• Expectation maximization versus Kelly criterion

    From Timothy Chow@21:1/5 to All on Sat Oct 16 21:21:10 2021
    Here's something that I've wondered about for some time but have
    never managed to make precise.

    The conventional wisdom for backgammon money play is that you
    should make the play that maximizes expectation (or "equity" if
    you like).

    Now, it is well known that in some situations, maximizing
    expectation is the wrong idea. The Wikipedia article on the
    Kelly criterion mentions this example:

    In a study, each participant was given $25 and asked to place
    even-money bets on a coin that would land heads 60% of the
    time. Participants had 30 minutes to play, so could place about
    300 bets, and the prizes were capped at $250.

    Without the $250 cap, the strategy that maximizes expectation is
    to bet your entire bankroll each time. But even if there were no
    cap, it seems crazy to adopt a strategy that is virtually certain
    to leave you bankrupt. The Kelly strategy of betting 20% of your
    stake each time seems much more sensible.

    The thing that I've wondered is whether it's possible to come up
    with a backgammon scenario where we can make a plausible argument
    that expectation maximization is the wrong thing to do.

    Of course, in a match, or in a situation with table stakes, we know
    that there are plenty of situations where you wouldn't just "make
    the money-game play." But what I have in mind are those funny
    positions that demonstrate that when there truly is no limit to the
    cube value, then money-game equity is undefined. Can we somehow
    take advantage of those positions to come up with an argument against expectation maximization?

    I know this is vague; as I said, I've never managed to make this
    idea precise...

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Sun Oct 17 05:11:19 2021
    On Sunday, October 17, 2021 at 2:21:14 AM UTC+1, Tim Chow wrote:
    Here's something that I've wondered about for some time but have
    never managed to make precise.

    The conventional wisdom for backgammon money play is that you
    should make the play that maximizes expectation (or "equity" if
    you like).

    Now, it is well known that in some situations, maximizing
    expectation is the wrong idea. The Wikipedia article on the
    Kelly criterion mentions this example:

    In a study, each participant was given $25 and asked to place
    even-money bets on a coin that would land heads 60% of the
    time. Participants had 30 minutes to play, so could place about
    300 bets, and the prizes were capped at $250.

    Without the $250 cap, the strategy that maximizes expectation is
    to bet your entire bankroll each time. But even if there were no
    cap, it seems crazy to adopt a strategy that is virtually certain
    to leave you bankrupt. The Kelly strategy of betting 20% of your
    stake each time seems much more sensible.

    The thing that I've wondered is whether it's possible to come up
    with a backgammon scenario where we can make a plausible argument
    that expectation maximization is the wrong thing to do.

    Of course, in a match, or in a situation with table stakes, we know
    that there are plenty of situations where you wouldn't just "make
    the money-game play." But what I have in mind are those funny
    positions that demonstrate that when there truly is no limit to the
    cube value, then money-game equity is undefined. Can we somehow
    take advantage of those positions to come up with an argument against expectation maximization?

    I know this is vague; as I said, I've never managed to make this
    idea precise...

    It's utility theory. You maximise the expected utility of your wealth.
    The reality is that money backgammon should never be played to maximise
    your equity or expected value, but should be played to maximise the expected utility.

    The (false) model assumes that players are staking at a low enough level that they
    maximise their equities by maximising their expected utilities.

    Suppose players are playing for $10 a point. At this level, they are unlikely to be comfortable with losing $1000.
    So suppose the cube is at 64.
    And suppose you own the cube and it's your final roll, and you have 19 winning rolls and 17 losing rolls. Equity-wise, it's clearly D/T.
    This would give you an expected utility of 19/36(utility(wealth + 1280) ) + 17/36 (utility(wealth - 1280))
    This is likely to be less than holding to give an expected utility of 19/36(utiity(wealth + 640)) + 17/36(utility(wealth-640)).

    So you hold to maximise your expected utility.
    How do you know your expected utility? That's a personal matter which reflects your preferences
    for wealth outcomes -- for example a preference for a sure $1000 as opposed to a 50% chance of
    gaining $2001.
    If you google "utility function" you will probably find a realistic function to justify the above calculations.
    Bg players have an intuitive feeling for the above, which is why they probably wouldn't double in the above
    situation. They would try to settle and hold if no settlement was available.

    As a practical matter, the fair settlement is (obviously) that the on-roll player gets 1/18 of $1280 = very slightly more
    than $71.11. Since paying is psychologically unpleasant, I would round down and suggest that the underdog
    pays $71.11.
    However, the underdog might not agree to this. Any settlement where the favourite receives a non-negative
    amount is likely to be better than just rolling so a settlement of zero should be available.
    But you can't force your opponent to settle, so if you can't agree, hold instead of double.

    But you may say "But maybe the players can easily afford the loss." That's a bit unlikely else they'd be playing for higher
    stakes in the first place.

    Paul

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  • From MK@21:1/5 to Tim Chow on Mon Oct 18 00:58:03 2021
    On October 16, 2021 at 7:21:14 PM UTC-6, Tim Chow wrote:

    I know this is vague; as I said, I've never managed to make this
    idea precise...

    Yup, sure sounds like optimun delirium equilibrium...

    MK

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Mon Oct 18 22:13:31 2021
    On 10/17/2021 8:11 AM, peps...@gmail.com wrote:
    It's utility theory. You maximise the expected utility of your wealth.

    Well, yes, that's one way to argue against maximizing equity.
    But the Kelly criterion isn't derived from maximizing utility.

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to peps...@gmail.com on Tue Oct 19 00:58:05 2021
    On Tuesday, October 19, 2021 at 8:55:15 AM UTC+1, peps...@gmail.com wrote:
    On Tuesday, October 19, 2021 at 3:13:33 AM UTC+1, Tim Chow wrote:
    On 10/17/2021 8:11 AM, peps...@gmail.com wrote:
    It's utility theory. You maximise the expected utility of your wealth.
    Well, yes, that's one way to argue against maximizing equity.
    But the Kelly criterion isn't derived from maximizing utility.
    The Kelly criterion maximises utility if you assume that the utiity of your wealth
    is the log of your wealth. Since (for example) log (500) > (log 250 + log 750)/2,
    the Kelly criterion results in an anti-gambling bias (risk-aversion).
    This is because the log function is concave.
    Other concave increasing functions could be chosen, which might be less conservative
    than Kelly, which is very conservative.
    And, re your capped-at-250 game:
    If you really can do 300 coin tosses, then you can mail me my cheque for $250 right now.
    If I gamble 5 each time, then I'm extremely unlikely to run out of cash.
    My expected number of wins is 180 with the standard deviation being sqrt(300 * 0.24) = approx 8.5
    I make my 250 with 150 wins. I can only fail this (assuming I don't run out of cash) if I get more than 3.5
    standard deviations less than the mean -- something we can pretty much rule out.
    Since I'm generous, I'll pay the postage cost myself and accept a cheque for only $240.

    I messed up the computations trying to do them too fast -- 150 wins out of 300 only breaks even.
    But the first part of my post might be correct.

    Paul

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Tue Oct 19 00:55:14 2021
    On Tuesday, October 19, 2021 at 3:13:33 AM UTC+1, Tim Chow wrote:
    On 10/17/2021 8:11 AM, peps...@gmail.com wrote:
    It's utility theory. You maximise the expected utility of your wealth.
    Well, yes, that's one way to argue against maximizing equity.
    But the Kelly criterion isn't derived from maximizing utility.

    The Kelly criterion maximises utility if you assume that the utiity of your wealth
    is the log of your wealth. Since (for example) log (500) > (log 250 + log 750)/2,
    the Kelly criterion results in an anti-gambling bias (risk-aversion).
    This is because the log function is concave.
    Other concave increasing functions could be chosen, which might be less conservative
    than Kelly, which is very conservative.
    And, re your capped-at-250 game:
    If you really can do 300 coin tosses, then you can mail me my cheque for $250 right now.
    If I gamble 5 each time, then I'm extremely unlikely to run out of cash.
    My expected number of wins is 180 with the standard deviation being sqrt(300 * 0.24) = approx 8.5
    I make my 250 with 150 wins. I can only fail this (assuming I don't run out of cash) if I get more than 3.5
    standard deviations less than the mean -- something we can pretty much rule out.
    Since I'm generous, I'll pay the postage cost myself and accept a cheque for only $240.

