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...
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.
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:The Kelly criterion maximises utility if you assume that the utiity of your wealth
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.
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.
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.
On 10/19/2021 3:55 AM, peps...@gmail.com wrote:
The Kelly criterion maximises utility if you assume that the utiity of your wealthThat is true, but that is not how Kelly *derived* his criterion.
is the log of your wealth.
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.
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.The problem I have with your attempted answer is that it
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.
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 asI 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.
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).
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.
On 10/19/2021 4:32 PM, peps...@gmail.com wrote:
You asked a question. I attempted to answer.The problem I have with your attempted answer is that it
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.
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).
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.The problem I have with your attempted answer is that it
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.
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.
I just looked back at your OP.The fact that one can also interpret the Kelly criterion asI understand all that, but aren't you just answering your own question fully?
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).
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.
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.
Again, I reiterate that I am quite expert in this subject
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.
I am watching this closely to see if Chow will call Paul
"Red Big"...? :)
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 highAh, this is getting closer to addressing my question. But I'm
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.
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.
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.
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.
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 plausibleBut doesn't that sneak in another arbitrariness? And one that cries out
class of strategies.
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
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.
"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 muchWell, some keep the stakes constant and set some limit for their maximum loss.
more convincing than the second because keeping the staking proportion constant is an intuitive idea, and one that many people would do in practice.
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).
On 10/21/2021 4:36 PM, peps...@gmail.com wrote:
Utility theory is clearly the correct theoretical framework to address Tim'sPlease show my how utility theory answers the question below, which
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).
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
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.
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.
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:
I don't see why this is true. I want B to have a >50% probability ofQuestion: Is there some strategy that B can adopt which will giveSince A and the house are adopting the same strategy, the probability that A makes a profit
B a >50% probability of being ahead of A at the end of the sessions?
must be <= 50%.
So any strategy by B that makes a profit with more than 50% probability solves your problem.
*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.
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:
I don't see why this is true. I want B to have a >50% probability ofQuestion: Is there some strategy that B can adopt which will giveSince A and the house are adopting the same strategy, the probability that A makes a profit
B a >50% probability of being ahead of A at the end of the sessions?
must be <= 50%.
So any strategy by B that makes a profit with more than 50% probability solves your problem.
*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 theI 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.
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.
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.
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:
I don't see why this is true. I want B to have a >50% probability ofQuestion: Is there some strategy that B can adopt which will giveSince A and the house are adopting the same strategy, the probability that A makes a profit
B a >50% probability of being ahead of A at the end of the sessions?
must be <= 50%.
So any strategy by B that makes a profit with more than 50% probability solves your problem.
*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.
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.
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.
On 10/23/2021 5:29 PM, MK wrote:No, that isn't my complaint.
Chow's example is a typical failed effort to apply a simplistic coinThe questions I'm discussing evaporate in any non-contact race, because
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.
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
On 10/23/2021 5:29 PM, MK wrote:
Chow's example is a typical failed effort to apply a simplistic coinThe questions I'm discussing evaporate in any non-contact race, because
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.
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.
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.
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.I'm okay with the assumption that the stake for both players
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.
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.
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 thenNo, I'm not insisting on i.i.d. I don't mind deviating from that.
revert to expected value play.
I didn't realise iid was another unstated assumption.
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.
On Wednesday, October 27, 2021 at 3:21:33 AM UTC+1, Tim Chow wrote: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
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 thenNo, I'm not insisting on i.i.d. I don't mind deviating from that.
revert to expected value play.
I didn't realise iid was another unstated assumption.
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 exampleOk, I'll try again but I'll cheat by having a new strategy.
is wrong but it would help if you would spell out the argument in some
more detail.
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
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 thenNo, I'm not insisting on i.i.d. I don't mind deviating from that.
revert to expected value play.
I didn't realise iid was another unstated assumption.
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 exampleOk, I'll try again but I'll cheat by having a new strategy.
is wrong but it would help if you would spell out the argument in some more detail.
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]
PaulI 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.
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%.
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.Okay, I think I see what you're trying to do...this is interesting.
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%.
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.
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