The [0,1] Game: part 9
Jerrod Ankenman
10 to follow soon. For new readers: You can read previous installments
of this series at:
As usual, direct any questions or comments to me at
jerroda...@yahoo.com, or if you know my or Bill's other addresses,
that's fine too.]
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The [0,1] game and Derivatives: A Study of Some Poker-Like Games
Bill Chen/Jerrod Ankenman
Part 9: A comparison of game values and proving these strategies are
optimal.
This is part 9 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 Parts 7 and 8, we solved simple
introductory games to the finite pot cases.
In Part 9, we'll take a little break from equation-writing and
equilibrium-finding to look at a couple of interesting issues that
arise.
We left the values of games 7 and 8 as an exercise for the reader at
that time. But what do we know about these values? Clearly, we know
that the value of game 8 for player X (where both players can bet)
cannot be less than the value of game 7 (where X is forced to check
dark) - if it were less, X could simply check dark and achieve his
game 7 equity.
But how much value is there to being able to bet? We can find out by
calculating the equities of games 7 and 8 and comparing them.
Game 7:
Referring back to Part 7, we have the following solution to the game:
y1 = p(p+1)/(2p+1)(p+2)
x*_1 = 2p(p+1)/(2p+1)(p+2)
y0 = 1 - y1/(p+1) = 1 - (p/(2p+1)(p+2))
So what does this mean in real life?
X is forced to check dark. Y bets if his hand is less than the y1
expression above, and bets if his hand is greater than the y0
expression above. X calls if his hand is less than the x*_1 expression
above.
So what is the value of the game for X? Well, when Y value bets, X
calls with twice as many hands, right? So half the time when X calls,
Y wins, and half the time, they have the same range of hands and break
even. So Y wins 1/2 a bet when he value bets and gets called, which is
y1*x*_1 of the time. Y also loses a bet when he bluffs and gets
called, which occurs (1-y0)*x*_1 of the time. But Y wins the pot when
he bluffs successfully, which occurs when X is below y0 but above
x*_1. And Y wins half the pot when he bluffs out an equal hand, so
when X is in the same range as Y (1-y0), Y will bluff him out of his
half of the pot.
So Y's equity is:
y1x*_1/2 - (1-y0)x*_1 + p(y0-x*_1)(1-y0) + p(1-y0)^2/2
For pot size 1, this is:
(2/9)(4/9)/2 - (1-(8/9))(4/9) + (1)((8/9)-(4/9))(1-(8/9)) +
(1)(1-(8/9))^2/2
4/81 - 4/81 + 4/81 + 1/162 = 9/162 = 1/18
We can probably simplify the above expression anyway...
(using a = 1/(p+1))
y1x*_1/2 - (1-y0)x*_1 + p(y0-x*_1)(1-y0) + p(1-y0)^2/2
y1^2 - (ay1)2y1 + p(1-ay1-2y1)(ay1) + p(ay1)^2/2
y1^2 - 2ay1^2 + pay1 - p(ay1)^2 - 2pa(y1^2) + p(ay1)^2/2
y1^2 - 2ay1^2 + (1-a)y1 - (1-a)(a)(y1)^2 - (2)(1-a)(y1^2) +
(1-a)(a)(y1^2)/2
y1^2 - 2ay1^2 + y1 - ay1 - (1/2)(1-a)(a)(y1)^2 - 2y1^2 + 2ay1^2
-y1^2 + y1 - ay1 - 1/2ay1^2 + 1/2a^2y1^2
-y1^2 - 1/2ay1^2 + 1/2a^2y1^2 + y1 - ay1
y1^2(1/2a^2 - 1/2a - 1) + y1(1-a)
y1^2(1/2a - 1)(a + 1) + y1(1-a) [*]
Now, we can take the formula for y1 and swap it around to use a
instead of p:
y1 = p(p+1)/(2p+1)(p+2)
2-a = 2 - 1/(p+1) = (2p+2-1)/(p+1) = (2p+1)/(p+1)
so 1/(2-a) = (p+1)/(2p+1)
1+a = 1 + 1/(p+1) = (p+2)/(p+1)
1/(1+a) = (p+1)/(p+2)
1-a = p/(p+1)
y1 = (1-a)/(1+a)(2-a)
Plugging back into [*]:
y1(1-a)(1/2a - 1)(a + 1)/(1+a)(2-a) + y1(1-a) [*]
-y1(1-a)/2 + y1(1-a)
y1(1-a)/2
Some values from this expression:
p=1 1/18
p=2 1/10
p=3 9/70
p=5 15/77
p=inf (a=0) 1/4
By the way, it bears discussing that this "a" expression is a critical
one that appears often in our work. It of course represents the ratio
of value bets/raises to bluffs. In our finite pot cases, we will have
this bluffing region that corresponds to our value betting regions,
and it's size will be a*the size of the betting region.
Of course, a is a function of the pot size, and as we increase the
index of the bet we are considering, a's value moves as well. We will
see this in detail when we analyze games with multiple bets allowed.
Now, Game 8:
Our solution to this game was:
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).