    Paul

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Tue Oct 19 12:19:26 2021
    On 10/19/2021 3:55 AM, peps...@gmail.com wrote:

    The Kelly criterion maximises utility if you assume that the utiity of your wealth
    is the log of your wealth.

    That is true, but that is not how Kelly *derived* his criterion.
    Taking your suggestion above, we are left with the question, why
    take the utility to be the log of your wealth? Why not the square
    root, or some function derived from empirical observation of actual
    human behavior?

    Instead, suppose that we decide to adopt the strategy of betting
    some fraction f of our current wealth every time. This is a plausible
    class of strategies. Kelly then derived the value of f that would,
    with probability 1, get ahead and stay ahead of any other value of f.
    It's therefore optimal in a rather strong sense that does not depend
    on some arbitrary assumption about utility functions.

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Tue Oct 19 13:32:02 2021
    On Tuesday, October 19, 2021 at 5:19:30 PM UTC+1, Tim Chow wrote:
    On 10/19/2021 3:55 AM, peps...@gmail.com wrote:

    The Kelly criterion maximises utility if you assume that the utiity of your wealth
    is the log of your wealth.
    That is true, but that is not how Kelly *derived* his criterion.
    Taking your suggestion above, we are left with the question, why
    take the utility to be the log of your wealth? Why not the square
    root, or some function derived from empirical observation of actual
    human behavior?

    Instead, suppose that we decide to adopt the strategy of betting
    some fraction f of our current wealth every time. This is a plausible
    class of strategies. Kelly then derived the value of f that would,
    with probability 1, get ahead and stay ahead of any other value of f.
    It's therefore optimal in a rather strong sense that does not depend
    on some arbitrary assumption about utility functions.

    You asked a question. I attempted to answer.
    Please explain in what ways my attempted answer falls short and
    fails to answer your question (perhaps by rephrasing your question).
    I consider myself an expert in this domain.

    Paul

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Tue Oct 19 21:04:29 2021
    On 10/19/2021 4:32 PM, peps...@gmail.com wrote:
    You asked a question. I attempted to answer.
    Please explain in what ways my attempted answer falls short and
    fails to answer your question (perhaps by rephrasing your question).
    I consider myself an expert in this domain.

    The problem I have with your attempted answer is that it
    just raises the question, how do you choose a utility function
    in a canonical way? Sure, I can pick an arbitrary utility
    function and point out that if I maximize the utility function
    then it will generally not maximize equity. By by itself,
    this is not a compelling argument against maximizing equity.

    Kelly's derivation illustrates one way to argue for a specific
    strategy that is different from expectation maximization. As
    I said, one considers the one-parameter family of strategies
    that bets a fraction f of one's stake, and shows that there is
    a specific value of f that is optimal.

    The fact that one can also interpret the Kelly criterion as
    maximizing the log wealth is interesting, but sort of irrelevant
    in my book. Just about *any* strategy can be interpreted as
    maximizing *some* utility function. What's interesting about
    Kelly's derivation is that the specific value of f beats any
    other value of f *even if you're just concerned about wealth*
    (and not some other utility function that is a nonlinear
    function of wealth).

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to peps...@gmail.com on Wed Oct 20 00:44:16 2021
    On Wednesday, October 20, 2021 at 8:34:39 AM UTC+1, peps...@gmail.com wrote:
    On Wednesday, October 20, 2021 at 2:04:33 AM UTC+1, Tim Chow wrote:
    On 10/19/2021 4:32 PM, peps...@gmail.com wrote:
    You asked a question. I attempted to answer.
    Please explain in what ways my attempted answer falls short and
    fails to answer your question (perhaps by rephrasing your question).
    I consider myself an expert in this domain.
    The problem I have with your attempted answer is that it
    just raises the question, how do you choose a utility function
    in a canonical way? Sure, I can pick an arbitrary utility
    function and point out that if I maximize the utility function
    then it will generally not maximize equity. By by itself,
    this is not a compelling argument against maximizing equity.

    Kelly's derivation illustrates one way to argue for a specific
    strategy that is different from expectation maximization. As
    I said, one considers the one-parameter family of strategies
    that bets a fraction f of one's stake, and shows that there is
    a specific value of f that is optimal.

    The fact that one can also interpret the Kelly criterion as
    maximizing the log wealth is interesting, but sort of irrelevant
    in my book. Just about *any* strategy can be interpreted as
    maximizing *some* utility function. What's interesting about
    Kelly's derivation is that the specific value of f beats any
    other value of f *even if you're just concerned about wealth*
    (and not some other utility function that is a nonlinear
    function of wealth).
    I understand all that, but aren't you just answering your own question fully? You seem to have asked a question and answered it fully yourself.
    But you seem to be leaving the question open without acknowledging that
    it has now been answered (by yourself).
    Now you've answered the question as well as asking it, how can we help you further?
    Again, I reiterate that I am quite expert in this subject -- it's just a communication issue
    of knowing what you want.

    I just looked back at your OP.
    Here's how I see the discussion:
    Tim: Please show me an example of where correct backgammon play doesn't maximise equity?
    Tim: I understand the reasoning behind the Kelly criterion and I accept that this is a good staking strategy.
    Paul: Well then you've just answered your own question, haven't you? Suppose the cube is very high
    and the position is D/T. Vary the backgammon rules so that you can stake what you want (not necessarily
    the value of the cube). Then the Kelly criterion value would be much closer to zero than it would be to
    the current cube value.
    So, by the Kelly criterion, holding is better, because a stake of zero is better than
    a stake of current-cube-value, even though the expectation approach leads to doubling.

    Now, of course, this is not at all a rigorous proof of anything. But I don't see how you can expect an
    extremely precise answer to a question that you admit is vague.

    Paul

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Wed Oct 20 00:34:38 2021
    On Wednesday, October 20, 2021 at 2:04:33 AM UTC+1, Tim Chow wrote:
    On 10/19/2021 4:32 PM, peps...@gmail.com wrote:
    You asked a question. I attempted to answer.
    Please explain in what ways my attempted answer falls short and
    fails to answer your question (perhaps by rephrasing your question).
    I consider myself an expert in this domain.
    The problem I have with your attempted answer is that it
    just raises the question, how do you choose a utility function
    in a canonical way? Sure, I can pick an arbitrary utility
    function and point out that if I maximize the utility function
    then it will generally not maximize equity. By by itself,
    this is not a compelling argument against maximizing equity.

    Kelly's derivation illustrates one way to argue for a specific
    strategy that is different from expectation maximization. As
    I said, one considers the one-parameter family of strategies
    that bets a fraction f of one's stake, and shows that there is
    a specific value of f that is optimal.

    The fact that one can also interpret the Kelly criterion as
    maximizing the log wealth is interesting, but sort of irrelevant
    in my book. Just about *any* strategy can be interpreted as
    maximizing *some* utility function. What's interesting about
    Kelly's derivation is that the specific value of f beats any
    other value of f *even if you're just concerned about wealth*
    (and not some other utility function that is a nonlinear
    function of wealth).

    I understand all that, but aren't you just answering your own question fully? You seem to have asked a question and answered it fully yourself.
    But you seem to be leaving the question open without acknowledging that
    it has now been answered (by yourself).
    Now you've answered the question as well as asking it, how can we help you further?
    Again, I reiterate that I am quite expert in this subject -- it's just a communication issue
    of knowing what you want.

    Paul

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  • From pepstein5@gmail.com@21:1/5 to peps...@gmail.com on Wed Oct 20 00:46:38 2021
    On Wednesday, October 20, 2021 at 8:44:17 AM UTC+1, peps...@gmail.com wrote:
    On Wednesday, October 20, 2021 at 8:34:39 AM UTC+1, peps...@gmail.com wrote:
    On Wednesday, October 20, 2021 at 2:04:33 AM UTC+1, Tim Chow wrote:
    On 10/19/2021 4:32 PM, peps...@gmail.com wrote:
    You asked a question. I attempted to answer.
    Please explain in what ways my attempted answer falls short and
    fails to answer your question (perhaps by rephrasing your question).
    I consider myself an expert in this domain.
    The problem I have with your attempted answer is that it
    just raises the question, how do you choose a utility function
    in a canonical way? Sure, I can pick an arbitrary utility
    function and point out that if I maximize the utility function
    then it will generally not maximize equity. By by itself,
    this is not a compelling argument against maximizing equity.

