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Thread: Underlying Reason for Floating Advantage

  1. #27


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    Quote Originally Posted by DSchles View Post
    FA does NOT apply to a BS player!! It has no effect whatsoever on a BS player. Fewer decks in play, after starting with a greater number, does NOT help a BS player! You have to know the count in order for FA to make sense.

    If you shuffle six decks and play BS off the top, do you think your edge instantly becomes greater if you throw five decks on the floor and play BS with the one deck that remains??!!

    Don
    Yes i do. Are you saying that a single deck game the house edge is the same as in a 6 deck game?

    Are you saying that if you randomly take 52 cards out of a 6 deck shoe and then play 1 hand off the top your chance on average of getting a blackjack is exactly the same as if you dealt the hand from a shoe with 6 decks?

  2. #28


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    Quote Originally Posted by DSchles View Post
    If you shuffle six decks and play BS off the top, do you think your edge instantly becomes greater if you throw five decks on the floor and play BS with the one deck that remains??!!

    Don
    Ping! Lightbulb just went on - I get now why B.S. edge (lack thereof) doesn't change as the shoe depletes. Having, say, 2 decks remaining in a 6-deck shoe IS different than playing a 2-deck game because those two remaining decks in the shoe are almost certainly NOT the same composition as a freshly shuffled 2-deck game. Removing, say, an A during the final two decks of a shoe has the same effect on the odds of getting BJ as it would at any other point in the shoe.

    So that theory is toast... back to original question: what DOES cause FA?

  3. #29


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    Quote Originally Posted by Optimus Prime View Post
    Having, say, 2 decks remaining in a 6-deck shoe IS different than playing a 2-deck game because those two remaining decks in the shoe are almost certainly NOT the same composition as a freshly shuffled 2-deck game
    Of course its not the same. Because it has more potential variation in terms of the possible composition of cards. For example the 2 decks remaining could be loaded with 90% high cards or 90% low cards or anything in between.

    But for a given ratio of 10s and aces there will be slightly more blackjacks dealt from a smaller stock of cards than from a larger stock and the house edge will not be the same.

  4. #30


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    Quote Originally Posted by angle_sh00ter View Post
    Yes i do. Are you saying that a single deck game the house edge is the same as in a 6 deck game?
    Of course he's not saying that. See my other post above. If you shuffle 6 decks together and then remove a sequence of 52 cards from somewhere in the shoe, and play with those 52 cards, that's NOT the same as playing with a single deck.

  5. #31


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    Quote Originally Posted by angle_sh00ter View Post
    Hmm i may have poorly expressed my thoughts.
    Your thoughts are clear. They're simply wrong!

    Quote Originally Posted by angle_sh00ter View Post
    Off course your edge doesn't change just by flipping the cards upside down. You are still playing from 6 standard decks. Your edge will be exactly what the house edge is for a 6 deck game with whatever the rules are.
    It's strange that you understand this, but not the next part.

    Quote Originally Posted by angle_sh00ter View Post
    But are you saying that if you randomly shuffle the 6 decks and then randomly take 52 cards and then play a hand from just those 52 cards that your edge will be the same as it would be playing a hand dealt from the entire 6 deck shoe?
    Of course! What's the difference between randomly taking 52 cards or taking the bottom 52??

    Quote Originally Posted by angle_sh00ter View Post
    I'm saying it would not and i gave my reasons for why i think that.
    And I'm saying that you're palpably wrong.

    Don

  6. #32


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    Quote Originally Posted by angle_sh00ter View Post
    Yes i do. Are you saying that a single deck game the house edge is the same as in a 6 deck game?

    If you aren't counting,
    grabbing the bottom deck of six is NOT a single deck game!! One minute, you understand this, and then the next minute, you don't!

    Quote Originally Posted by angle_sh00ter View Post
    Are you saying that if you randomly take 52 cards out of a 6 deck shoe and then play 1 hand off the top your chance on average of getting a blackjack is exactly the same as if you dealt the hand from a shoe with 6 decks?
    Precisely the same! Why stop at one deck from six? Reductio ad absurdum, let's just take a "deck" of two cards. Do you understand now?

    Don
    Last edited by DSchles; 11-19-2018 at 01:52 PM.

  7. #33


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    Quote Originally Posted by Optimus Prime View Post
    Ping! Lightbulb just went on - I get now why B.S. edge (lack thereof) doesn't change as the shoe depletes. Having, say, 2 decks remaining in a 6-deck shoe IS different than playing a 2-deck game because those two remaining decks in the shoe are almost certainly NOT the same composition as a freshly shuffled 2-deck game. Removing, say, an A during the final two decks of a shoe has the same effect on the odds of getting BJ as it would at any other point in the shoe.
    Correct.

    Quote Originally Posted by Optimus Prime View Post
    So that theory is toast... back to original question: what DOES cause FA?
    Asked and answered by me, Griffin, and by Wong, pp. 70-71.

    Don

  8. #34


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    Quote Originally Posted by Optimus Prime View Post
    Ping! Lightbulb just went on - I get now why B.S. edge (lack thereof) doesn't change as the shoe depletes. Having, say, 2 decks remaining in a 6-deck shoe IS different than playing a 2-deck game because those two remaining decks in the shoe are almost certainly NOT the same composition as a freshly shuffled 2-deck game. Removing, say, an A during the final two decks of a shoe has the same effect on the odds of getting BJ as it would at any other point in the shoe.

