The [0,1] Game: Part 8
Jerrod Ankenman
Bill Chen/Jerrod Ankenman
Part 8: The finite pot (continued)
This is part 8 in a many-part investigation of what we will refer to as
the [0,1] game, which can in some ways be thought of as analogous to
poker and is actually interesting in its own right.
Recap to this point: We spent Parts 1-6 studying the [0,1] game with an
important stipulation, that the pot was "infinite." We later defined
this as a rule that "no player may fold." We created optimal strategies
for games with and without check-raise, culminating in our creation of
optimal strategies for the heads-up game with unlimited raises and
check-raising allowed. We calculated the ex-showdown value of these
games, and gained insight into the methods that we will use to solve
even more complex games. In Part 7, we solved a simple introduction to
the finite pot cases; one of the original problems of the emerging
discipline of game theory.
In Part 8, we will continue looking at simpler finite pot games, working
toward more and more complicated rules.
Game #8: Finite pot of size p, one bet left.
In Part 7 we solved this game for the case when Player X checked dark.
Now we'll allow both players to bet and solve the game again.
The new points we have are:
x1 and x*_0 = x0, the betting and bluffing thresholds for X.
y*_1, the call/fold threshold for Y.
Creating our game diagram:
x1 x*_1 x0
|------------------|----------|----------|----------|----|----|
y1 y*_1 y0
Now, it turns out that we don't know beforehand whether x*_1 or y*_1 is
left of the other, but it winds up not mattering, because we don't have
to relate them to each other in constructing the equations.
All of these equations are derived by familiar mechanisms, so I'm going
to just provide the interval tables and equations and group them again
at the bottom:
At x0:
Y's hand X bets X checks
[0,y*_1] -1 0
[y*_1,x0] +p 0
[x0,1] +p 0
(if x checks at x0, y will bluff him out!)
y*_1 = p(1 - y*_1)
y*_1(p+1) = p
y*_1 = (p/(p+1)) [1]
At y0:
X's hand Y bets Y checks
[0,x1] <X already bet>
[x1,x*_1] -1 0
[x*_1,y0] +p 0
[y0,x0] p p
[x0,1] <X already bet>
x*_1 - x1 = p(y0-x*_1)
x*_1 + px*_1 = py0 + x1
x*_1 = (py0 + x1)/(p+1) [2]
At x*_1:
Y's hand X calls X folds
[0,y1] -1 0
[y0,1] +(p+1) 0
y1 = (1-y0)(p+1)
y1/(p+1) = 1 - y0
y0 = 1 - y1/(p+1) [3]
At y*_1:
X's hand Y calls Y folds
[0,x1] -1 0
[x0,1] +(p+1) 0
x1 = (1-x0)(p+1)
x1/(p+1) = 1 - x0
x0 = 1 - x1/(p+1) [4]
At x1:
Y's hand X bets X checks
[0,x1] -1 -1
[x1,y1] +1 +1
[y1,y*_1] +1 0
[y*_1,y0] 0 0
[y0,1] 0 +1
y*_1 - y1 = 1 - y0 [5]
At y1:
X's hand Y bets Y checks
[0,x1] <x already bet>
[x1,y1] -1 0
[y1,x*_1] +1 0
[x*_1,1] 0 0
y1 - x1 = x*_1 - y1
y1 = (x*_1 + x1)/2 [6]
So our six unknowns are x1,y1,x*_1,y*_1,x0,y0.
And our six equations are:
y*_1 = x0 * (p/(p+1)) [1]
x*_1 = (py0 + x1)/(p+1) [2]
y0 = 1 - y1/(p+1) [3]
x0 = 1 - x1/(p+1) [4]
y*_1 - y1 = 1 - y0 [5]
y1 = (x*_1 + x1)/2 [6]
For ease of solving these equations, we're going to designate the
quantity (1/(p+1)) as a constant, a.