    Kelly's derivation illustrates one way to argue for a specific
    strategy that is different from expectation maximization. As
    I said, one considers the one-parameter family of strategies
    that bets a fraction f of one's stake, and shows that there is
    a specific value of f that is optimal.

    The fact that one can also interpret the Kelly criterion as
    maximizing the log wealth is interesting, but sort of irrelevant
    in my book. Just about *any* strategy can be interpreted as
    maximizing *some* utility function. What's interesting about
    Kelly's derivation is that the specific value of f beats any
    other value of f *even if you're just concerned about wealth*
    (and not some other utility function that is a nonlinear
    function of wealth).
    I understand all that, but aren't you just answering your own question fully?
    You seem to have asked a question and answered it fully yourself.
    But you seem to be leaving the question open without acknowledging that
    it has now been answered (by yourself).
    Now you've answered the question as well as asking it, how can we help you further?
    Again, I reiterate that I am quite expert in this subject -- it's just a communication issue
    of knowing what you want.
    I just looked back at your OP.
    Here's how I see the discussion:
    Tim: Please show me an example of where correct backgammon play doesn't maximise equity?
    Tim: I understand the reasoning behind the Kelly criterion and I accept that this is a good staking strategy.
    Paul: Well then you've just answered your own question, haven't you? Suppose the cube is very high
    and the position is D/T. Vary the backgammon rules so that you can stake what you want (not necessarily
    the value of the cube). Then the Kelly criterion value would be much closer to zero than it would be to
    the current cube value.
    So, by the Kelly criterion, holding is better, because a stake of zero is better than
    a stake of current-cube-value, even though the expectation approach leads to doubling.

    Now, of course, this is not at all a rigorous proof of anything. But I don't see how you can expect an
    extremely precise answer to a question that you admit is vague.

    Correction: I should have said "A stake of the current-cube-value would be closer to the Kelly criterion value than
    a stake of twice the current cube value" -- staking zero is not an option.

    Paul

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  • From MK@21:1/5 to peps...@gmail.com on Wed Oct 20 01:48:04 2021
    On October 20, 2021 at 1:46:39 AM UTC-6, peps...@gmail.com wrote:

    Again, I reiterate that I am quite expert in this subject

    I am watching this closely to see if Chow will call Paul
    "Red Big"...? :)

    But as they say, dogs don't byte dogs. I bet they will
    settle for both being right depending on how it's worded.

    Tim, lick the shit off of Paul's ass and assert that you
    are right.

    Paul, lick the shit off of Tim's ass and assert that you
    are right.

    It's just a simple case of coin toss as usual...

    MK

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Wed Oct 20 21:43:38 2021
    On 10/20/2021 3:44 AM, peps...@gmail.com wrote:
    Paul: Well then you've just answered your own question, haven't you? Suppose the cube is very high
    and the position is D/T. Vary the backgammon rules so that you can stake what you want (not necessarily
    the value of the cube). Then the Kelly criterion value would be much closer to zero than it would be to
    the current cube value.
    So, by the Kelly criterion, holding is better, because a stake of zero is better than
    a stake of current-cube-value, even though the expectation approach leads to doubling.

    Ah, this is getting closer to addressing my question. But I'm
    still having trouble seeing how to flesh out your proposal. Let
    me start trying to spell it out and maybe you can finish it off
    for me.

    We imagine A and B playing very long (but not infinitely long)
    money-game sessions---let's say against the house rather than
    against each other, because I think that's easier to think about.
    The maximum cube value is large (but not infinitely large).
    A follows the usual strategy of maximizing expectation. What
    exactly is B supposed to do in order to be virtually certain,
    at the end of the session, to have more money than A?

    In your sketch above, you suppose that the cube value is very
    high, but is that important? The Kelly criterion doesn't depend
    on how much money is currently at stake. Maybe you're just arguing
    that once we get close to the maximum cube value then we are in a
    "table stakes" situation so the cube action will change. I can see
    that, but in that case, the Kelly criterion is irrelevant.

    ---
    Tim Chow

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  • From Timothy Chow@21:1/5 to All on Wed Oct 20 21:44:24 2021
    On 10/20/2021 4:48 AM, MK wrote:
    I am watching this closely to see if Chow will call Paul
    "Red Big"...? :)

    Only if Paul starts arguing that the adjective "red" should
    precede the adjective "big" in some circumstances.

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Thu Oct 21 02:01:09 2021
    On Thursday, October 21, 2021 at 2:43:43 AM UTC+1, Tim Chow wrote:
    On 10/20/2021 3:44 AM, peps...@gmail.com wrote:
    Paul: Well then you've just answered your own question, haven't you? Suppose the cube is very high
    and the position is D/T. Vary the backgammon rules so that you can stake what you want (not necessarily
    the value of the cube). Then the Kelly criterion value would be much closer to zero than it would be to
    the current cube value.
    So, by the Kelly criterion, holding is better, because a stake of zero is better than
    a stake of current-cube-value, even though the expectation approach leads to doubling.
    Ah, this is getting closer to addressing my question. But I'm
    still having trouble seeing how to flesh out your proposal. Let
    me start trying to spell it out and maybe you can finish it off
    for me.

    We imagine A and B playing very long (but not infinitely long)
    money-game sessions---let's say against the house rather than
    against each other, because I think that's easier to think about.
    The maximum cube value is large (but not infinitely large).
    A follows the usual strategy of maximizing expectation. What
    exactly is B supposed to do in order to be virtually certain,
    at the end of the session, to have more money than A?

    In your sketch above, you suppose that the cube value is very
    high, but is that important? The Kelly criterion doesn't depend
    on how much money is currently at stake. Maybe you're just arguing
    that once we get close to the maximum cube value then we are in a
    "table stakes" situation so the cube action will change. I can see
    that, but in that case, the Kelly criterion is irrelevant.

    If B observes A's play, then there are trivial strategies.
    For example, B should simply stop playing when B observes that B
    is greatly outperforming A, and B should play bigger than usual if B is underperforming A.
    If B is not allowed to observe A's play, and B's total wealth is not a parameter under
    consideration, then I don't see a reason not to maximise expected value.
    If B is not allowed to observe A's play and we take your question very literally,
    then clearly no such strategy exists because A could just be extremely lucky. If we want to maximise B's probability of ending the session ahead, then clearly this is a very different problem which you haven't asked.
    If you want to deviate from the expected value theory and to ignore B's wealth, the "Max probability of ending the session ahead" is probably the most interesting
    question for recreational players. But I'll ignore that question unless you (or someone else)
    shows an interest in it.

    The Kelly criterion is about your stake in relation to your bankroll.
    In the case of backgammon, the stake is the cube value muliplied by an integer between 1 and 3 inclusive.
    The value of the cube is important because a person's bankroll is fixed, so the higher the cube,
    the higher the stake/bankroll proportion.
    No, I'm not arguing that once we get close to the maximum cube value, [....]. Money backgammon doesn't have a "maximum cube value" although it certainly should.

    I don't think you're formulating your questions clearly enough.
    Good questions might be general as well as narrow in scope.
    For example, it's a good question to ask "How would we solve x ^ 2 + 5x + 3 = 0"?,
    and it's also a good question to ask "I'm 15 years old in a state school in New York. I get a lot of math problems
    which have numbers involving letters like x, y, and z. I can't get those problems right.
    What are some resources for learning about this type of mathematics?"

    But asking question(s) that are just completely unclear is not a valuable activity in my opinion.

    Paul

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  • From Axel Reichert@21:1/5 to Timothy Chow on Thu Oct 21 19:46:38 2021
    Timothy Chow <tchow12000@yahoo.com> writes:

    why take the utility to be the log of your wealth? Why not the square
    root, or some function derived from empirical observation of actual
    human behavior?

    Instead, suppose that we decide to adopt the strategy of betting some fraction f of our current wealth every time. This is a plausible
    class of strategies.

    But doesn't that sneak in another arbitrariness? And one that cries out
    for ending up as a log/exponential thing in the broad sense? So I do not
    see a qualitative difference here: Instead of the arbitrary assumption
    about utility functions we have an arbitrary assumption about a
    "plausible" betting strategy.