    So that theory is toast... back to original question: what DOES cause FA?
    Quote Originally Posted by angle_sh00ter View Post
    Of course its not the same. Because it has more potential variation in terms of the possible composition of cards. For example the 2 decks remaining could be loaded with 90% high cards or 90% low cards or anything in between.

    But for a given ratio of 10s and aces there will be slightly more blackjacks dealt from a smaller stock of cards than from a larger stock and the house edge will not be the same.
    Saying the same thing over and over again, thinking we'll change our minds, doesn't suddenly make you right; it just makes you stubborn.

    Don

  9. #35


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    Quote Originally Posted by Optimus Prime View Post
    Of course he's not saying that. See my other post above. If you shuffle 6 decks together and then remove a sequence of 52 cards from somewhere in the shoe, and play with those 52 cards, that's NOT the same as playing with a single deck.
    Exactly correct.

    Don

  10. #36
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    Don I understand your point and the way you explain it by removing one deck is powerful but I am not convinced, though I do think you may be right, that BS has the same edge with any number of decks remaining. My skepticism is from the fact that whether they know it or not they are playing a much wider array of TC late in the shoe than early in the shoe. We know TC frequency is close to neutral or near neutral TCs and more extreme TCs are nonexistent early in the shoe. And late in the shoe the range of TC increases greatly and the frequency of TCs skews away from neutral TCs. This should have some kind of affect on the BS player even though he doesn't change his strategy and may be clueless about what the deck composition is. If someone has simmed this out of curiosity, then I will accept that. But to say playing one deck off the top when you know the deck is near neutral is the same as playing one deck late in the shoe when the range of TC and frequency of extreme TC is higher just doesn't seem to be a given. Add in the greater effect of removing one card and that has to have more of an effect.

    I am wondering that since an CSM increases advantage for the BS player slightly if they don't have a decreased advantage late in the shoe. After all by your logic continually playing the first round of a shoe should have no effect on advantage but the BS player has a slightly higher advantage when playing against a CSM. That has been researched and documented. It seems to defy the logic of it doesn't matter what part of the deck you play a BS strategist always have the same advantage. Is the only reason you are convinced that advantage for BS must be the same is because of your logical argument that you can throw 5 decks on the floor ..., or are you basing it on actual research? If it is research, give us the lowdown on the research. The CSM research seems to show it does matter what part of the shoe you are playing, and might be used to argue that since advantage is slightly higher on the first round for BS it must be slightly lower for the rest of the shoe. I just don't think your argument works for the reasons stated. Some counters play BS as their playing strategy and just use bet variations to get an advantage. They would never bet big with no cards removed.

    So what we know is on the first round out of the shoe the BS player is playing a deck composition that is neutral and that it has a bigger advantage than the average of all following deck composition later in the shoe. And late in the shoe a BS player may have a huge advantage or a huge disadvantage based on the current deck composition for the remaining cards, or anywhere in between. But we don't know how that averages out late in the shoe. I would be amazed if it averages out to the HE. I suspect it may average out to be more of a disadvantage than the HE or it may be an improvement over the HE from the effect on the remains deck composition from removing one card when it comes to resolving BJ matchups that take more than one card to resolve. I don't think this matters much to counters but it is an interesting aside. Your argument assumes that having a neutral deck at all times is the same as having a wide variety of deck compositions with increased frequency for extreme TC deck compositions. Your argument is very logical but I can't make it fit with the reality that I know early in the shoe the BS is always playing against a pack that is neutral or very close to it. And late in the shoe, even though he doesn't know it, the range of deck compositions is greatly varied with an increasing frequency of really whacky deck compositions as you approach no remaining cards. Those two situations can't be equivalent. Or perhaps I just think too much like a counter. Remember a BS player has a lower HE against a CSM.
    Last edited by Three; 11-19-2018 at 02:46 PM.

  11. #37


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    Quote Originally Posted by DSchles View Post

    If you aren't counting,
    grabbing the bottom deck of six is NOT a single deck game!! One minute, you understand this, and then the next minute, you don't!
    No i understand fully that playing with 52 random cards would not be the same as playing with a fresh single deck.

    However for arguments sake assume the 52 random cards happen to contain 4 aces and 16 10 value cards. Now we would have a situation where both the 5 decks remaining in the shoe and the 1 deck taken out both have the same % of aces and tens. Making for an easier comparison.

    If there are better odds of being dealt a BJ from 1 random deck than from the 5 random decks (when the ratio of 10s and aces are equal for both decks) then to my mind at least that demonstrates that one is inherently more advantageous, however slight the difference may be.

  12. #38


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    Lets have 6 single deck games. Whats the chance of a heads up player getting a blackjack? What is the house edge?

    Now lets take the 6 single decks and mix them together and play a shoe game with same rules. Now what is the chance of a heads up player getting a blackjack? What is the house edge?

    Now tell me why doing the reverse doesnt apply in the same way. Taking a 6 deck game and splitting it up into 6 single decks. Yes all the cards still add up to the same total but it plays different when split into 6 1 deck games

  13. #39


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    "Those two situations can't be equivalent."

    Those two situations are precisely equivalent. It's thinking otherwise that is completely aberrational. See Griffin's comment, bottom of page 70: "But the overall basic strategy edge must remain constant." Is there something about that statement that isn't clear?

    Don

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