Then:
1 - a = p/(p+1)
y*_1 = (1-a) [1]
x*_1 = a(py0 + x1) [2]
y0 = 1 - ay1 [3]
x0 = 1 - ax1 [4]
y*_1 - y1 = 1 - y0 [5]
y1 = (x*_1 + x1)/2 [6]
y*_1 - y1 = 1 - (1 - ay1)
y*_1 - y1 = ay1
y*_1 = (1+a)y1
1-a = (1+a)y1
y1 = (1-a)/(1+a)
y0 = 1 - ay1
y0 = 1 - a(1-a)/(1+a)
y0 = 1+a - a + a^2/(1+a)
y0 = (1+a^2)/(1+a)
y1 = (x*_1 + x1)/2
2(1-a)/(1+a) = x*_1 + x1
x*_1 = a(py0 + x1)
2(1-a)/(1+a) - x1 = a(py0 + x1)
2(1-a)/(1+a) - x1 = (1-a)y0 + ax1
2(1-a)/(1+a) - (1-a)y0 = x1 + ax1
2(1-a)/(1+a) - (1-a)(1-ay1) = (1+a)x1
(2(1-a) - (1-a)(1+a)(1-ay1))/(1+a) = (1+a)x1
(2(1-a) - (1-a)(1+a)(1-ay1)) = (1+a)^2x1
(2(1-a) - (1-a)(1+a^2)) = (1+a)^2x1
((1-a)(2 - 1 - a^2)) = (1+a)^2x1
(1-a)(1-a^2) = (1+a)^2x1
(1-a)^2(1+a) = (1+a)^2x1
x1 = (1-a)^2/(1+a)
2(1-a)/(1+a) = x*_1 + (1-a)^2/(1+a)
x*_1 + (1 - 2a + a^2)/(1+a) = (2-2a)/(1+a)
x*_1 = (2-2a-1+2a-a^2)/(1+a)
x*_1 = (1-a^2)/(1+a)
x*_1 = (1-a)
x0 = 1 - ax1
x0 = 1 - a(1-a)^2/(1+a)
x0 = 1 - a(1-2a+a^2)/(1+a)
x0 = (1 + a - a+2a^2-a^3)/(1+a)
x0 = (1+2a^2-a^3)/(1+a)
OK, so this is our solution:
y*_1 and x*_1 are both 1-a, or p/(p+1).
x1 is (1-a)^2/(1+a)
y1 is (1-a)/(1+a)
and the bluffing thresholds are both 1-a*(corresponding betting region).
Again we'll leave the game value as an exercise to the reader. It's
a(x1)/2.
As you can see, the finite pot cases are significantly more complex and
tedious. As we move forward with these, we'll be introducing even MORE
notation and finding new and simpler ways to write the indifference
equations at different points.
Next: A comparison of game values and proving these strategies are
optimal.
Barbara Yoon
> Finite pot of size p, one bet left.
>
> y*_1, the call/fold threshold for Y.
>
> Creating our game diagram:
> x1 x*_1 x0
> |------------------|----------|----------|----------|----|----|
> y1 y*_1 y0
The "x*_0" there is confusing...where is that a factor in anything...?!
> At x1:
>
> Y's hand X bets X checks
> [0,x1] -1 -1
> [x1,y1] +1 +1
> [y1,y*_1] +1 0
> [y*_1,y0] 0 0
> [y0,1] 0 +1
>
> y*_1 - y1 = 1 - y0 [5]
Of your five scenarios shown above, I understand your "[0,x1] -1 -1"
-- but then would you please explain to us more thoroughly about
your other four scenarios, what happens with the initial pot, "p"...?!
Thanks...
Jerrod Ankenman
> > Finite pot of size p, one bet left.
> >
> > x1 and x*_0 = x0, the betting and bluffing thresholds for X.
> > y*_1, the call/fold threshold for Y.
> >
> > Creating our game diagram:
> > x1 x*_1 x0
> > |------------------|----------|----------|----------|----|----|
> > y1 y*_1 y0
>
> The "x*_0" there is confusing...where is that a factor in anything...?!
x*_0 and x0 are the same; x0 is actually just shorthand for x*_0. We
haven't yet introduced the raise-bluff notation, which really helps to
make the * points clear, but basically the x*_n point is the threshold
between calling the nth bet and taking the next weaker action (which
is often bluffing or bluff-raising). 0 is a somewhat special case, but
we simply define x*_0 as the first-bet bluffing threshold. Then we
just call it x0 for short.
> > At x1:
> >
> > Y's hand X bets X checks
> > [0,x1] -1 -1
> > [x1,y1] +1 +1
> > [y1,y*_1] +1 0
> > [y*_1,y0] 0 0
> > [y0,1] 0 +1
> >
> > y*_1 - y1 = 1 - y0 [5]
>
> Of your five scenarios shown above, I understand your "[0,x1] -1 -1"
> -- but then would you please explain to us more thoroughly about
> your other four scenarios, what happens with the initial pot, "p"...?!
