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Thread: Need help to understand Throp's paper.

  1. #1
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    Need help to understand Throp's paper.

    http://www.edwardothorp.com/wp-conte...rEachRound.pdf

    In page 6, [Proof] section...How does the mapping prove the probability distributions for the two segments are identical?

    +++

    Proof. Consider two games played on the same pack of n cards. Game
    1, or G1, has m rounds with strategy sets S1, · · · , Sm. On round k, player i
    plays simple strategy S. A typical ordering is x1, · · · , xn. In Game 2, or G2,
    there is just one round with only one player. This player also uses strategy
    S. A typical ordering is y1, · · · , yn. Let f : x = x1, · · · , xn ? y = y1, · · · , yn
    be the mapping which selects the cards used by player i, round k of G1, and
    those cards which the dealer needs to use in order to settle the hand of player
    i in G1, and maps them into A = y1, · · · , yt. If player i busts, the dealer in
    G1 doesn’t have to draw for player i, but may have to draw for other players
    who don’t bust. These latter cards are not selected for A by f . Also if player
    i has a natural the dealer settles with player i without drawing any cards.
    Any cards he might draw because of other players are not selected for A.
    The order of cards in A is the same as the order in which they were used in
    G1, so that they exactly suffice and the player has the same result in G2 as
    in G1. Preserve the order of the remaining cards in x and map them into
    B = yt+1, · · · , yn.

    This mapping is defined on every x because of (A3m) and the fact the
    strategy sets are deterministic. It is well-defined, i.e. each x maps onto one
    and only one y. Further, f is onto, i.e. every y corresponds to some x. To
    see this, pick any y = (y1, · · · , yn). Use game 2 to select an initial segment
    A. Call the rest B. Now use A,B to play game 1 as follows. On round k draw
    cards in order from A whenever the S player or the dealer need cards prior to
    settling the S player’s hand. Also draw cards in order from B whenever other
    players or the dealer on round k, or anyone on other rounds, need cards.

    Since f is onto and the number of x’s and y’s are the same, namely n!,
    then f is one-to-one whence the set of f (x) = y are simply a shuffling or
    rearrangement of the x’s. Therefore the list of sub orderings in G1 which
    map into A’s in G2 is identical to the list of A’s except for rearrangement.
    Therefore the probability distributions for S are identical in G1 and G2. Thus
    the probability distribution for S is always the same as that for round 1 with
    only one player, using S, versus the dealer. This proves the theorem.

  2. #2


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    The key is where he says
    Quote Originally Posted by peterlee View Post
    Since f is onto and the number of x’s and y’s are the same, namely n!,
    then f is one-to-one whence the set of f (x) = y are simply a shuffling or
    rearrangement of the x’s
    Because the mapping is one-to-one, then each card from the first segment maps to the same card in the second segment and vice versa.

  3. #3
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    Quote Originally Posted by Gronbog View Post
    The key is where he says

    Because the mapping is one-to-one, then each card from the first segment maps to the same card in the second segment and vice versa.
    (1)..Let f : x = x1, · · · , xn ? y = y1, · · · , yn be the mapping which selects the cards used by player i, round k of G1?

    (2)..The order of cards in A is the same as the order in which they were used in G1, so that they exactly suffice and the player has the same result in G2 as in G1. ?

    (3)..each x maps onto one and only one y. Further, f is onto, i.e. every y corresponds to some x.?

    (4)..Since f is onto and the number of x’s and y’s are the same, namely n!,
    then f is one-to-one whence the set of f (x) = y are simply a shuffling or
    rearrangement of the x’s.?

    (1) is the assumption.
    Because of (1), it should be (2)
    Because of (2), it should be (3)
    Because of (3), it should be (4), and the reason is n!.

    +++
    Should I read as this way? (If it is, I still don't understand)

    Is it because there are n! sequents, the frequency of what appear in the first round, is the same frequency that appear in the second round or any round?

  4. #4


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    Start with a full shoe. You are player X.

    Consider 3 scenarios. X plays with same strategy in each.

    1. All cards dealt from a full shoe.
    Code:
         X plays versus dealer
         Probability of each possible hand and EV can be computed
    2. Start with full shoe & remove some cards and throw them on the floor. If too many cards are thrown on the floor it is necessary to draw needed cards from floor instead of shoe.
    Code:
         X plays versus dealer
         This is the same as scenario 1 except some cards are in shoe and some on floor
    3. Start with full shoe & remove some cards and throw them on the table. Several other players grab the cards on the table and fight amonst themselves to construct valid hands out of them. If a hand is not valid or 1 or more of these additional players wants more cards, additional cards would come from shoe.
    Code:
         X and several other players play versus dealer
         This is the same as scenario 1 for player X except some cards are in shoe and some on table
    All scenarios are identical relative to player X versus dealer. X only knows his cards and dealer's up card whether unseen cards are in shoe, on floor, or on table.

    k_c

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    Quote Originally Posted by k_c View Post
    Start with a full shoe. You are player X.