Let's again calculate the value of the game directly (remember this is
ex-showdown value):
X's hand Y's hand Y's EV Frequency
[0,y1] [0,y1] 0
[y1,x*_1] [0,y1] 1 ay1^2
[x*_1,x0] [0,y1] 0
[x0,1] [0,y1] 1 a(1-a)y1^2 [*]
[0,x1] [y1,y*_1] -1 a(1-a)y1^2 [*]
[x1,x0] [y1,y*_1] 0
[x0,1] [y1,y*_1] 1 a^2(1-a)y1^2
[0,x1] [y*_1,y0] 0
[x1,x0] [y*_1,y0] 0
[x0,1] [y*_1,y0] -p 2a^3y1^2
[0,x1] [y0,x0] 0
[x1,x*_1] [y0,x0] -1 2a^3y1^2 [1a]
[x*_1,y0] [y0,x0] p 2a^4y1/(1+a) [1b]
[y0,x0] [y0,x0] p/2 a^4y1^2
[x0,1] [y0,x0] -p a^3(1-a)y1^2
[0,x1] [x0,1] 0
[x1,x*_1] [x0,1] -1 2a^2(1-a)y1^2
[x*_1,x0] [x0,1] p a^2(1-x1)(1-a)y1
[x0,1] [x0,1] -p/2 a^2(1-a)^2y1^2
Now there are some equivalences available (used in calculating the
frequencies):
x*_1 - y1 = 1-a - (1-a)/(1+a) = a(1-a)/(1+a) = ay1
y*_1 - y1 = 1-a - (1-a)/(1+a) = a(1-a)/(1+a) = ay1
y0 - x*_1 = 1 - ay1 - 1 + a = a - ay1 = a(1-y1)
1 - x0 = a(1-a)y1
1 - y0 = ay1
y0 - y*_1 = 1 - ay1 - 1 + a = a - ay1 = a(1-y1) = a(2a/(a+1))
x*_1 - x1 = 1-a - (1-a)^2/(1+a) = (1-a^2 - 1 + 2a - a^2)/(1+a) =
2a(1-a)/(1+a) = 2ay1
x0 - y0 = 1 - ax1 - 1 + ay1 = a(1-a-1+2a-a^2)/(1+a)=
a^2(1-a)/(1+a) = a^2y1
x0 - x*_1 = 1-ax1 - 1 + a = a - ax1 = a(1-x1) =
1-y1 = 1 - (1-a)/(1+a) = 2a/(1+a)
So our EV expression is (pre-cancelling the second and third items and
the other items marked 1a and 1b and substituting x1 for (1-a)y1
sometimes):
ay1^2 + a^2x1y1 - 2pa^3(y1^2) + pa^4y1^2/2 - pa^3x1y1 - 2a^2x1y1 +
pa^2(1-x1)(1-a)y1
- pa^2(1-a)x1y1/2
Since pa = 1-a,
ay1^2 + a^2x1y1 - 2a^2x1y1 + a^3x1y1/2 - a^2(1-a)x1y1 - 2a^2x1y1 +
(a-ax1)(1-a)^2y1
- a(1-a)^2x1y1/2
ay1^2 - 3a^2x1y1 + a^3x1y1/2 - a^2(1-a)x1y1 + (a-ax1)(1-a)^2y1 -
a(1-a)^2x1y1/2
ay1^2 - 3a^2x1y1 + a^3x1y1/2 - a^2(1-a)x1y1 + a(1-a)^2y1 -
3/2a(1-a)^2x1y1
a(y1^2 - 4ax1y1 + 3/2a^2x1y1/2 + (1-a)^2y1 - 3/2x1y1 + 3ax1y1 -
3/2a^2x1y1)
a(y1^2 - ax1y1 + (1-a)^2y1 - 3/2x1y1)
a(y1^2 - a(1-a)y1^2 + (1-a)^2y1 - 3/2(1-a)y1^2)
a(y1^2 - ay1^2 + a^2y1^2 + y1-2ay1+a^2y1 - 3/2y1^2 + 3/2ay1^2)
a(-1/2y1^2 + 1/2ay1^2 + a^2y1^2 + y1 - 2ay1 + a^2y1)
ay1(-1/2y1 + 1/2ay1 + a^2y1 + 1 - 2a + a^2)
ay1(y1(a-1/2)(a+1) + (1-a)^2)
ay1((1-a)(a-1/2) + (1-a)^2)
ay1(1-a)((a-1/2) + (1-a))
ay1(1-a)(1/2)
ax1/2
So the equity of this game simplifies all the way down to ax1/2.
Some values of this function:
p=1 1/24
p=2 1/18
p=3 9/160
p=5 25/504
p=inf (a=0) 0
Comparing these to the values we achieved for Game 7, we can see that
at pot size = 1, Y gives up a third of his edge when X is allowed to
bet. This fraction continues to increase (p=2, it's 2/5) and as the
pot size goes to infinity, Y eventually surrenders his entire edge. Of
course, the infinite pot one-bet game isn't very interesting. :)
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Now let's take a brief look at something that was brought up in some
of the side conversation we've had about our work on the [0,1] game,
which is important but we've tended to gloss over.