>
> Thanks...
What we're concerned about here is the difference in value between the
two actions. As a result, we tend to work in "ex-showdown" values -
that is, the values of actions independent of the pot size. When X is
at x1, no actions will cause X or Y to fold the best hand, so the p
units in the pot becomes irrelevant. We could create this table like
this:
Y's hand X bets X checks
[0,x1] -1 -1
[x1,y1] p+1 p+1
[y1,y*_1] p+1 p
[y*_1,y0] p p
[y0,1] p p+1
But since we're only concerned with the differences here, we simply
shortcut by ignoring the size of the pot when it doesn't matter.
Note that when there is bluffing, as for example at y*_1:
X's hand Y calls Y folds
[0,x1] -1 0
[x0,1] +(p+1) 0
Now Y is getting bluffed out of a pot that he would have won, so he
actually loses the pot in addition to losing the bet he could have won
by calling. This is equivalent to:
X's hand Y calls Y folds
[0,x1] -1 0
[x0,1] +1 -p
What's important is the difference in value between the two actions.
Jerrod Ankenman
Barbara Yoon
>>> Finite pot of size p, one bet left.
>>>
>>> x1 and x*_0 = x0, the betting and bluffing thresholds for X.
>>> y*_1, the call/fold threshold for Y.
>>>
>>> Creating our game diagram:
>>> x1 x*_1 x0
>>> |------------------|----------|----------|----------|----|----|
>>> y1 y*_1 y0
me:
>> The "x*_0" there is confusing...where is that a factor in anything...?!
Jerrod Ankenman:
> x*_0 and x0 are the same; x0 is actually just shorthand for x*_0.
Jerrod...as you know, this is rather complicated and difficult material
(at least it is for ME, anyway), and having two different 'names' for the
same variable, for no apparent reason -- I mean, the "x*_0" doesn't
even appear anywhere else in the rest of your post -- only serves to
make it even more difficult to follow your notation and logic...
>>> At x1:
>>>
>>> Y's hand X bets X checks
>>> [0,x1] -1 -1
>>> [x1,y1] +1 +1
>>> [y1,y*_1] +1 0
>>> [y*_1,y0] 0 0
>>> [y0,1] 0 +1
>>>
>>> y*_1 - y1 = 1 - y0 [5]
>> Of your five scenarios shown above, I understand your "[0,x1] -1 -1"
>> -- but then would you please explain to us more thoroughly about
>> your other four scenarios, what happens with the initial pot, "p"...?!
Jerrod Ankenman:
> What we're concerned about here is the difference in value between
> the two actions. As a result, we tend to work in "ex-showdown"
> values - that is, the values of actions independent of the pot size.
> When X is at x1, no actions will cause X or Y to fold the best hand,
> so the p units in the pot becomes irrelevant. We could create this
> table like this:
>
> Y's hand X bets X checks
> [0,x1] -1 -1
> [x1,y1] p+1 p+1
> [y1,y*_1] p+1 p
> [y*_1,y0] p p
> [y0,1] p p+1
>
> But since we're only concerned with the differences here, we simply
> shortcut by ignoring the size of the pot when it doesn't matter.
Jerrod...my advice to you here then would be for you not to take any
such "shortcuts" in your attempts to communicate such complicated
and difficult material to us readers who are not already familiar with
your notation and logic...OK?!
Mark Rafn
>The [0,1] game and Derivatives: A Study of Some Poker-Like Games
> Bill Chen/Jerrod Ankenman
Are these articles available in html or pdf (or printed book) form?
Usenet plain-text is a bit hard to read.
--
Mark Rafn da...@dagon.net <http://www.dagon.net/>
Barbara Yoon
>
> |------------------|----------|----------|----------|----|----|
> y1 y*_1 y0
> ...our six equations are:
> y*_1 = x0 * (p/(p+1)) [1]
> x*_1 = (py0 + x1)/(p+1) [2]
> y0 = 1 - y1/(p+1) [3]
> x0 = 1 - x1/(p+1) [4]
> y*_1 - y1 = 1 - y0 [5]
> y1 = (x*_1 + x1)/2 [6]
>
> For ease of solving these equations, we're going to designate the
> quantity (1/(p+1)) as a constant, a. Then: 1 - a = p/(p+1)
> OK, so this is our solution:
> y*_1 and x*_1 are both 1-a, or p/(p+1).