    Best regards

    Axel

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Thu Oct 21 13:29:51 2021
    On 10/21/2021 5:01 AM, peps...@gmail.com wrote:

    If we want to maximise B's probability of ending the session ahead, then clearly this is a very different problem which you haven't asked.
    If you want to deviate from the expected value theory and to ignore B's wealth,
    the "Max probability of ending the session ahead" is probably the most interesting
    question for recreational players. But I'll ignore that question unless you (or someone else)
    shows an interest in it.

    Since you want a precise question, let's go with this one.

    A and B are playing money-game sessions against the house, which
    we can assume is adopting an equilibrium strategy. There is a
    maximum cube value; let's say 2^20. The sessions last for some
    fixed amount of time; let's say 2^20 dice rolls. A and B start
    with the same amount of money. A adopts an equilibrium strategy.
    A and B have no visibility into each other's sessions until the
    very end.

    Question: Is there some strategy that B can adopt which will give
    B a >50% probability of being ahead of A at the end of the sessions?

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Axel Reichert on Thu Oct 21 11:21:59 2021
    On Thursday, October 21, 2021 at 6:46:40 PM UTC+1, Axel Reichert wrote:
    Timothy Chow <tchow...@yahoo.com> writes:

    why take the utility to be the log of your wealth? Why not the square
    root, or some function derived from empirical observation of actual
    human behavior?

    Instead, suppose that we decide to adopt the strategy of betting some fraction f of our current wealth every time. This is a plausible
    class of strategies.
    But doesn't that sneak in another arbitrariness? And one that cries out
    for ending up as a log/exponential thing in the broad sense? So I do not
    see a qualitative difference here: Instead of the arbitrary assumption
    about utility functions we have an arbitrary assumption about a
    "plausible" betting strategy.

    Best regards

    Axel

    Axel,

    Despite my disagreements with Tim on this thread [actually a disagreement with the
    way he's expressing himself here rather than anything he's actually said], here I
    totally agree with Tim's point.

    Suppose you want to justify Kelly's criterion (betting algorithm).
    There are two (pedagogical) strategies for justifying this algorithm.
    One is to say that it is in some sense (Tim's sense actually) best among all the strategies
    which involve betting a constant fraction of your wealth each time.
    Another is to say that it maximises the expected utility of your wealth if your utility function
    is logarithmic.
    Tim's opinion (and mine, too) is that the first justification is much more convincing than the second because
    keeping the staking proportion constant is an intuitive idea, and one that many people would do in practice.
    On the other hand, the utility function approach has an obvious rejoinder -- "Well, why are you using the log
    function instead of any other of the vast number of concave increasing functions which are just as plausible?"

    Where I have a disagreement with Tim is that he now appears to have convinced himself that the Kelly criterion
    makes good sense. So he knows it now. Why does he keep on asking questions?? It's very obvious that the Kelly approach has bg implications. According to Kelly, you don't usually max the stake
    even if the odds are in your favour. We readily translate this to bg by observing that a Kelly staker might not
    double in a last-roll position with 19 winning rolls and 17 losing rolls.

    I'm not sure why but, in maths circles, if someone makes an interesting observation, it's slightly taboo or unconventional
    to say "Here's an interesting fact/result that I've discovered..." Perhaps it's unusual because it makes a person
    vulnerable to a "So what?" reply.

    So rather than just assert the discovery (that the Kelly criterion has implications for backgammon for example), a person
    can disguise the discovery as a question so as not to be vulnerable to a so-what attack.
    An example of this rhetorical style is to say "I've just written a best-selling novel about backgammon that got reviewed
    favourably in the New York Times. Does anyone know any other popular novels about backgammon?"

    Actually, knowing Tim (online), I don't think he is using this trick, but he's certainly reminding me of this
    technique.

    Paul

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  • From Axel Reichert@21:1/5 to peps...@gmail.com on Thu Oct 21 21:27:53 2021
    "peps...@gmail.com" <pepstein5@gmail.com> writes:

    One is to say that it is in some sense (Tim's sense actually) best
    among all the strategies which involve betting a constant fraction of
    your wealth each time. Another is to say that it maximises the
    expected utility of your wealth if your utility function is
    logarithmic.

    Tim's opinion (and mine, too) is that the first justification is much
    more convincing than the second because keeping the staking proportion constant is an intuitive idea, and one that many people would do in
    practice.

    Well, some keep the stakes constant and set some limit for their maximum
    loss.

    But I agree that the first justification might seem easier/more
    intuitive, even if maybe only because few people grasp the concept of
    utility functions. (-:

    Thanks for the clarification!

    Axel

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  • From pepstein5@gmail.com@21:1/5 to Axel Reichert on Thu Oct 21 13:36:59 2021
    On Thursday, October 21, 2021 at 8:27:55 PM UTC+1, Axel Reichert wrote:
    "peps...@gmail.com" <peps...@gmail.com> writes:

    One is to say that it is in some sense (Tim's sense actually) best
    among all the strategies which involve betting a constant fraction of
    your wealth each time. Another is to say that it maximises the
    expected utility of your wealth if your utility function is
    logarithmic.

    Tim's opinion (and mine, too) is that the first justification is much
    more convincing than the second because keeping the staking proportion constant is an intuitive idea, and one that many people would do in practice.
    Well, some keep the stakes constant and set some limit for their maximum loss.

    But I agree that the first justification might seem easier/more
    intuitive, even if maybe only because few people grasp the concept of
    utility functions. (-:

    Thanks for the clarification!


    Utility theory is clearly the correct theoretical framework to address Tim's questions on this thread and it's surprising (to me) that he doesn't acknowledge this.
    (Perhaps the reason is that he doesn't realise this).
    He is an established mathematician with quite a good track record of publication.
    I'd guess that he's (by far) the best mathematician who posts here regularly, but I don't know this.

    Paul

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Thu Oct 21 22:25:43 2021
    On 10/21/2021 4:36 PM, peps...@gmail.com wrote:
    Utility theory is clearly the correct theoretical framework to address Tim's questions on this thread and it's surprising (to me) that he doesn't acknowledge this.
    (Perhaps the reason is that he doesn't realise this).

    Please show my how utility theory answers the question below, which
    I repeat for your convenience.

    A and B are playing money-game sessions against the house, which
    we can assume is adopting an equilibrium strategy. There is a
    maximum cube value; let's say 2^20. The sessions last for some
    fixed amount of time; let's say 2^20 dice rolls. A and B start
    with the same amount of money. A adopts an equilibrium strategy.
    A and B have no visibility into each other's sessions until the
    very end.

    Question: Is there some strategy that B can adopt which will give
    B a >50% probability of being ahead of A at the end of the sessions?

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Fri Oct 22 02:56:15 2021
    On Friday, October 22, 2021 at 3:25:46 AM UTC+1, Tim Chow wrote:
    On 10/21/2021 4:36 PM, peps...@gmail.com wrote:
    Utility theory is clearly the correct theoretical framework to address Tim's
    questions on this thread and it's surprising (to me) that he doesn't acknowledge this.
    (Perhaps the reason is that he doesn't realise this).
    Please show my how utility theory answers the question below, which
    I repeat for your convenience.
    A and B are playing money-game sessions against the house, which
    we can assume is adopting an equilibrium strategy. There is a
    maximum cube value; let's say 2^20. The sessions last for some
    fixed amount of time; let's say 2^20 dice rolls. A and B start
    with the same amount of money. A adopts an equilibrium strategy.
    A and B have no visibility into each other's sessions until the
    very end.

    Question: Is there some strategy that B can adopt which will give
    B a >50% probability of being ahead of A at the end of the sessions?

    ---
    Tim Chow

    Utility theory is not needed to answer this question. But, of course, this is not a repetition
    of a question you've asked before.
    The answer to your question is an absolutely trivial "Yes!!" leading me to think you're again
    not stating what you mean.
    Since A and the house are adopting the same strategy, the probability that A makes a profit
    must be <= 50%.
    So any strategy by B that makes a profit with more than 50% probability solves your problem.
    B's strategy is this:
    Accept any cube from the house so long as B's winning probability is > 0.
    B doubles whenever the theoretical cube action is either D/P or TG but never doubles otherwise.
    B stops playing after the first game.
    B is clearly more likely to win than not because the house can only win by removing all its checkers
    whereas B has an additional winning route of reaching D/P or TG while owning the cube.
    All you're asking is for strategies that sacrifice equity to increase the probability of staying ahead
    -- an easy thing to do in both theory and practice which is sometimes called "steaming".