    Consider 3 scenarios. X plays with same strategy in each.

    1. All cards dealt from a full shoe.
    Code:
         X plays versus dealer
         Probability of each possible hand and EV can be computed
    2. Start with full shoe & remove some cards and throw them on the floor. If too many cards are thrown on the floor it is necessary to draw needed cards from floor instead of shoe.
    Code:
         X plays versus dealer
         This is the same as scenario 1 except some cards are in shoe and some on floor
    3. Start with full shoe & remove some cards and throw them on the table. Several other players grab the cards on the table and fight amonst themselves to construct valid hands out of them. If a hand is not valid or 1 or more of these additional players wants more cards, additional cards would come from shoe.
    Code:
         X and several other players play versus dealer
         This is the same as scenario 1 for player X except some cards are in shoe and some on table
    All scenarios are identical relative to player X versus dealer. X only knows his cards and dealer's up card whether unseen cards are in shoe, on floor, or on table.

    k_c
    I understand the 3 scenarios are the same, just like after shuffling, dealer let the player to cut away some cards and put them at the bottom, there is no different from where to cut, the overall EV is the same.

    There is another scenario, the player can choose a point there is a picture and cut it away, the dealer will put those cards including the picture to the bottom of the shoe, so there will be a picture less for the first round(is it?).

    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.
    Thorp proves there is the same EV from his example, even a red card is shorted after the first round.
    But how to prove it is the same as a blackjack game?
    Throp's paper explained it, but I don't understand the mapping one to one method.

  6. #6


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    Quote Originally Posted by peterlee View Post
    I understand the 3 scenarios are the same, just like after shuffling, dealer let the player to cut away some cards and put them at the bottom, there is no different from where to cut, the overall EV is the same.

    There is another scenario, the player can choose a point there is a picture and cut it away, the dealer will put those cards including the picture to the bottom of the shoe, so there will be a picture less for the first round(is it?).
    I may be wrong but I think Thorp's paper is saying that unseen cards cannot alter probabilities because no information is known about them. If you tell me you have seen a card and it is a ten then unseen doesn't apply.

    The only player that "sees" cards and can apply this information to his strategy is the player I have designated as player X. I think what Thorp is saying is that any other cards that may or may not be seen after the fact are just random "noise."

    Also dealer "sees" cards and applies this information to his strategy. player X EV depends on how well his strategy does statistically versus dealer strategy.

    k_c

  7. #7
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    Quote Originally Posted by k_c View Post
    I may be wrong but I think Thorp's paper is saying that unseen cards cannot alter probabilities because no information is known about them. If you tell me you have seen a card and it is a ten then unseen doesn't apply.

    The only player that "sees" cards and can apply this information to his strategy is the player I have designated as player X. I think what Thorp is saying is that any other cards that may or may not be seen after the fact are just random "noise."

    Also dealer "sees" cards and applies this information to his strategy. player X EV depends on how well his strategy does statistically versus dealer strategy.

    k_c
    I use a seen card in the example, just try to simplify, because I think a seen ten cut, and a first round played, both have the same result, that is, the running count will drop to negative on average before the next hand.
    Let's forget about the seen card, go back to the blackjack game, we are quite sure the first round use more tens on average. When Don brought the problem to Thorp, I guess Don has also mentioned about this.
    Thorp replied by an example, shows that even black cards are more than red cards, the second round is still ev=0.
    Doesn't Thorp want to say, even the first round uses more tens in blackjack, the ev of the second round is still =0 ?

    In the paper: Theorem 1. (Fundamental invariance theorem for simple strategies.)
    Given (A1), (A2) and (A3m), any specified simple strategy S has a probability distribution on segments which is invariant,...
    Blckjack rules and BS should be a simple strategy.
    So rules and BS apply to the first round and then turns out more tens are used, this is not an additional information.
    I am not saying it is wrong in the paper, I just don't understand the proof.

    +++
    (I tried Thorp's example using more cards, with different rules, the results are the same ev=0 after the first round. I did not know about this before, and still don't know the reasons how to make them unchanged..one thing in my mind, for all the combination lists, no matter what happen at the beginnings of the game, the rest of the sequents in the lists keep unchanged...??)

  8. #8


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    Quote Originally Posted by peterlee View Post
    I use a seen card in the example, just try to simplify, because I think a seen ten cut, and a first round played, both have the same result, that is, the running count will drop to negative on average before the next hand.
    Let's forget about the seen card, go back to the blackjack game, we are quite sure the first round use more tens on average. When Don brought the problem to Thorp, I guess Don has also mentioned about this.
    Thorp replied by an example, shows that even black cards are more than red cards, the second round is still ev=0.
    Doesn't Thorp want to say, even the first round uses more tens in blackjack, the ev of the second round is still =0 ?