> x1 is (1-a)^2/(1+a)
> y1 is (1-a)/(1+a)
> Again we'll leave the game value as an exercise to the reader.
> It's a(x1)/2.
Interesting...looking at the 'pot-limit' case of p=1, how with both
players playing optimally in this, the action would go 'check-check'
(1 - 1/6 - 1/12) x (1 - 1/3 - 1/6) = 3/8 of the time... And what about
if one of the players were to deviate from these strategies -- for
example, if the second player, Player Y, were to 'value bet' ("y1")
too often -- then in what ways might Player X adjust his own strategy
so as to best take advantage of Player Y's errant ways...?!
Barbara Yoon
Mark Rafn:
> Are these articles available in html or pdf (or printed book) form?
> Usenet plain-text is a bit hard to read.
Card Player magazine had a number of 'game theory' articles
some years ago, including one on this exact "[0,1] Game: Part 8"
(for p=1) -- Card Player's treatment didn't go into so much fine
calculation detail as Chen and Ankenman do here, but Card Player
did relate the 'bottom line' to actual poker hands, and if I recall,
also went further into playing strategies and counter-strategies...
And so do you have any old back issues of Card Player...??
Bill chen
>
> Interesting...looking at the 'pot-limit' case of p=1, how with both
> players playing optimally in this, the action would go 'check-check'
> (1 - 1/6 - 1/12) x (1 - 1/3 - 1/6) = 3/8 of the time... And what about
> if one of the players were to deviate from these strategies -- for
> example, if the second player, Player Y, were to 'value bet' ("y1")
> too often -- then in what ways might Player X adjust his own strategy
> so as to best take advantage of Player Y's errant ways...?!
Well, the short answer is that X doesn't have to adjust. If Y bets
too much, then he will simply lose equity against otpimal play. Say Y
decided to bet the pot at 2/5=0.4 if checked to. X will check with
[1/6, 11/12] and check-call with [1/6, 1/2]. By betting instead of
taking the free showdown at 0.4, Y will lose 6.67 equity.
Note that X can gain more by playing exploitively. Suppose Y bets a
little too much but doesn't up his number of bluffs. Then X can
safely fold some of his worst calling hands (near 0.5), and
furthermore he should check and sand-bag with his good hands from
[0,1].
Bill
Jerrod Ankenman
As a general rule, when a player plays suboptimally against you and
you wish to exploit that, you should adjust your strategy to give him
more opportunities to make the suboptimal play. So for example if Y
moves y1 to the right (value bets more often), X should actually move
x1 to the left (check-call more often), as this enables Y to make his
mistake.
As a corollary X should move x0 to the right, preserving the a*x1
relationship between these two regions.
In addition, if Y is not offsetting his increased value bet frequency
with a corresponding number of bluffs, then X should move x*_1 to the
left, calling less often, because Y is more likely to have his
value-betting hands when he bets.
Jerrod Ankenman
Barbara Yoon
>>> Finite pot of size p, one bet left.
>>>
>>> x1 x*_1 x0
>>> |------------------|----------|----------|----------|----|----|
>>> y1 y*_1 y0
>>> ...
>>> Then: 1 - a = p/(p+1)
>>> ...
>>> OK, so this is our solution:
>>> y*_1 and x*_1 are both 1-a, or p/(p+1).
>>> x1 is (1-a)^2/(1+a)
>>> y1 is (1-a)/(1+a)
>>> ...bluffing thresholds are both 1-a*(corresponding betting region).
me:
>> And what about if one of the players were to deviate from these
>> strategies -- for example, if the second player, Player Y, were to
>> 'value bet' ("y1") too often -- then in what ways might Player X
>> adjust his own strategy so as to best take advantage of Player Y's
>> errant ways...?!
Bill chen:
> Well, the short answer is that X doesn't have to adjust. If Y bets
> too much, then he will simply lose equity against otpimal play.
> Say Y decided to bet the pot at 2/5=0.4 if checked to. X will check
> with [1/6, 11/12] and check-call with [1/6, 1/2]. By betting instead
> of taking the free showdown at 0.4, Y will lose 6.67 equity. Note
> that X can gain more by playing exploitively. Suppose Y bets a little
> too much but doesn't up his number of bluffs. Then X can safely
> fold some of his worst calling hands (near 0.5), and furthermore
> he should check and sand-bag with his good hands from [0,1].