    Ok, you could object to my solution on the grounds that a single game is unlikely to last as long as 2 ^ 20
    dice rolls, but it's very easy to adapt. For example, instead of stopping after game 1, revert to an equilibrium
    strategy after game 1.

    Paul

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  • From MK@21:1/5 to Tim Chow on Fri Oct 22 20:03:47 2021
    On October 22, 2021 at 8:23:01 PM UTC-6, Tim Chow wrote:

    50% chance that A is -2.
    50% chance that A is +2.

    25% chance that B is -3.
    75% chance that B is +1.

    Half the time, A is +2, and is ahead of B no matter what. Half the
    time, A is -2, and has a 25% chance of being ahead of B. So A comes
    out ahead of B 5/8 of the time.

    Good evening Ladies and Assholes,
    heeere comesss Chowie the match phd...

    Hah hah hah! :))

    MK

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Fri Oct 22 22:22:58 2021
    On 10/22/2021 5:56 AM, peps...@gmail.com wrote:
    On Friday, October 22, 2021 at 3:25:46 AM UTC+1, Tim Chow wrote:

    Question: Is there some strategy that B can adopt which will give
    B a >50% probability of being ahead of A at the end of the sessions?

    Since A and the house are adopting the same strategy, the probability that A makes a profit
    must be <= 50%.
    So any strategy by B that makes a profit with more than 50% probability solves your problem.

    I don't see why this is true. I want B to have a >50% probability of
    *being ahead of A*. Suppose we have the following probabilities, and
    A and B are independent.

    50% chance that A is -2.
    50% chance that A is +2.

    25% chance that B is -3.
    75% chance that B is +1.

    Half the time, A is +2, and is ahead of B no matter what. Half the
    time, A is -2, and has a 25% chance of being ahead of B. So A comes
    out ahead of B 5/8 of the time.

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Sat Oct 23 02:45:33 2021
    On Saturday, October 23, 2021 at 3:23:01 AM UTC+1, Tim Chow wrote:
    On 10/22/2021 5:56 AM, peps...@gmail.com wrote:
    On Friday, October 22, 2021 at 3:25:46 AM UTC+1, Tim Chow wrote:

    Question: Is there some strategy that B can adopt which will give
    B a >50% probability of being ahead of A at the end of the sessions?
    Since A and the house are adopting the same strategy, the probability that A makes a profit
    must be <= 50%.
    So any strategy by B that makes a profit with more than 50% probability solves your problem.
    I don't see why this is true. I want B to have a >50% probability of
    *being ahead of A*. Suppose we have the following probabilities, and
    A and B are independent.

    50% chance that A is -2.
    50% chance that A is +2.

    25% chance that B is -3.
    75% chance that B is +1.

    Half the time, A is +2, and is ahead of B no matter what. Half the
    time, A is -2, and has a 25% chance of being ahead of B. So A comes
    out ahead of B 5/8 of the time.

    I think I can solve the problem with a suitable strategy if we add the simplifying assumptions that the stake (for both players) is fixed and is
    small enough that gambler's ruin is not an issue.

    Now B's strategy is to always double when the double is legal (according to the house rules, not traditional bg rules),
    so long as B's winning probability is > 0, and to accept every cube if B's winning probability > 0.

    As before, B will sometimes win through D/P whereas the house only wins by removing all the checkers.
    So B's probability of winning is > 50%.
    Clearly the cube will be jacked up so high that the absolute value of the cube is highly likely to be higher than for A's games.
    So your counter-example wouldn't work.

    Paul

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  • From pepstein5@gmail.com@21:1/5 to peps...@gmail.com on Sat Oct 23 02:48:34 2021
    On Saturday, October 23, 2021 at 10:45:34 AM UTC+1, peps...@gmail.com wrote:
    On Saturday, October 23, 2021 at 3:23:01 AM UTC+1, Tim Chow wrote:
    On 10/22/2021 5:56 AM, peps...@gmail.com wrote:
    On Friday, October 22, 2021 at 3:25:46 AM UTC+1, Tim Chow wrote:

    Question: Is there some strategy that B can adopt which will give
    B a >50% probability of being ahead of A at the end of the sessions?
    Since A and the house are adopting the same strategy, the probability that A makes a profit
    must be <= 50%.
    So any strategy by B that makes a profit with more than 50% probability solves your problem.
    I don't see why this is true. I want B to have a >50% probability of
    *being ahead of A*. Suppose we have the following probabilities, and
    A and B are independent.

    50% chance that A is -2.
    50% chance that A is +2.

    25% chance that B is -3.
    75% chance that B is +1.

    Half the time, A is +2, and is ahead of B no matter what. Half the
    time, A is -2, and has a 25% chance of being ahead of B. So A comes
    out ahead of B 5/8 of the time.
    I think I can solve the problem with a suitable strategy if we add the simplifying assumptions that the stake (for both players) is fixed and is small enough that gambler's ruin is not an issue.

    Now B's strategy is to always double when the double is legal (according to the house rules, not traditional bg rules),
    so long as B's winning probability is > 0, and to accept every cube if B's winning probability > 0.

    As before, B will sometimes win through D/P whereas the house only wins by removing all the checkers.
    So B's probability of winning is > 50%.
    Clearly the cube will be jacked up so high that the absolute value of the cube is highly likely to be higher than for A's games.
    So your counter-example wouldn't work.

    Correction -- Stake fixed for both players and equal.

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Sat Oct 23 02:34:02 2021
    On Saturday, October 23, 2021 at 3:23:01 AM UTC+1, Tim Chow wrote:
    On 10/22/2021 5:56 AM, peps...@gmail.com wrote:
    On Friday, October 22, 2021 at 3:25:46 AM UTC+1, Tim Chow wrote:

    Question: Is there some strategy that B can adopt which will give
    B a >50% probability of being ahead of A at the end of the sessions?
    Since A and the house are adopting the same strategy, the probability that A makes a profit
    must be <= 50%.
    So any strategy by B that makes a profit with more than 50% probability solves your problem.
    I don't see why this is true. I want B to have a >50% probability of
    *being ahead of A*. Suppose we have the following probabilities, and
    A and B are independent.

    50% chance that A is -2.
    50% chance that A is +2.

    25% chance that B is -3.
    75% chance that B is +1.

    Half the time, A is +2, and is ahead of B no matter what. Half the
    time, A is -2, and has a 25% chance of being ahead of B. So A comes
    out ahead of B 5/8 of the time.

    Thanks. This is a good counter-example.
    So, now, my current thoughts (more tentative since I've been shown to be wrong) lead me to a response of "No, no such strategy exists."

    I can't quite get to a proof but the fact that B has no idea how much A is betting
    is a problem for B.
    If A is betting large, B must also bet large.
    But B doesn't know how much A is betting, so B must bet large in any case,
    and then B is subject to gambler's ruin. However, if A is betting small, A has no gambler's ruin problem.

    I admit that this is far from rigorous, even though your question is now precise enough.
    Can the players vary the stakes between the game? (I think that it's a better question if the answer is no).

    But I think the answer is "No, no such strategy exists" whether the stakes vary during the session or not.
    And the (very vague and informal) reasoning is as above. B must bet large becaue B can't assume A isn't
    betting large. However, B is then hit by additional gambler's ruin problems in the case that A is betting small.

    With backgammon being a hugely complicated game (in the context of attempting rigorous mathematical proofs),
    I'm not sure this vagueness can be improved on too easily although Doug Zare and Curtis McMullen might like to give
    it a try.

    Paul

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  • From MK@21:1/5 to peps...@gmail.com on Sat Oct 23 14:29:52 2021
    On October 23, 2021 at 3:45:34 AM UTC-6, peps...@gmail.com wrote:

    On October 23, 2021 at 3:23:01 AM UTC+1, Tim Chow wrote:

    50% chance that A is -2.
    50% chance that A is +2.
    25% chance that B is -3.
    75% chance that B is +1.
    ..... So A comes out ahead of B 5/8 of the time.

    I think I can solve the problem with a suitable strategy
    if .... gambler's ruin is not an issue.

    Now B's strategy is to always
    double ..... so long as B's winning probability is > 0,
    and to accept every cube if B's winning probability > 0.

    As before, B will sometimes win through D/P
    whereas the house only wins by removing all the checkers.
    So B's probability of winning is > 50%.
    Clearly the cube will be jacked up so high that the absolute
    value of the cube is highly likely to be higher than for A's games.

