Thanks for the calculation!Originally Posted by ericfarmer
Sincerely,
Cac
No. I certainly expect that your statement will hold for the simplified model if it holds for the general real-world model.
I just want to make sure you agree that this simplified model is an sufficient test of the principle you are espousing.
I can imagine an even simpler model where the dealer and the player both stand. Then, it is clear to me that the EV of every round is the same. The basic strategy does not change the cards that are used for successive rounds.
In a model where the dealer has a basic strategy that depends on the card flow (draw until a total of 17), the dealer eats more small cards than big cards in the first round. There is no need to have the player also using a complicated BS to cause the number of cards used per round to fluctuate even more. Therefore, the simplified model tests the fundamental issue at hand. I hope this will shed light on the OP's question.
I don't know offhand how many cards we need from a single deck to guarantee that we get two rounds completed, but I know it is easy to do a complete CA for the first 12 cards in a reasonable amount of time because I have done it before.
I think it is evident that Don's conjecture is correct for the simplified model I outlined above. Since the player always gets the first and third card in the deck and then stands in the first round, then a sequence of cards off the top such as '49354' will complete a round. The dealer stands after drawing to '954' for a total of 18. A possible sequence for the second round is 'T8TT' with the dealer standing on a total of 18. The player loses the first round and wins the second round.
Every possible shuffle of the deck starting with the sequence '49354T8TT' will consume 5+4 cards in the two rounds and have the same scoring results for the two rounds. But, the transposed sequence 'T8TT49354' is equally likely to come out of the shuffle. This shuffle will consume 4+5 cards and the scoring results will be reversed.
So, it is obvious that the first and second rounds will always have the same EV. There is no outcome of the two rounds that is not equally likely to come out in the reverse order. This will still be true if we give the player a fixed BS for the two hands. However, it will take longer to do a complete combinatorial analysis as we make the model more complex.
As soon as we add a cut card, the symmetry is broken and it is less likely to get some permutations of the different rounds than other permutations because some rounds have more cards than other rounds.
Last edited by OnlineAP; 05-16-2025 at 02:04 PM.
I was impressed that that you had a method to solve for the minimum number of hands needed to reach the cut card even if it wasn't perfect for SPL3. This seems like a really complicated problem that I would never even attempt to solve.
Thanks for sharing all your code and for your very clear explanations.
I don't agree with your wording or analysis, above, of what I wrote. Thorp mentions nothing whatsoever about 18 rounds. That was your finding, not his. And, yet again, my only error--despite your implication to the contrary--was in thinking that the CCE--which, again, was NOT the original discussion--doesn't manifest itself until we get near the cut card. And that belief was simply due to the fact that any or all previous studies of the CCE never went out to the, what, tenth decimal place?--to show that there was any difference.
I wouldn't change a syllable of what I wrote here: ""Here's what's true: With no cut card, if you play a fixed number of rounds, with a guarantee that you can never run out of cards for the final round, the BS EV for a hand off the top of the shoe and the BS EV for the entire shoe, playing all hands, are identical." Is there something you would change in that statement?
Don
Now I am confused as to what you mean by 'doesn't manifest itself until we get near the cut card.' The cut card has no effect on any shoe if we never hit the cut card. There is no 'proximity fuse' in the cut card. Eric is correctly pointing out that you can hit the cut card in some shoes way earlier than we would expect, given that the median number of rounds is out in the 40s. You are right to say that it is entirely unimportant from an EV perspective. I have never spent much time thinking about the CCE. Thanks to Eric's work, I think I understand the CCE better now thanks to this discussion.
I don't know if anyone has looked at it in the following way. Put the cut card anywhere you want, but at a fixed spot with relatively shallow penetration. Now play a fixed number of rounds such that you hit the cut card on some modest fraction of all rounds. Retroactively cancel all bets and payoffs for any shoe where you hit the cut card. Now, calculate the EV of all the rounds in the remaining shoes. I expect the EV to be the same as the EV for the first round. This should be true even though many of the shoes are played until very near the cut card or even exactly up to the cut card.
Sorry if my wording was confusing. If the cut card is at 75% of a six-deck shoe (234 cards), and, at the 19th round (maybe 100 cards, playing alone), the BS EV changes, this is what I meant by "'doesn't manifest itself until we get near the cut card." Stating that another hand might be involved is not what I meant.
Don
For the single-deck game, the EV of the first round was derived by summing the results of all 52! possible sequences.
The second-round EV is calculated in the same way. The two rounds have the same number of sequences with the same outcomes, that is why both EV are identical. This is the core insight of the paper.
If we exclude any of the results from the 52! sequences in the second round, the sum will differ.
In the paper, A1 to A6 and D1 to D10, some of these rules must be followed, while others cannot be violated. Otherwise, the expected value (EV) for later rounds will no longer remain the same.
If we only mention three of them and treat this subset as the complete framework, it may mislead others into assuming nothing else applies.
The case of 'lij45o6' and youself illustrate this issue.
A card counter will alter the BS player's EV, along with many other factors.
No, there is nothing I would change in your statement above. I'm noting your comment before this, where you refer to "namely, Thorp's paper and, hence, NO CCE (your emphasis)". The word "hence" means "by implication", suggesting that (you think that) your original topic of discussion being Thorp's paper implies an assumption of no CCE, or equivalently, that if we go off in the weeds talking about the CCE, then Thorp's paper stops being relevant or applicable. This isn't true; see Section 3, p. 34-38. He even presents a specific "small example" game (Example 6, "Woolworth blackjack") that can be analyzed exactly, very similarly to the more complex-but-still-analytically-tractable "lose a dollar for each non-ten" example game that I presented earlier in this thread.
Again, this simply isn't true, or is at least pretty selective reading. See his assumption (A3m) ("Assume that m rounds can always be played"), and the Example 5 p. 33-34 where he works through an admittedly more back-of-the-envelope specific calculation of a maximum applicable value of m for 8 decks and SPL1. Okay, he doesn't explicitly say, "6D at 75% pen yields m=18", but it's interesting and pretty cool that even Thorp's simple counting argument works pretty well as a non-rigorous *guess* at m even for a shallower cut card like our 75% running example: 85 pips per suit x 4 suits x 6 decks, reduced to 75% penetration, is a total of 1530 pips, divided by a maximum of 82 pips per round (for SPL1)... yields at least m=18 rounds that can be completed, and thus "only" the first 18 rounds are guaranteed to have the same EV. The only thing missing is the explicit calculation that we conducted here.
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