Jerrod Ankenman:
> As a general rule, when a player plays suboptimally against you and
> you wish to exploit that, you should adjust your strategy to give him
> more opportunities to make the suboptimal play. So for example
> if Y moves y1 to the right (value bets more often), X should actually
> move x1 to the left (check-call more often), as this enables Y to
> make his mistake. As a corollary X should move x0 to the right,
> preserving the a*x1 relationship between these two regions.
> In addition, if Y is not offsetting his increased value bet frequency
> with a corresponding number of bluffs, then X should move x*_1
> to the left, calling less often, because Y is more likely to have his
> value-betting hands when he bets.
Hmm...Bill says...
"Suppose Y bets a little too much... Then X can safely fold some
of his worst calling hands (near 0.5)"...
...and Jerrod says...
"...if Y moves y1 to the right (value bets more often)... ...then X
should move x*_1 to the left, calling less often"...
...and that certainly makes good sense -- but youse guys' wordings
here (especially Bill's) tend to give the impression (to me, anyway)
that with Player Y value-betting 'a little' too much, Player X's best
calling counter-strategy would be to call similarly only just 'a little'
less often... Is that the impression that you want to convey...?!
Jerrod Ankenman
> Jerrod Ankenman <jerroda...@yahoo.com> wrote:
> >The [0,1] game and Derivatives: A Study of Some Poker-Like Games
> > Bill Chen/Jerrod Ankenman
>
> Are these articles available in html or pdf (or printed book) form?
> Usenet plain-text is a bit hard to read.
I do have a sort of far-reaching plan to html-ize or potentially
publish all these articles, perhaps in revised form. However, right
now I'm just trying to catch up the documented material with the
things we're talking about and working on right now, without rushing
through the complications too much.
So the short answer to your question is: not right now, but probably
later.
Jerrod Ankenman
Steve Blay
Congratulations on your thoughtful analysis of this game. I too have
been working with the "0,1" game for a while now, independently of
your work. Just a couple days ago I was discussing my results with
someone on 2+2, and they said "you should really check out rgp...
Jerrod A. is analyzing the exact same game..."
Anyway, I thought it might interest you to see how I was going about
my analysis. (Actually, I prefer your way of solving it with the
equilibrium equations. I never considered doing it that way.) I just
mainly used algebra which was a royal pain. For example, for what you
call "game 7", my profit equation for player 2 was:
P = x[(1-y)(p) + y[(x/2y)(-b)+(1-(x/2y))(b+p)]] + xz[(1-y)(p)+y(-b)] +
w[(xz)(p)+ w(p/2)]
where:
P = profit this player makes per hand
p = pot size in dollars
b = bet size in dollars
z = b/(b+p) = bluffing frequency
x = legitimate hands this player will bet
xz = percentage of hands this player will bluff with
y = hands the other player will call with
w = 1-(x+xz) = percentage of hands that this player checks
simplifying arrives at...
P = -x^2(pz^2/2 + b) + x(pz) + p/2
then taking the derivative wrt x, and finding the maximum yields:
x=pz/(pz^2+2b)
which, assuming b=1 (as you did), this can actually be simplified
into:
x=p(p+1)/(2p+1)(p+2)
which is the exact same answer you came up with for y1 in game #7!
I was happy to see our answers matched up.
btw, I've also written a program that analyzes the [0,1] game for a
given strategy by each player. Basically it assigns each player 1000
equally spaced real numbers on the interval [0,1], and plays the game
using the given strategy (1000^2 total iterations). I tried plugging
in the values you arrive at for game #8, with a pot size of 1, and got
the result:
player B wins .541709 per play
Subtracting the .50 he is "supposed to win" (given the pot size),
yields a profit of .041709, which is roughly equal to the value you
came up with in post #8:
a(x1)/2 = (1/2)(1/6)/2 = .0416666
Anyway, keep up the good work and I look forward to hearing from
you...
Steve
Barbara Yoon
>>> Finite pot of size p, one bet left.
>>>
>>> x1 x*_1 x0
>>> |------------------|----------|----------|----------|----|----|
>>> y1 y*_1 y0
>>> ...
>>> Then: 1 - a = p/(p+1)
>>> ...
>>> OK, so this is our solution:
>>> y*_1 and x*_1 are both 1-a, or p/(p+1).