    So your counter-example wouldn't work.

    I took the liberty to quote you by shortening in order to highlight
    the essential parts of your argument, which may make it easier
    to understand for others as well as myself.

    Even though at a much slower pace than I would like, I'm pleased
    to see that progress is being made in the cube skill debate. There
    is no progress without some people sticking their necks out, even
    if at the risk of damaging their reputations by possibly being wrong.
    I hope that more mathematicians and others will find the courage
    to participate and contribute.

    Chow's example is a typical failed effort to apply a simplistic coin
    toss (or imaginary games that he invents and such) argument to
    backgammon because he is in denial of the "cube skill" fallacy. If
    he tries to give an actual backgammon example, all he can do is
    refer to last few rolls positions.

    His above example would somewhat work for coin tossing with
    more evenly distributed/corresponding bets but even so the best
    he can claim is that "A will come out ahead of B, in anywhere from
    4/8 to 5/8 of the time".

    Although in a different context and in different words, I suggested
    many times in the past an expriment with a "tweaked" version of
    XG or Gnubg playing against XG or Gnubg, by doubling when it's
    winning chance is more than 50% and by always taking when it's
    winning chance is more than 50%, thus forcing as many games as
    possible to be played out to the very end, and I had offered my own
    prediction that the tweaked versions of bots would come out ahead.

    For the developers of bots, it would be such a trivial effort to create
    a tweaked version of their bots and run this (and similar) experiments
    that I'm overly surprised and disappointed that they have never tried.

    As I often repeat: it's not a matter of if but just a matter of time...

    MK

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  • From Timothy Chow@21:1/5 to All on Sat Oct 23 22:05:55 2021
    On 10/23/2021 5:29 PM, MK wrote:
    Chow's example is a typical failed effort to apply a simplistic coin
    toss (or imaginary games that he invents and such) argument to
    backgammon because he is in denial of the "cube skill" fallacy. If
    he tries to give an actual backgammon example, all he can do is
    refer to last few rolls positions.

    The questions I'm discussing evaporate in any non-contact race, because
    we can easily calculate an upper bound on how large the cube gets, and
    work out how many times you need to play out the game in order to drive
    the variance down.

    I think that potential trouble can occur only when the cube values grow
    "too fast" so that the number of trials needed to drive down the
    variance also grows "too fast." But Paul will complain that I'm being
    too vague here.

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Sun Oct 24 03:10:35 2021
    On Sunday, October 24, 2021 at 3:06:00 AM UTC+1, Tim Chow wrote:
    On 10/23/2021 5:29 PM, MK wrote:
    Chow's example is a typical failed effort to apply a simplistic coin
    toss (or imaginary games that he invents and such) argument to
    backgammon because he is in denial of the "cube skill" fallacy. If
    he tries to give an actual backgammon example, all he can do is
    refer to last few rolls positions.
    The questions I'm discussing evaporate in any non-contact race, because
    we can easily calculate an upper bound on how large the cube gets, and
    work out how many times you need to play out the game in order to drive
    the variance down.

    I think that potential trouble can occur only when the cube values grow
    "too fast" so that the number of trials needed to drive down the
    variance also grows "too fast." But Paul will complain that I'm being
    too vague here.

    ---
    Tim Chow
    No, that isn't my complaint.
    And, in fact, in relation to your posting immediately before this one, I explicitly said
    that you were (finally) being clear enough.
    In my opinion, more clarity is called for when you ask others to contribute to your discussion.
    For example, if someone claims "Wealthy nations tend to be more secular than poor nations,"
    then the claim makes sense, despite the lack of definitions.
    [I suspect it's true but the point I'm making is that it makes sense, not that it's true.]
    However, if someone says "I'm doing a study on the correlation between secularity and wealth. Please can you provide examples of secular wealthy nations?"
    then the onus is on that person to be more explicit about what they mean by "wealthy" and "secular".

    With regard to the matter at hand, no matter how clear you attempt to be (and your final attempt at clarity was good), there will always be unstated
    assumptions which the reader will have to guess at.
    My guess (after further thought) is that one of your unstated assumptions is that the unit stake is constant (because we look at simple problems first)
    and the same for both players (because we look at simple problems first and also because finding the required strategy looks hopeless otherwise).

    If you mean this, then I think my previous post is a fine answer although I didn't provide a rigorous proof (but a rigorous proof isn't really feasible in any case).
    Basically, jack the cube super-high and never pass anything. Because you never pass, your probability of winning is higher than A's 50%. Because the cube
    is super-high, the absolute value of your games will be higher than A's. So you satisfy the problem constraints.
    The main gap here is that ("jacking the cube super-high") is not clearly allowed by the rules of backgammon (because players take it in turns to double).
    But experience of steaming players leads me to think that this approach satisfies the demands of your problem, whether it's provable or not.

    Paul

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Sun Oct 24 03:23:10 2021
    On Sunday, October 24, 2021 at 3:06:00 AM UTC+1, Tim Chow wrote:
    On 10/23/2021 5:29 PM, MK wrote:
    Chow's example is a typical failed effort to apply a simplistic coin
    toss (or imaginary games that he invents and such) argument to
    backgammon because he is in denial of the "cube skill" fallacy. If
    he tries to give an actual backgammon example, all he can do is
    refer to last few rolls positions.
    The questions I'm discussing evaporate in any non-contact race, because
    we can easily calculate an upper bound on how large the cube gets, and
    work out how many times you need to play out the game in order to drive
    the variance down.

    I think that potential trouble can occur only when the cube values grow
    "too fast" so that the number of trials needed to drive down the
    variance also grows "too fast." But Paul will complain that I'm being
    too vague here.

    Your phrase "drive down the variance" gives further evidence that either
    you aren't saying what you mean, or I'm not understanding what you're saying. [But this is a common problem when communicating about complex topics.]
    When you created a precise question out of your topic, the easy (to me) strategies for answering your problem involve driving the variance up, not down.

    Paul

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Sun Oct 24 08:07:39 2021
    On 10/23/2021 5:45 AM, peps...@gmail.com wrote:
    I think I can solve the problem with a suitable strategy if we add the simplifying assumptions that the stake (for both players) is fixed and is small enough that gambler's ruin is not an issue.

    Now B's strategy is to always double when the double is legal (according to the house rules, not traditional bg rules),
    so long as B's winning probability is > 0, and to accept every cube if B's winning probability > 0.

    As before, B will sometimes win through D/P whereas the house only wins by removing all the checkers.
    So B's probability of winning is > 50%.
    Clearly the cube will be jacked up so high that the absolute value of the cube is highly likely to be higher than for A's games.
    So your counter-example wouldn't work.

    I'm okay with the assumption that the stake for both players
    is fixed (if by that you mean, in traditional backgammon
    language, a point is always worth $C where C is a constant).
    And I'm also okay with ignoring gambler's ruin---I'll let
    either player go arbitrarily far into debt.

    By "house rules, not traditional bg rules" are you just
    referring to the 2^20 cap on the cube value?

    I understand why my counterexample no longer applies, but I
    don't follow your argument at the end, where you say "Clearly."

    Maybe we should spell out a few obvious things. What we're
    looking for, essentially, is a violation of the law of large
    numbers. That is, suppose that what A is doing is accumulating
    the sum of a bunch of i.i.d. draws of a random variable X, and
    B is accumulating the sum of a bunch of i.i.d. draws of a random
    variable Y. Suppose that X and Y have finite mean and variance,
    that the expected value E[Y] of Y is less than the expected value
    E[X] of X, and that A and B both make the same number n of draws.

    Now pick some e > 0 that is much smaller than E[X] - E[Y]. Then
    the weak law of large numbers tells us that the probability that
    A's final total is less than n*(E[X] - e) tends to zero as n goes
    to infinity, and similarly the probability that B's final total
    is greater than n*(E[Y] + e) tends to zero as n goes to infinity.
    So that means that there isn't going to be any way for B to come
    out ahead when n is large.

    So what are the loopholes in this argument? Well, the way I set
    up the problem, n isn't necessarily the same for A and B. But
    this doesn't look too promising to me; I don't think that B has
    a lot of control over how many rolls each game takes.

