-10 card to each half of the split). Let the random variable X1 be the final outcome of the first half of this split, and X2 be the outcome of the second half. Then is it true that E[X1]=E[X2]?

I have had second thoughts on this. Consider a shoe composition of 1 five, 1 six, and a bunch of tens and 10-10 vs 10 with possible resplits. EVn given an n has been drawn is clearly not equal to EVn from the starting condition. However the way you asked the question is that an n is drawn on each hand, in which case it seems EVs are equal. I based my "repair the EV" algorithm on there being no need to "repair" EV if an n is drawn on both hands and also if a p is drawn on one and an n on the other. Only in the case where a p is drawn on both hands does the EV need to be "repaired" (at least for SPL2 and SPL3 anyway.)

This is the point MGP was making that there is more than one path to the final EV.

k_c

[ericfarmer]

I have had second thoughts on this. Consider a shoe composition of 1 five, 1 six, and a bunch of tens and 10-10 vs 10 with possible resplits. EVn given an n has been drawn is clearly not equal to EVn from the starting condition. However the way you asked the question is that an n is drawn on each hand, in which case it seems EVs are equal. I based my "repair the EV" algorithm on there being no need to "repair" EV if an n is drawn on both hands and also if a p is drawn on one and an n on the other. Only in the case where a p is drawn on both hands does the EV need to be "repaired" (at least for SPL2 and SPL3 anyway.)

This is the point MGP was making that there is more than one path to the final EV.

k_c

I realize that I may have used misleading language in my phrasing of my question, or rather, by phrasing it as a question at all. That is, those two expected values *are* equal-- I was "asking" to clarify whether the disagreement on this point, that existed back in that 2003 bjmath.com thread, still exists today.

And we can even take this further. For example, instead of conditioning on the case where we split exactly two hands, instead condition on the case where we split exactly *three* hands (so that the pattern of dealing to the splits, using notation that I think MGP started with, is the combination of either NPNN or PNNN). Let random variables X_1, X_2, X_3 be the outcome of the first, second, and third of these hands, respectively. Then E[X_1]=E[X_2]=E[X_3]. Note that this isn't a computational simplification, it's a correct mathematical statement. (Extending this further, in the context of your algorithm description, the "repairing" at the endpoint is essentially really "renaming" of the computed values to reflect an expectation conditioned on a different subset of possible outcomes.(*) This renaming is what seemed to me to be the heart of the bjmath.com discussion. By conditioning as described here, we can skip the repairing, so to speak, and furthermore, generalize easily to arbitrarily large maximum numbers of split hands, using the two Catalan-ish summations discussed back in that bjmath.com thread.)

(*) I should emphasize, as I tried to do back then as well, that you're right that "there is more than one path to the final EV." I'm not trying to argue that anyone's algorithm is incorrect. But I did argue then about the interpretation/understanding of the mathematics-- the "labeling," so to speak-- underlying these algorithms.

E

[iCountNTrack]

There are several interesting wrinkles here, I'll just mention a few. First, the phrase "maximizes expected value" almost always requires clarification. There are two senses in which we can rarely actually claim to *maximize* achievable expected return. One is when restricting attention to total-dependent (vs. composition-dependent) strategy. That is, although composition-dependent strategy is more complex to *specify*, it's generally easier to actually *optimize*, i.e. to compute a strategy that actually maximizes the expected value of some appropriately-defined random variable. That's because composition-dependent expected value can be defined and computed recursively, while total-dependent strategy can have circular dependencies that require a "global" optimization approach that no CA I know of actually implements. For example, should I stand or hit with 10,3 vs. 10? That's a total of hard 13 vs. 10; resolving to hit requires ensuring that the EV of hitting is greater than that of standing... but the EV of hitting depends on making EV-maximizing decisions after the hit as well, e.g. 10,3,3 vs. 10, and the *probability* of even encountering that hand depends on having already resolved all decisions that *could* lead to that hand.

(Granted, I'm willing to let Nairn pass on this here, since he sort of sidesteps the issue by explicitly specifying a fixed playing strategy in his Appendix A. It simply would have been clearer/more correct to remove the claim of EV maximality altogether.)

Computing truly EV-maximizing total-dependent strategy is technically hard even if pair splitting were not a thing. But here we're focusing on pair splits, where things get even more complicated. This second problem is that any explicit strategy, whether total- or composition-dependent, computed by almost any CA (including mine), does not actually maximize the expected value of any easy-to-describe random variable, because of the complexity of accounting for pair splitting. For example, what should you do with 2,6 vs. dealer 5? If we temporarily ignore that pair splitting is a thing, we might resolve that the best strategy is to hit... but what if we encounter this hand as the result of splitting 2s against the dealer 5? Knowing that we're in half of the split might suggest that doubling down is optimal... but worse, if we now "retroactively" change our "zero memory" strategy to *always* double down 2,6 vs. 5, whether we are splitting or not, can *raise* the overall EV for the round.

(Various CAs expose various options for configuring how sophisticated our splitting strategy can be to deal with this sort of thing. For example, my CA's "default" strategy option, referred to elsewhere in this forum as CDZ-, is to stick with the "split-agnostic" strategy of hitting, applied after pair splits as well. There are other options like CDP1 or CDP, that allow us to "know" if we are in a split hand to modify strategy... but the holy grail referred to as CDZ (without the minus) that computes the "overall" EV-maximizing strategy that depends only on the dealer's up card and the composition of the player's *current* (possibly-split) hand, is beyond our current ability to compute AFAIK.)

There is more, but that's hopefully a conversation starter.

I was re-reading the thread to better understand the discussion, and forgot that CDZ meant optimal post-split strategy where strategy is always recomputed based on the dealer's up card and current hand composition of all split hands. My CA is only capable of computing the EV for player's 2 cards vs dealer up card, still not possible to compute an overall EV.