>>> x1 is (1-a)^2/(1+a)
>>> y1 is (1-a)/(1+a)
>>> ...bluffing thresholds are both 1-a*(corresponding betting region).
me:
>> And what about if one of the players were to deviate from these
>> strategies -- for example, if the second player, Player Y, were to
>> 'value bet' ("y1") too often -- then in what ways might Player X
>> adjust his own strategy so as to best take advantage of Player Y's
>> errant ways...?!
Jerrod Ankenman:
> As a general rule, when a player plays suboptimally against you
> and you wish to exploit that, you should adjust your strategy
> to give him more opportunities to make the suboptimal play.
Bill chen:
> Well, the short answer is that X doesn't have to adjust. If Y bets
> too much, then he will simply lose equity against otpimal play.
> Note that X can gain more by playing exploitively.
Bill...while the clear directness of your "short answer" there
is appreciated, the tone of it does tend to give the impression
(to me, anyway) that you might be stating it as something of
"a general rule" -- so is that what you really do mean in this,
that if either player deviates in any way from "otpimal play,"
"then he will simply lose equity"...?!
Barbara Yoon
> Jerrod and Bill -- Congratulations on your thoughtful analysis of
> this game. I too have been working with the "0,1" game for a while
> now, independently of your work. Just a couple days ago I was
> discussing my results with someone on 2+2, and they said "you
> should really check out rgp... Jerrod A. is analyzing the exact same
> game..." Anyway, I thought it might interest you to see how I was
> going about my analysis. (Actually, I prefer your way of solving it
> with the equilibrium equations. I never considered doing it that way.)
> I just mainly used algebra which was a royal pain. For example,
> matched up.
Steve...not to take anything away from the good work done here by
you, and Bill Chen and Jerrod Ankenman, but Card Player magazine
had a number of 'game theory' articles some years ago, including
on these exact "8" and "7" games -- Card Player's treatment didn't
go into so much fine calculation detail as you guys, but Card Player
did relate the 'bottom lines' to actual poker hands, and if I recall,
also went further into playing strategies and counter-strategies...
...and so do you have any old back issues of Card Player...??
Bill chen
Morgenstern and Von Neumann did game 7 50 years ago. We are not
claiming that games 7 and 8 are that groundbreaking, just that the
methods we have developed will get us the solution to far more complex
games which we will introduce later.
Bill
"Barbara Yoon" <by...@erols.com> wrote in message news:<b4gk3m$88p$1...@bob.news.rcn.net>...
Bill chen
No, and this is an important point. There are times where a player
can deviate from optimal without hurting his equity if his opponent
continues to play optimally and does not adjust.
In the [0,1] game this happens precisely where one player can move his
threshold point around without it affecting his equity against nearby
points. an example will clear this up. Let's look at x1, the
threshold between X betting and check-calling. Now if Y is near x1,
say in [x1-e, x1+e], then it doesn't matter what X does--since 1 bet
will go into the pot regardless, if X bets, Y will call and if X
check-calls, Y will bet. So unless Y changes some other parts of his
strategy (away from x1) X will continue to enjoy the same equity he
has playing optimally. Now if he moves the threshold too much (into
one of the other Y thresholds, he will start to lose money.
However, this is not so for y1. If X has a hand a little worse than
y1, then betting at y1 instead of checking will gain one bet (and lose
1 bet if it's a little better). So the threshold for y1 is
rigid--deviating from it loses against optimal play.
This will have an effect once we start looking at real game cases,
because as you look at the optimal solutions you realize that x is
indifferent to bluffing in [y0,1]. X has to bluff a=1/(1+P) of his
value bets but he gets to choose where to put them. If the cards are
correlated, X will bluff with precisely the hands which make it less
likely for Y to have a "good" calling/raising hand that has positive
equity calling instead of one that is simply ambivalent to calling).
BTW to answer your other post, what should X do if Y values bets too
much (moves y1 higher)? Well there are two ways Y can do this. One
way would be to leave everything else in Y's strategy the same,
including the bluffing frequency. Now, this would seem a poor way to
try to play aggressively since X can just not call with hands worse
than y1 (since Y is not making the bet/bluff ratio correct). However
in real life I find just the opposite adjustment made. In the Oaks
Lowball game there is a player who will bluff approximately correctly
(top pair is approx. correct) but will value bet way too many hands
like 9's and T's. Many of the other players adjust by calling him
with kings and such, while the correct exploitive strategy is actually
to call with close to the range of hands he would bet (actually a
little tighter)
If Y adjusts both his bluffing and betting frequencies to be more
aggreesive, then this is harder to play against. Many aggressive
players do this and enjoy a modicum of success. Best exploitive
strategy is for X to check all good hands to trap since even with the
nuts trapping a bet is easier than inducing a call. X will still be
indifferent to calling between y0 and y1, but X can tighten up his
calling requirements here (because X is checking more good hands, he
doesn't have to call with as many mediocre hands after he checks).