    The more natural loophole to try to exploit is the assumption of
    finite mean and variance. This is why, at the outset, I alluded
    to those funny positions where the equity is undefined. Now by
    putting a cap of 2^20 on the cube, we ensure that the mean and
    variance are finite, but we are simultaneously putting a cap on
    the value of n. That is, I'm not capping the cube and allowing
    n to go to infinity, which would run afoul of the law of large
    numbers; I'm allowing the cube value to be comparable to n. This
    should give us some leeway to play with.

    We're not necessarily home free, though. The weak law sometimes
    holds even when the mean is undefined. So any "counterexample"
    should be calculated carefully.

    I agree with you that actual backgammon is too complicated to
    analyze completely rigorously, but I'd be satisfied with some
    caricature that at least mimics the way the doubling cube works.

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Sun Oct 24 08:17:49 2021
    On Sunday, October 24, 2021 at 1:07:44 PM UTC+1, Tim Chow wrote:
    On 10/23/2021 5:45 AM, peps...@gmail.com wrote:
    I think I can solve the problem with a suitable strategy if we add the simplifying assumptions that the stake (for both players) is fixed and is small enough that gambler's ruin is not an issue.

    Now B's strategy is to always double when the double is legal (according to the house rules, not traditional bg rules),
    so long as B's winning probability is > 0, and to accept every cube if B's winning probability > 0.

    As before, B will sometimes win through D/P whereas the house only wins by removing all the checkers.
    So B's probability of winning is > 50%.
    Clearly the cube will be jacked up so high that the absolute value of the cube is highly likely to be higher than for A's games.
    So your counter-example wouldn't work.
    I'm okay with the assumption that the stake for both players
    is fixed (if by that you mean, in traditional backgammon
    language, a point is always worth $C where C is a constant).
    And I'm also okay with ignoring gambler's ruin---I'll let
    either player go arbitrarily far into debt.

    By "house rules, not traditional bg rules" are you just
    referring to the 2^20 cap on the cube value?

    I understand why my counterexample no longer applies, but I
    don't follow your argument at the end, where you say "Clearly."

    Maybe we should spell out a few obvious things. What we're
    looking for, essentially, is a violation of the law of large
    numbers. That is, suppose that what A is doing is accumulating
    the sum of a bunch of i.i.d. draws of a random variable X, and
    B is accumulating the sum of a bunch of i.i.d. draws of a random
    variable Y. Suppose that X and Y have finite mean and variance,
    that the expected value E[Y] of Y is less than the expected value
    E[X] of X, and that A and B both make the same number n of draws.

    Now pick some e > 0 that is much smaller than E[X] - E[Y]. Then
    the weak law of large numbers tells us that the probability that
    A's final total is less than n*(E[X] - e) tends to zero as n goes
    to infinity, and similarly the probability that B's final total
    is greater than n*(E[Y] + e) tends to zero as n goes to infinity.
    So that means that there isn't going to be any way for B to come
    out ahead when n is large.

    So what are the loopholes in this argument? Well, the way I set
    up the problem, n isn't necessarily the same for A and B. But
    this doesn't look too promising to me; I don't think that B has
    a lot of control over how many rolls each game takes.

    The more natural loophole to try to exploit is the assumption of
    finite mean and variance. This is why, at the outset, I alluded
    to those funny positions where the equity is undefined. Now by
    putting a cap of 2^20 on the cube, we ensure that the mean and
    variance are finite, but we are simultaneously putting a cap on
    the value of n. That is, I'm not capping the cube and allowing
    n to go to infinity, which would run afoul of the law of large
    numbers; I'm allowing the cube value to be comparable to n. This
    should give us some leeway to play with.

    We're not necessarily home free, though. The weak law sometimes
    holds even when the mean is undefined. So any "counterexample"
    should be calculated carefully.

    I agree with you that actual backgammon is too complicated to
    analyze completely rigorously, but I'd be satisfied with some
    caricature that at least mimics the way the doubling cube works.

    ---
    Tim Chow

    My idea was for B's strategy to be radically different from iid. Play the first game to maximise winning probability and then
    revert to expected value play.
    I didn't realise iid was another unstated assumption.

    Paul

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Tue Oct 26 22:21:29 2021
    On 10/24/2021 11:17 AM, peps...@gmail.com wrote:
    My idea was for B's strategy to be radically different from iid. Play the first game to maximise winning probability and then
    revert to expected value play.
    I didn't realise iid was another unstated assumption.

    No, I'm not insisting on i.i.d. I don't mind deviating from that.
    But your analysis of your example is still too terse for me to follow.

    We already established that having B have a >50% probability of being
    positive doesn't by itself mean very much. I'm not saying your example
    is wrong but it would help if you would spell out the argument in some
    more detail.
    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Wed Oct 27 06:42:13 2021
    On Wednesday, October 27, 2021 at 3:21:33 AM UTC+1, Tim Chow wrote:
    On 10/24/2021 11:17 AM, peps...@gmail.com wrote:
    My idea was for B's strategy to be radically different from iid. Play the first game to maximise winning probability and then
    revert to expected value play.
    I didn't realise iid was another unstated assumption.
    No, I'm not insisting on i.i.d. I don't mind deviating from that.
    But your analysis of your example is still too terse for me to follow.

    We already established that having B have a >50% probability of being positive doesn't by itself mean very much. I'm not saying your example
    is wrong but it would help if you would spell out the argument in some
    more detail.

    Ok, I'll try again but I'll cheat by having a new strategy.
    B moves the checkers exactly as if it was DMP, thereby giving B a greater winning probability than A.
    B's cube play is such that B's only goal (in relation to the cube) is to maximise the cube value.
    Since B assumes the house is using an expected value strategy, B will cube when behind to get beavered.
    With backgammon being so complex, we can only make guesses at the results. Since B never gets cubed out, I'll hypothesise that B's winning probability is >= 55% (but this type of thing is impossible to prove).
    Since B is maxing the cube value and A isn't, I'll hypothesise that the probability that [the absolute value of B's result > the absolute value of A's result]
    (in a single game) is >= 90%.
    Therefore, (over a single game), assuming my hypotheses are correct, B outperforms A with a probability >= 55% * 95% > 52%.
    Above calculation is the probability that ( [B wins and A loses] or [B wins and A wins and A's absolute value is less than B's absolute value]

    Paul

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  • From pepstein5@gmail.com@21:1/5 to peps...@gmail.com on Wed Oct 27 10:05:49 2021
    On Wednesday, October 27, 2021 at 2:42:14 PM UTC+1, peps...@gmail.com wrote:
    On Wednesday, October 27, 2021 at 3:21:33 AM UTC+1, Tim Chow wrote:
    On 10/24/2021 11:17 AM, peps...@gmail.com wrote:
    My idea was for B's strategy to be radically different from iid. Play the first game to maximise winning probability and then
    revert to expected value play.
    I didn't realise iid was another unstated assumption.
    No, I'm not insisting on i.i.d. I don't mind deviating from that.
    But your analysis of your example is still too terse for me to follow.

    We already established that having B have a >50% probability of being positive doesn't by itself mean very much. I'm not saying your example
    is wrong but it would help if you would spell out the argument in some
    more detail.
    Ok, I'll try again but I'll cheat by having a new strategy.
    B moves the checkers exactly as if it was DMP, thereby giving B a greater winning probability than A.
    B's cube play is such that B's only goal (in relation to the cube) is to maximise the cube value.
    Since B assumes the house is using an expected value strategy, B will cube when behind to get beavered.
    With backgammon being so complex, we can only make guesses at the results. Since B never gets cubed out, I'll hypothesise that B's winning probability is >= 55% (but this type of thing is impossible to prove).
    Since B is maxing the cube value and A isn't, I'll hypothesise that the probability that [the absolute value of B's result > the absolute value of A's result]
    (in a single game) is >= 90%.
    Therefore, (over a single game), assuming my hypotheses are correct, B outperforms A with a probability >= 55% * 95% > 52%.
    Above calculation is the probability that ( [B wins and A loses] or [B wins and A wins and A's absolute value is less than B's absolute value]

    Paul
    I think the beavering possibility makes a real difference to my argument. 90% to get the absolute value as higher than A's might be an overestimate, which
    would destroy my argument. But I think I've given a template which could readily be made to work if you tinker with both the strategy and the percentages.
    I think that, if B doubles at B's very first opportunity, it's guaranteed to be a beaver -- leading to the cube being on 4 right away. If you allow raccoons, my strategy
    works even better but I don't think raccoons are particularly standard so I would assume "no raccoons" unless stated otherwise.
    Having said that 90% is too high, I think that 55% for B's winning chances is much too low so I'm optimistic that I'm basically solving your problem about as
    well as a person can, given that (almost) nothing is provable in such a complex game.