    In the paper: Theorem 1. (Fundamental invariance theorem for simple strategies.)
    Given (A1), (A2) and (A3m), any specified simple strategy S has a probability distribution on segments which is invariant,...
    Blckjack rules and BS should be a simple strategy.
    So rules and BS apply to the first round and then turns out more tens are used, this is not an additional information.
    I am not saying it is wrong in the paper, I just don't understand the proof.

    +++
    (I tried Thorp's example using more cards, with different rules, the results are the same ev=0 after the first round. I did not know about this before, and still don't know the reasons how to make them unchanged..one thing in my mind, for all the combination lists, no matter what happen at the beginnings of the game, the rest of the sequents in the lists keep unchanged...??)

    I have suggested an algorithm which would use data from a sim to try and enumerate statistical probabilities of each rank for a game of 1 player versus dealer on round 2 using basic strategy following round 1 using basic strategy (no cut card influence) in other thread. I think 1 player alongside of several other players would turn out the same as indicated by Thorp.

    I do not think Gronbog incorporated number of cards in his data. If all card combinations were possible combinatorial analysis could be used to compute probability of a round consisting of n cards but instead this could come from sim.

    If n = 4 cards probability of tens is high
    As n increases probability of tens lessens and probability of low cards increases

    Sim would accumulate number of 4 card, 5 card, 6 card,....., n card rounds
    From that compute prob 4 card round, 5 card round, 6 card round,.......,n card round

    Sim would also accumulate comp of each rank after 4 cards, comp after 5 cards,.....,comp after n cards
    From that compute prob of each rank after 4 cards, prob after 5 cards,......,prob after n cards (divide by 52*decks - n)

    From this compute statistical prob of each rank
    let r = rank
    let n = number of cards in round 1
    let max = max number of cards needed to complete round 1
    let p[n] = prob of n cards dealt on round 1 (from above)
    let pRank[n][r] = prob of each rank after n cards dealt on round 1 (from above)
    let statistical_pRank[r] = statistical prob of each rank after round 1 is dealt

    Code:
    for r = 1 to 10
    for n = 4 to max
         statistical_pRank[r] += p[n] * pRank[n][r]
    next n
    next r
    I'm not sure if I explained any better than before. The idea is that statistical_pRank should be weighted by the probability of the number of cards drawn on round 1.

    Without making any claims, if everything falls into place it may be turn out that
    statistical_pRank could approximate {1/13,1/13,1/13.1/13,1/13,1/13,1/13.1/13,1/13,4/13}

    k_c

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    Chatgpt translated:


    This is how I can visualize the situation and better understand why the expected values of the two rounds are the same:

    Imagine a spreadsheet with 52 columns and 52! rows. Each row contains a unique sequence of the 52 cards.


    If the first round uses the cards #1234, this "card type" has 48! sequences starting from the first card.
    If the second round uses the cards starting from the fifth card, there will also be 48! sequences with the card type #1234.
    Thus, each card type in the first and second rounds has the same number of sequences, and therefore the expected value is the same.

    If the first round uses five cards, such as #56789, it means that #5678 is not enough to complete the game and the fifth card is needed. This results in 48 card types replacing #5678, consisting of #5678+1, #5678+2, and so on up to #5678+9,...+52. Each of these 48 card types has 47! sequences, giving a total of 48*47!=48! sequences. Therefore, #5678 and the 48 card types that replace it have the same 48! sequences.

    No matter how many additional cards are drawn in the first round, the second round will still have the same card types with the same number of sequences.


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    Hi PeterLee,

    I read through the paper as well as the proof of the theorem you mentioned several times. I paid particular attention to the mapping argument used by Thorp to justify the invariance theorem. Here is my takeaway insofar as it relates to your confusion about why his mapping argument works:

    Imagine I have a biased D6 die with probability distribution P(X=1) =/= ... =/= P(X=6). It is not strictly necessary that the probabilities are all distinct, but it does allow us to follow the logic more closely. The expected value of the number rolled on the upppermost face can easily be computed as E(X) = 1*P(X=1) + ... + 6*P(X=6). Permute the listing of the values of your random variable in any one of 6! ways. For example, instead of listing X=1,2,3,4,5,6 and P(X=x) = 1/21,2/21,3/21,4/21,5/21,6/21 in that order, write X=5,2,3,1,6,4 and P(X=x) = 5/21,2/21,3/21,1/21,6/21,4/21. This can be done either at random or according to a deterministic decision criterion; either way, there exists some rule that maps the ordered listings of values and attendant probabilities with the rearranged listings. Call this rule f. f is guaranteed to be one-to-one and onto, so that every value of my random variable along with its probability of occurrence occurs once and only once post-arrangement. Thus, after rearranging everything in the listing of my random variable I'm still able to write out the probability mass function and compute the correct expected value, even if it may be slightly inconvenient for me to list out everything now. Using my example earlier, E(X) = 5*5/21+2*2/21+...+4*4/21 is still the correct expected value E(X) = 1*1/21+2*2/21+...+6*6/21 thanks to the commutative property of addition of real numbers. Again, the crux of the argument is that the conclusion only works because the rule f by which I rearrange the listings of values and probabilities of my random variable X is one-to-one and onto, so nothing essential is lost and nothing inessential is gained post-arrangement.