To this point, we've been creating equilibrium equation based on
diagrams we've drawn - we've stipulated that "X check-raises his best
hands, bets the next group, checks in the middle, and bluffs on the
end." more or less without proof. In fact, this is only one
paramterization of the solution, and our technique of solving
equilibrium equations will in fact only find the optimal solution to
the parameterization we have chosen.
However, we can in fact prove that the solutions we are finding are in
fact at least co-optimal. A pair of strategies are optimal if, when
played against each other, neither player can unilaterally change
his/her strategy and improve his/her expected value. If we can
demonstrate that neither player can improve on the EV given by the
candidate strategies, we have shown that the parameterization we chose
was the correct one, and that the strategies are optimal.
Let's take Game 8 as an example.
The solution we found for Game 8 was as follows:
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).
Now, we'll just consider the case where p=1, for simplicity's sake.
However, this holds true for any valid value of p.
So the solution to our game when p=1 is:
x1 = 1/6
y1 = 1/3
x*_1 = y*_1 = 1/2
y0 = 5/6
x0 = 11/12
First, let's consider Y. We'll hold X's strategy constant and see if
we can improve on Y's EV by changing his strategy.
Case 1: X bets.
Y's strategy indicates that Y should call with hands to the left of
y*_1, and fold with hands to the right of y*_1.
We know that X has one of 2 types of hands: a hand on [0,1/6] or a
hand on [11/12,1]. First let's consider Y's hands between 1/6 and
11/12. Clearly all of these hands are equivalent given that X bet.
Either Y has a hand that beats a bluff, or he doesn't. Since X is
bluffing with 1/12 of his hands and betting with 1/6 of his hand, he
is a 2:1 favorite to have a real hand. Since Y is getting 2:1 odds to
call from the pot, his EV of calling with any hand between 1/6 and
11/12 is 0.
Now let's consider hands from 0 to 1/6. Y's strategy calls for calling
with these hands. The only alternative strategy Y can employ is to
fold instead. Y's EV from calling with hands in this range must be
positive, as he will now beat X's bluffs (1/3) and any hands X might
have that are between Y's hand and 1/6. Since Y is getting 2:1 from
the pot, his call will have positive EV. The alternative, folding, has
0 EV. Therefore, calling must be correct.
Last, let's consider hands from 11/12 to 1. Y's strategy calls for
folding with these hands. The only alternative strategy Y can employ
is to call instead. Y's EV from calling with hands in this range must
be negative, as he will now lose to X's value bets (2/3) and also lose
to the best of X's bluffs. Since Y is getting only 2:1 from the pot,
his call will have negative EV. The alternative, folding, has 0 EV.
Therefore, folding must be correct.
Case 2: X checks.
When X checks, he has a hand between 1/6 and 11/12. Y's strategy calls
for 3 different actions:
Bet with hands from 0 to 1/3.
Check hands from 1/3 to 5/6.
Bet hands from 5/6 to 1.
If Y bets, X will call with hands from 1/6 to 1/2, per his strategy.
Y's EV from betting is actually pretty easy to calculate if you know
his hand:
Let's look at hands to the left of 1/2 first. As long as Y is < 1/2,
he will never force a better hand to fold by betting. So the following
are true:
EV(check): 0
EV(bet): <wins from getting called by worse hands> - <losses from
getting called by better hands>
EV(bet): (1/2-Y) - (Y-1/6) = 2/3 - 2Y
EV(bet) is greater than 0 as long as Y is < 1/3. So for any hand to
the left of 1/3, Y will do better by betting and for hands to the
right of 1/3, Y will do better by simply checking.
Next let's look at hands to the right of 1/2. When Y is > 1/2, he will
never get called by a worse hand, but might be able to force a better
hand to fold by betting. So the following are true:
EV(check): 0
EV(bet): <wins from forcing a better hand to fold> - <losses from
getting called by better hands>
EV(bet): (Y-1/2) - (1/2-1/6) = Y - 5/6
EV(bet) is greater than 0 as long as Y is > 5/6. So for any hand to
the left of 5/6, betting loses money, but for hands to the right of
5/6, betting has positive EV.
So as you can see, the strategy specified for Y:
(y1 = 1/3, y*_1 = 1/2, y0 = 5/6) cannot be improved by changing
strategies, assuming we hold X's strategy constant.
This, however, is not enough. To conclude that these strategies are
optimal, we must also show that X cannot improve his EV by playing
differently given Y playing an optimal strategy.
This demonstration (along the same lines as the above) is left as an
exercise to the reader (no really! I mean it this time!).
By showing that neither player can improve unilaterally, we show that
the strategies are optimal, and that our "guess" as to the
parameterization is correct.
In the next installment, we'll introduce the "raise-bluff" notation,
which is the last critical piece of notation to be introduced that
will allow us to specify the solutions to these games in a much more
generalized way.
Next: The two-bet finite pot game.