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Thread: True Count & Number of Players

  1. #53


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    Quote Originally Posted by peterlee View Post
    Hi k_c, isn't it your bjstrat program can find out all combinations of the first round? Then we can know the exact ratios of cards are dealt in the first round.
    Let me try to explain the problem relative to 1 player versus dealer.

    1. Start with a full shoe and always use basic strategy

    Code:
    Round 1
       No matter where cut card is placed BSEV for first round = full shoe BSEV
       First round can consist of 4,5,6......x number of cards
       Some card combinations such as 4 cards consisting of 2,2,2,2 are impossible because round is incomplete
    
    Round 2
       Cut card needs to be placed such that it is never encountered on round 1 regardless of number of cards dealt on round 1
       If this condition is met then round 2 EV = round 1 EV? I say yes.
    If cut card is encountered on the first round at all this changes round 2 either more or less. For the series of sims I ran cut card was not encountered even once for 10,000,000 round 1 rounds when placed after the 15th card. The closer to the top cut card was placed the greater effect on round 2. Of course if it is placed after 1st, 2nd, 3rd, or 4th card there is no round 2.

    Having said this I don't know the effect on rank probabilities but I think the above parameters need to be considered in any sim to determine them. Position of cut card will certainly effect round 2 rank probs. Round 1 rank probs are just those of a full shoe. Would round 2 rank probs = round 1 rank probs if cut card is placed such that it can never ever be encountered in round 1? If not what would they be?

    k_c

  2. #54


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    Forget about the entire cut-card discussion. Just don't go there. Play a fixed number of rounds against one (or more) BS players. Don't confuse the issue by introducing the cut card. It isn't relevant to what we're trying to prove, namely that, if you don't run out of cards, the BS EV for all rounds is identical.

    Don

  3. #55
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    Quote Originally Posted by DSchles View Post
    Randomly shuffled deck of 2 red and two black cards. There are six equally probable sequences.
    Draw until first red card appears.
    Payoff is +1 for red, -1 for black.
    E round 1 is 0.
    After first round, six partial decks remain and we know they have exactly one red card.
    3 are red card poor, two are neutral and one is red card rich.
    Second round: repeat.
    Result: E=0 again.
    Bring the idea to blackjack: even the remaining cards is bias on average, the chance of having a blackjack in the second round can still be the same as the first round, so as to all the other combinations(?), that is why the two EVs are identical... This is beyond my knowledge and imagination, so I need some time to digest.

    Thanks Don

  4. #56
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    Quote Originally Posted by k_c View Post
    Assume basic strategy EV = 0.0%
    Assume deal from top of shoe for round 1 so round 1 EV = 0.0%
    Assume 2 possibilities for shoe comp & EV using basic strategy for round 2
    60% of the time more positive cards played on round 1 causes round 2 EV = -1.0%
    40% of the time more negative cards played on round 1 causes round 2 EV = +1.5%

    Round 2 EV = 0.0%, same as round 1 EV.
    This should be the same as Thorp's example, I missed the idea.

  5. #57


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    I will throw in my two cents in here. It may not be as mathematical or sophisticated but it works! The basic premise of any card game is to play with the information you have or in the case of a Basic Strategy player information you chose to have (since you decided to ignore the cards already seen). That being said, based solely on the information of the dealer's upcard and your cards your expectation value will always be the same at any point in the shoe. Think of it the dealer takes a slug from the top of the shoe or the middle of the shoe or the end of the shoe, your expectation is the same because you have no knowledge whatsoever about the slug and the results will average. This reminds of casino scams: Come and play our 2D game where they take 104 cards from an 8 deck shoe and claim it's a 2-deck game (but I digress). As Norm pointed out Combinatorial Analysis cannot be used to prove this that's because slug frequencies cannot be accurately computed (the reason being cards are not drawn randomly because of dealer and player strategy). The only way to show this to run a simulation using flat bet Basic Strategy with different number of rounds which will match the EV value computed using a Combinatorial Analyzer.
    Chance favors the prepared mind

  6. #58


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    I'll restate the claim made by Thorp, k_c, and me. The act of playing blackjack, even if it introduces a last-card bias on the first round for more tens than random, CANNOT change the BS EV from one round to the next, or even from the first round to the second round. BS EV is invariant over rounds, provided there is no cut card and you don't run out of cards.

    I'm certain that this has to be true.

    Don

  7. #59


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    I've read some of Thorp's correspondence with Don on this and I am now also convinced.

  8. #60


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    Algorithm: Compute statistical rank probs (no cut card) from sim

    Statistical probability of each rank after dealing n number of cards from a full shoe could be computed from a sim. n is determined by using basic strategy from the top of a freshly shuffled shoe, done for a large number of shoes. If deal is 1 player versus dealer and player busts round could be recorded without dealer draw or dealer could draw before recording. This may or may not make a difference but I am inclined to think it wouldn't.

    This is similar to a combinatorial analysis approach but CA wouldn't apply because not all card combinations are possible on a round. I don't have the time to do this right now. I think it's reasonably simple if someone else wants to implement it and post results.

    Hopefully I haven't made any logical errors.

    let max - maximum number of cards that can possibly be played on round 1
    let n = number of cards played on round 1 (range = 4 to max)

    let round1[n] = accumulation of number of round 1 rounds of n cards (determined by sim for each possible n)
    let p[n] = prob of n cards on round 1
    Code:
    p[n] = round1[n] / (round1[4] + round1[5] + .... + round1[max])
    let r = ranks (range 1 to 10)
    let postRound1Comp[n][r] = accumulation of comp of each rank post round 1 where round 1 consists of n cards (sim)
    let avePostRound1Comp[n][r] = average post round 1 number of each rank where round 1 consists of n cards
    let avePostRound1_pRank[n][r] = average prob of rank r where round 1 consists of n cards
    Code:
    avePostRound1Comp[n][r] = postRound1Comp[n][r] / round1[n] for each n = 4 to max
    avePostRound1_pRank = avePostRound1Comp[n][r] / (52 * decks - n) for each n = 4 to max
    let statistical_pRank[r] = statistical prob of each rank for all post round 1 (round 2) rounds (initialize statistical_pRank[r] to 0)
    Code:
    for r = 1 to 10
    for n = 4 to max
         statistical_pRank[r] += p[n] * avePostRound1_pRank[n][r]
    next n
    next r
    k_c
    Last edited by k_c; 01-21-2023 at 04:24 PM. Reason: add avePostRound1_pRank[n][r] as a parameter

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