    Paul

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  • From pepstein5@gmail.com@21:1/5 to peps...@gmail.com on Wed Oct 27 11:11:58 2021
    On Wednesday, October 27, 2021 at 6:05:51 PM UTC+1, peps...@gmail.com wrote:
    On Wednesday, October 27, 2021 at 2:42:14 PM UTC+1, peps...@gmail.com wrote:
    On Wednesday, October 27, 2021 at 3:21:33 AM UTC+1, Tim Chow wrote:
    On 10/24/2021 11:17 AM, peps...@gmail.com wrote:
    My idea was for B's strategy to be radically different from iid. Play the first game to maximise winning probability and then
    revert to expected value play.
    I didn't realise iid was another unstated assumption.
    No, I'm not insisting on i.i.d. I don't mind deviating from that.
    But your analysis of your example is still too terse for me to follow.

    We already established that having B have a >50% probability of being positive doesn't by itself mean very much. I'm not saying your example
    is wrong but it would help if you would spell out the argument in some more detail.
    Ok, I'll try again but I'll cheat by having a new strategy.
    B moves the checkers exactly as if it was DMP, thereby giving B a greater winning probability than A.
    B's cube play is such that B's only goal (in relation to the cube) is to maximise the cube value.
    Since B assumes the house is using an expected value strategy, B will cube when behind to get beavered.
    With backgammon being so complex, we can only make guesses at the results. Since B never gets cubed out, I'll hypothesise that B's winning probability is >= 55% (but this type of thing is impossible to prove).
    Since B is maxing the cube value and A isn't, I'll hypothesise that the probability that [the absolute value of B's result > the absolute value of A's result]
    (in a single game) is >= 90%.
    Therefore, (over a single game), assuming my hypotheses are correct, B outperforms A with a probability >= 55% * 95% > 52%.
    Above calculation is the probability that ( [B wins and A loses] or [B wins and A wins and A's absolute value is less than B's absolute value]

    Paul
    I think the beavering possibility makes a real difference to my argument. 90% to get the absolute value as higher than A's might be an overestimate, which
    would destroy my argument. But I think I've given a template which could readily be made to work if you tinker with both the strategy and the percentages.
    I think that, if B doubles at B's very first opportunity, it's guaranteed to be a beaver -- leading to the cube being on 4 right away. If you allow raccoons, my strategy
    works even better but I don't think raccoons are particularly standard so I would assume "no raccoons" unless stated otherwise.
    Having said that 90% is too high, I think that 55% for B's winning chances is much too low so I'm optimistic that I'm basically solving your problem about as
    well as a person can, given that (almost) nothing is provable in such a complex game.

    Tweaking the numbers a bit, all you need is a >= 60% chance of winning combined with a >=70% chance of having the absolute value of your game higher than A's.
    That leads to a probability of outperforming A that >= 60% * 85% = 51%.
    That has to be easily achievable by the types of strategies I've talked about. I think 90% for the absolute value component was too high.

    Paul

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  • From Timothy Chow@21:1/5 to peps...@gmail.com on Thu Oct 28 23:22:53 2021
    On 10/27/2021 9:42 AM, peps...@gmail.com wrote:
    On Wednesday, October 27, 2021 at 3:21:33 AM UTC+1, Tim Chow wrote:
    On 10/24/2021 11:17 AM, peps...@gmail.com wrote:

    B moves the checkers exactly as if it was DMP, thereby giving B a greater winning probability than A.
    B's cube play is such that B's only goal (in relation to the cube) is to maximise the cube value.
    Since B assumes the house is using an expected value strategy, B will cube when behind to get beavered.
    With backgammon being so complex, we can only make guesses at the results. Since B never gets cubed out, I'll hypothesise that B's winning probability is >= 55% (but this type of thing is impossible to prove).
    Since B is maxing the cube value and A isn't, I'll hypothesise that the probability that [the absolute value of B's result > the absolute value of A's result]
    (in a single game) is >= 90%.
    Therefore, (over a single game), assuming my hypotheses are correct, B outperforms A with a probability >= 55% * 95% > 52%.

    Okay, I think I see what you're trying to do...this is interesting.

    Here's how I would parse it. Consider the following scenario.

    Prob(A scores +1) = 0.5
    Prob(A scores -1) = 0.5

    Prob(B scores +2) = 0.5 + epsilon
    Prob(B scores -3) = 0.5 - epsilon

    Assume A and B are independent. If B scores +2 then B outscores A
    regardless of what A scores. This happens with probability > 0.5 so
    B outscores A with probability > 0.5, despite having a lower mean.

    The above scenario isn't exactly what you're aiming for but it's
    a similar idea. You're trying to jack up the cube so that whenever
    you win, you almost certainly win more than A does.

    I can see that this might work, although I'm not very confident that
    this can actually be achieved in backgammon. The trouble I see is
    that there will probably be a non-negligible fraction of your wins
    when you never fall behind. Imagine you quickly launch a successful
    blitz. A will frequently win a doubled gammon. It will be hard for
    B to win more than 4 points with your strategy, I think. So I don't
    think you're entitled to multiply your 55% wins by 95%.

    Nevertheless, I think your proposal is a good insight. I have to think
    about whether it can be pushed further.

    ---
    Tim Chow

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  • From pepstein5@gmail.com@21:1/5 to Tim Chow on Fri Oct 29 01:01:57 2021
    On Friday, October 29, 2021 at 4:22:57 AM UTC+1, Tim Chow wrote:
    On 10/27/2021 9:42 AM, peps...@gmail.com wrote:
    On Wednesday, October 27, 2021 at 3:21:33 AM UTC+1, Tim Chow wrote:
    On 10/24/2021 11:17 AM, peps...@gmail.com wrote:

    B moves the checkers exactly as if it was DMP, thereby giving B a greater winning probability than A.
    B's cube play is such that B's only goal (in relation to the cube) is to maximise the cube value.
    Since B assumes the house is using an expected value strategy, B will cube when behind to get beavered.
    With backgammon being so complex, we can only make guesses at the results. Since B never gets cubed out, I'll hypothesise that B's winning probability is >= 55% (but this type of thing is impossible to prove).
    Since B is maxing the cube value and A isn't, I'll hypothesise that the probability that [the absolute value of B's result > the absolute value of A's result]
    (in a single game) is >= 90%.
    Therefore, (over a single game), assuming my hypotheses are correct, B outperforms A with a probability >= 55% * 95% > 52%.
    Okay, I think I see what you're trying to do...this is interesting.

    Here's how I would parse it. Consider the following scenario.

    Prob(A scores +1) = 0.5
    Prob(A scores -1) = 0.5

    Prob(B scores +2) = 0.5 + epsilon
    Prob(B scores -3) = 0.5 - epsilon

    Assume A and B are independent. If B scores +2 then B outscores A
    regardless of what A scores. This happens with probability > 0.5 so
    B outscores A with probability > 0.5, despite having a lower mean.

    The above scenario isn't exactly what you're aiming for but it's
    a similar idea. You're trying to jack up the cube so that whenever
    you win, you almost certainly win more than A does.

    I can see that this might work, although I'm not very confident that
    this can actually be achieved in backgammon. The trouble I see is
    that there will probably be a non-negligible fraction of your wins
    when you never fall behind. Imagine you quickly launch a successful
    blitz. A will frequently win a doubled gammon. It will be hard for
    B to win more than 4 points with your strategy, I think. So I don't
    think you're entitled to multiply your 55% wins by 95%.

    Nevertheless, I think your proposal is a good insight. I have to think
    about whether it can be pushed further.

    Glad you found my insights interesting.
    I agree that 95% is too high, but I also think that 55% is too low.
    I did say this in a subsequent posting.
    As corrected, I think that 60% is a reasonable number of wins to aim at,
    and 70% is a reasonable probability for getting an absolute game value that is higher than A's.
    Assuming (and this may actually be a stretch) that the two probabilities are independent, this seems
    to satisfy your problem (60% * 50% (A losing) ) + (60% * (70% * 50%) (A winning but winning less than B)) = 51% > 50%.

    Paul

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