Bill
Barbara Yoon
>> you, and Bill Chen and Jerrod Ankenman, but Card Player magazine
>> had a number of 'game theory' articles some years ago, including
>> on these exact "8" and "7" games -- Card Player's treatment didn't
>> go into so much fine calculation detail as you guys, but Card Player
>> did relate the 'bottom lines' to actual poker hands, and if I recall,
>> also went further into playing strategies and counter-strategies...
>> ...and so do you have any old back issues of Card Player...??
Bill chen:
> Barbara, Morgenstern and Von Neumann did game 7 50 years ago.
> We are not claiming that games 7 and 8 are that groundbreaking,
> just that the methods we have developed will get us the solution to
> far more complex games which we will introduce later.
Bill...yes...I do 'get' what you're saying...and if Newton, or Pascal,
or Fermat, or others knew, and cared about poker, they would've
gotten it long before "Morgenstern and Von Neumann" too... But
my main point there was that even a superficial industry 'PR fluff rag'
like Card Player magazine, years ago, actually published some really
good quality technical material comparable to the good work that
you and Jerrod are now posting here in RGP...OK?!
Barbara Yoon
>>> Well, the short answer is that X doesn't have to adjust. If Y bets
>>> too much, then he will simply lose equity against otpimal play.
>>> Note that X can gain more by playing exploitively.
>> Bill...while the clear directness of your "short answer" there
>> is appreciated, the tone of it does tend to give the impression
>> (to me, anyway) that you might be stating it as something of
>> "a general rule" -- so is that what you really do mean in this,
>> that if either player deviates in any way from "otpimal play,"
>> "then he will simply lose equity"...?!
Bill chen:
> No, and this is an important point. There are times where a player
> can deviate from optimal without hurting his equity if his opponent
> continues to play optimally and does not adjust.
Bill...good...just a point that I think needed to be clarified...OK?!
Barbara Yoon
>> magazine, years ago, actually published some really good quality
>> technical material comparable to the good work that you and
>> Jerrod are now posting here in RGP...
And the irony of that is that in all likelihood, the publisher and editors
were totally ignorant and oblivious to the 'quality' of what they had...
Steve Blay
Even if one's work doesn't advance the "frontiers of science" doesn't
make it worthless. I spent the last few weeks analyzing this game,
and I wouldn't consider any of it wasted even though the game has
already been "solved".
I'm the kind of person who rarely accepts something without proving it
himself. These posts help people see the thought process behind it.
Barbara Yoon
>>>> magazine, years ago, actually published some really good quality
>>>> technical material comparable to the good work that you and
>>>> Jerrod are now posting here in RGP...
>> And the irony of that is that in all likelihood, the publisher and editors
>> were totally ignorant and oblivious to the 'quality' of what they had...
Steve Blay:
> Even if one's work doesn't advance the "frontiers of science" doesn't
> make it worthless. I spent the last few weeks analyzing this game,
> and I wouldn't consider any of it wasted even though the game has
> already been "solved". I'm the kind of person who rarely accepts
> something without proving it himself. These posts help people see
> the thought process behind it.
Steve...honestly, I was not trying to take anything away from the
good work done here by Bill Chen and Jerrod Ankenman, and you
(work worthy of a mathematics textbook, or journal) -- I was only
pointing out one other place where comparable work had been
published previously (in a POKER magazine, no less!), in case you
would want to look it up, to see somebody else's different approach,
or conclusions, or whatever... And I do thoroughly agree with what
you say here...because as it is with yourself, I too like to work out
these interesting and challenging questions for myself...OK?!
Bill chen
posts were worthless. Furthermore I think it's wonderful that you
worked through the same game and provided an independent confirmation
of our results and that Barbara is pointing out all of the places our
write-up might have problems. I
BTW we are not discouraged. We have gotten a few emails encouraging
us on the articles and I hope what we reveal in the future (when we
get to NL, multistreet games) will be eye-opening. So stay tuned!
Bill