The [0,1] Game: Part 7
Jerrod Ankenman
flipflopping we did about notation. I'll also be gone next week on
business and will likely have little time to work on the next few parts.
But they're coming...]
The [0,1] game and Derivatives: A Study of Some Poker-Like Games
Bill Chen/Jerrod Ankenman
Part 7: The finite pot
This is part 7 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.
However, in real life, our opponents fold. And we fold. And this forms a
critical part of our strategies. In infinite pot games, we bet and
raised in such a way as to make our opponents indifferent to different
actions at different points. As we look at finite pots, we will raise,
bet, and fold to create this same indifference.
We will need new notation to tackle these problems. In order to
establish this notation, we'll work through some very simple examples.
Note: While bet sizes are constant, we will always use the convention
that the bet size = 1 unit in describing the size of the pot.
Game #7a: one bet left, finite pot of size 1. *X checks dark*
First of all, we look at Player Y's strategy options. There are two:
1) Bet, in which case Player X can either call or fold.
2) Check, in which case the betting ends and there is a showdown.
We also know that Player Y can bet good hands (for value), as well as
bad hands (as a bluff).
Let's graph these regions along our familiar number line: (numbers are
of course not to scale)
|----------|---------------------|------------|
0 bet check bluff 1
Of course, we have a name already for the boundary between betting and
checking; that's y1.
What about the other threshold point? This is the threshold between
checking and bluffing. We're going to call this spot y*_0, or just plain
y0 for short.
y1 y0
|----------|---------------------|------------|
0 1
Now, what about X's strategy? The only decision X has is whether to call
Y's bet or to fold. We need a new notation for this point as well.
-----
Definition: u*_n, where n is an integer, and u is either x or y, is
defined as the threshold point between calling the nth bet and taking
the next weaker action (in this case folding) to the nth bet.
-----
Now as you can see, y*_0 or y0 is a special case of this more general
definition. The reason for this will become clear when we add raising.
y1 x*_1 y0
|----------|----------|----------|------------|
0 1
So let's summarize how this game plays:
The pot size is 1.
X checks dark (per the game formulation).
If Y is to the left of y1 (remember it's still lowball!) or Y is to the
right of y0, Y bets.
If Y bets, X calls with hands to the left of x*_1. Otherwise X folds.
Next we'll solve for the optimal strategies.
As usual, we'll write indifference equations at each point:
At y1, X will try to make Y indifferent between betting and checking:
(results are for Y)
X's hand Y checks Y bets
[0,y1] 0 -1
[y1,x*_1] 0 +1
[x*_1,1] 0 0
At y1, Y loses a bet by betting when X is to the left of y1, but gains a
bet when X is between y1 and x*_1. To the right of that point, X folds
but would have lost anyway.
The indifference equation:
y1 = (x*_1 - y1)
2y1 = x*_1 [1]
At x*_1, Y will try to make X indifferent between calling and folding:
Let's take a minute to think about something. We know from our poker
experience that Y's optimal bluffing ratio is going to be based on the
size of the pot - Sklansky's Theory of Poker is where many of us read
that for the first time. What is the ratio in this case? With a one unit
pot, it should be 1 to the (size of the pot + 1), or 1/2, right?
Y's hand X calls X folds
[0,y1] +1 0
[y1,y0] <y doesn't bet>
[y0,1] -1 +1 (pot)
y1 - (1 - y0) = 1 - y0
y1 = 2 - 2y0 [2]
Now let's solve for y0, the bluffing threshold.
2y0 = 2 - y1
y0 = 1 - y1/2
So the size of the bluffing region (1-y0) is actually y1/2. So we have a
match.
At y0, X will try to make Y indifferent to bluffing or checking.
Sklansky's Theory of Poker also said that Player X's calling frequency
should be (size of the pot)/(size of the pot + 1) of the hands Player X
holds than can beat a bluff. With a pot size of 1, that ratio is also
1/2.
X's hand Y checks Y bets
[0,x*_1] 0 -1
[x*_1,y0] 0 1 (pot)
[y0,1] 0 0
By bluffing at y0, Y loses a bet when X calls, but wins the pot when X
folds a hand better than y0.
x*_1 = y0 - x*_1
x*_1 = y0/2 [3]
y0 is the total number of X's hands that can beat a bluff. He calls with
half of them. Another match.
So we have three equations and three unknowns:
2y1 = x*_1 [1]
y1 = 2 - 2y0 [2]
x*_1 = y0/2 [3]
Solving:
2y1 = y0/2
2(2-2y0) = y0/2
4 - 4y0 = y0/2
8 - 8y0 = y0
y0 = 8/9
y1 = 2 - 2*(8/9) = 2/9
x*_1 = 2*(2/9) = 4/9
So that's our solution to this game: Y bets a total of 1/3 of the hands,
bluffing 1/3 of those. X calls with 4/9 of his hands. The ex-showdown
value of the game can be found as follows:
Y wins a bet when:
Y bluffs successfully. (half a bet when X is in Y's bluffing range)
Y bets a hand on [0,y1] and X calls on [y1,x*_1]
Y loses a bet when:
Y bluffs unsuccessfully.
Y will bluff 1/9 of the time. Of that time, he will lose a bet 4/9 of
the time, win the pot 4/9 of the time, and win a half a bet 1/9 of the
time.
(1/2)*(1/81) = 1/162
Y will value bet 2/9 of the time and get called 2/9 of that time.
(2/9) * (2/9) = 4/81
So Y's total edge is 9/162, or 1/18.
Throughout that whole solution, I marked all the places where a player's
win was +1 from the pot. So now let's go back and solve this game for an
arbitrary sized pot. We'll call the size of the pot....you guessed it,
p.
Game #7: finite pot of arbitrary size, one bet left, X checks dark.
I'm going to gloss over a little of the algebra here; the reader should
be able to follow what's happening (we're just substituting p for 1
where it's appropriate).
2y1 = x*_1 [1]
y1 - (1 - y0) = p(1 - y0)
y1 = (p+1)(1 - y0)
y1 - p - 1 = (p+1)(-y0)
(p+1-y1)/(p+1) = y0
1 - y1/(p+1) = y0 [2]
x*_1 = p(y0 - x*_1)
x*_1 + px*_1 = py0
(p+1)(x*_1) = py0
x*_1 = py0/(p+1) [3]
Notice [2] and [3] are the generalized forms of the bluffing and calling
ratios.
Solving these equations:
2y1 = py0/(p+1)
2y1(p+1) = p(1-(y1/(p+1)))
2y1(p+1) = p - (py1/(p+1))
2y1(p+1)^2 = p(p+1) - py1
2y1(p+1)^2 + py1 = p(p+1)
y1(2(p+1)^2 + p) = p(p+1)
y1(2p^2 + 4p + 4 + p) = p(p+1)
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))
And there you have it. An optimal solution to this simple one street
game for arbitrary pot sizes.
The calculation of the ex-showdown value of this game is left as an
exercise to the reader. This was a greatly simplified introduction to
the fairly complicated class of finite pot games. Next time we'll let
both players bet.
Next: The finite pot (continued).