    Thorp applied similar reasoning to the discrete random variable representing the player's net profit in this game where cards are dealt without replacement instead of this die roll where dice are rolled with replacement, and he had to address different cases depending on which cards were dealt initially. The underlying mapping argument still remains valid. Hope this helps!
    Last edited by JohnGalt007; 07-29-2023 at 12:46 PM.

  11. #11
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    Quote Originally Posted by JohnGalt007 View Post
    Hi PeterLee,

    I read through the paper as well as the proof of the theorem you mentioned several times. I paid particular attention to the mapping argument used by Thorp to justify the invariance theorem. Here is my takeaway insofar as it relates to your confusion about why his mapping argument works:

    Imagine I have a biased D6 die with probability distribution P(X=1) =/= ... =/= P(X=6). It is not strictly necessary that the probabilities are all distinct, but it does allow us to follow the logic more closely. The expected value of the number rolled on the upppermost face can easily be computed as E(X) = 1*P(X=1) + ... + 6*P(X=6). Permute the listing of the values of your random variable in any one of 6! ways. For example, instead of listing X=1,2,3,4,5,6 and P(X=x) = 1/21,2/21,3/21,4/21,5/21,6/21 in that order, write X=5,2,3,1,6,4 and P(X=x) = 5/21,2/21,3/21,1/21,6/21,4/21. This can be done either at random or according to a deterministic decision criterion; either way, there exists some rule that maps the ordered listings of values and attendant probabilities with the rearranged listings. Call this rule f. f is guaranteed to be one-to-one and onto, so that every value of my random variable along with its probability of occurrence occurs once and only once post-arrangement. Thus, after rearranging everything in the listing of my random variable I'm still able to write out the probability mass function and compute the correct expected value, even if it may be slightly inconvenient for me to list out everything now. Using my example earlier, E(X) = 5*5/21+2*2/21+...+4*4/21 is still the correct expected value E(X) = 1*1/21+2*2/21+...+6*6/21 thanks to the commutative property of addition of real numbers. Again, the crux of the argument is that the conclusion only works because the rule f by which I rearrange the listings of values and probabilities of my random variable X is one-to-one and onto, so nothing essential is lost and nothing inessential is gained post-arrangement.

    Thorp applied similar reasoning to the discrete random variable representing the player's net profit in this game where cards are dealt without replacement instead of this die roll where dice are rolled with replacement, and he had to address different cases depending on which cards were dealt initially. The underlying mapping argument still remains valid. Hope this helps!
    Thanks for the reply. I am not so clear about the mapping, partly due to the logic and partly due to the English, as I need to use translator to help me.

    [cards are dealt without replacement instead of this die roll where dice are rolled with replacement]
    How do I know the one-to-one mapping is still work on the card game without replacement?

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    Quote Originally Posted by peterlee View Post
    Thanks for the reply. I am not so clear about the mapping, partly due to the logic and partly due to the English, as I need to use translator to help me.
    Happy to help! I sympathize with your confusion. It happens to everyone at some point, especially if English or math are not your first languages. I recommend that you stay patient with the process and with yourself. If blackjack advantage play has taught us all anything, it's that you'll eventually understand it with persistence and dedication. The law of large numbers is in your favor in this case.

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    Quote Originally Posted by peterlee View Post
    [cards are dealt without replacement instead of this die roll where dice are rolled with replacement]
    How do I know the one-to-one mapping is still work on the card game without replacement?
    Any answer I could give here would either repeat Thorp's words exactly or else paraphrase his words to the point that such a simplification would be misleading. The argument is subtle and requires specialized notation to prove and to communicate. I recommend that you start from the beginning of the paper and follow the entire argument from the beginning. Make sure you understand each sentence and definition completely before moving on to the next sentence. Take notes everywhere he gives definitions, like rules to a game. And again, don't be too hard on yourself if you don't see it right away. Reading a mathematical paper is a separate skill from reading most other technical papers, but it can be learned and sharpened with practice.
    Last edited by JohnGalt007; 08-05-2023 at 04:48 PM.

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