The [0,1] Game: Part 4
Jerrod Ankenman
The [0,1] game and Derivatives: A Study of Some Poker-Like Games
Bill Chen/Jerrod Ankenman
Part 4: Check-raising and Game #4
This is part 4 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: In Parts 1 and 2, we analyzed a very simple [0,1]
game where the pot is infinite, but only one or two bets are allowed. We
found intuitively what the optimal strategy and value of the game were.
Then we created a method using indifference equations to solve the same
problems and introduced notation to help us solve more difficult
problems in a little harder way, and got the same answers. In Part 3, we
generalized to the case where unlimited bets were allowed. We discovered
that a single ratio governs all the raising after the first bet, and we
found its value and called it "r". We also solved the game with infinite
bets and found its value.
In this part: Back to simpler games, but we introduce check-raising.
Thus far, we have stipulated that check-raising was not allowed. As a
result, Y could value bet with impunity and X was forced to value bet
all his good hands. However, in poker, X can check to induce Y to bet
and then raise him. We add this feature to the game we analyzed in Part
#2 now.
Game #4: infinite pot, two bets allowed. Check-raise is allowed.
As usual, we'll draw our graph. This time, however, we see the
appearance of x2 for the first time, since we are now able to have a
scenario where X can put in the 2nd bet. What we'll find is that from 0
to x2, X will checkraise, from x2 to x1, X will bet, and from x1 to 1 X
will check-call. This is consistent with the way we've treated these
points previously.
x2 x1
|-----|----|---------|----------|----------|
0 y2 y1 1
There is, of course, no x3 or y3, as only two bets are allowed.
Another question that might come up is: "Why is x2 located to the left
of y2? Why couldn't it be somewhere else?"
The answer to this question is found in the definition of y2 - y2 is the
point where Y will raise a bet by X. If y2 were to the left of x2, there
would be no bet by X for y to raise, and so then y2 would be forced to
the same point as x2 so that it has some meaning.
But then the next question is: "Why does X check-raise with his best
hands, instead of betting his best hands and check-raising some lesser
hands?"
This is an important question. The answer is, honestly, that X doesn't
actually have to raise his best hands. In fact, what you'll find if you
think about it is: The only time that the rank of the hands that X
check-raises matters is when Y holds a hand between 0 and y2. X wins
when Y is on [x2,y2] and loses when Y is on [0,x2]. But it really
doesn't matter, does it? If X checkraises, Y will bet and two bets will
get in the pot. If X bets, then Y will raise anyway, so the same two
bets get into the pot either way.
So what we actually find is that it's only strictly important that X
check-raise SOME set of hands in the range [0,x2]. You can actually work
out the math and find that solving for the size of the interval X
check-raises on works out the same as assuming that X check-raises his
best hands. So we'll just do that, because it makes the algebra simpler.
Our first step is to write indifference equations.
First of all, let's write the equation for y1. At y1, player X wants to
make player Y indifferent to betting or checking, just like last time.
Now this equation has changed!
The values of actions at y1:
X's hand Bet Chk Diff
[0,x2] -2 0 -2
[x1,y1] -1 0 -1
[y1,1] 1 0 1
So our indifference equation is:
2x2 + y1 - x1 = 1 - y1
2y1 = x1 - 2x2 + 1 [1]
Next, let's write the equation for x1. Player Y wants to make player X
indifferent to betting or checking at x1.
The values of actions at x1:
Y's hand Bet Chk Diff
[0,y2] -1 -2 -1
[y1,1] 1 0 1
(note: there are other ranges with values, but they're the same, check
or bet)
So the indifference equation is:
y2 - 0 = 1 - y1 or
y2 = 1 - y1 [2]
Next, we'll write an indifference equation at y2. Player X wants to make
it indifferent for Y to call or raise at y2.
Values of actions of y2:
X's hand Call Rai Diff
[x2,y2] -1 -2 -1
[y2,x1] 1 2 1
The indifference equation, then is:
y2 - x2 = x1 - y2 or
2y2 = x2 + x1 [3]
Finally, we'll write an indifference equation at x2. Player Y wants to
make it indifferent for X to check-raise or bet at x2.
Values of actions at x2:
Y's hand Bet CR Diff
[y2,y1] 1 2 1
[y1,1] 1 0 -1
The indifference equation, then is:
y1 - y2 = 1 -y1
2y1 = y2 + 1 [4]
We've got four equations now, with four unknowns.
2y1 = x1 - 2x2 + 1 [1]
y2 = 1 - y1 [2]
2y2 = x2 + x1 [3]
2y1 = y2 + 1 [4]
Solving these equations:
2y1 = 1 - y1 + 1
y1 = 2/3
y2 = 1 - 2/3 = 1/3
4/3 = x1 - 2x2 + 1
1/3 = x1 - 2x2
2/3 = x1 + x2
1/3 = 3x2
x2 = 1/9
x1 = 5/9
So let's summarize this solution:
X checks from 0 to 1/9, bets from 1/9 to 5/9 and checks from 5/9 to 1.
If X bets, Y raises from 0 to 1/3 and calls the rest of the time.
If X checks, Y bets from 0 to 2/3 and checks from 2/3 to 1.
If Y bets, X raises from 0 to 1/9 and calls from 5/9 to 1.
Straightforward enough.
Now let's calculate the value of this game.
One straightforward way to calculate the value of any of these games is
to simply create a matrix and figure out who wins what where.
X's hand Y's hand XResult Frequency Net
[0,1/9] [0,1/9] 0 1/81 0
[0,1/9] [1/9,1/3] 2 2/81 4/81
[0,1/9] [1/3,5/9] 2 2/81 4/81
[0,1/9] [5/9,2/3] 2 1/81 2/81
[0,1/9] [2/3,1] 0 3/81 0
[1/9,1/3] [0,1/9] -2 2/81 -4/81
[1/9,1/3] [1/9,1/3] 0 4/81 0
[1/9,1/3] [1/3,5/9] 1 4/81 4/81
[1/9,1/3] [5/9,2/3] 1 2/81 2/81
[1/9,1/3] [2/3,1] 1 6/81 6/81
[1/3,5/9] [0,1/9] -2 2/81 -4/81
[1/3,5/9] [1/9,1/3] -2 4/81 -8/81
[1/3,5/9] [1/3,5/9] 0 4/81 0
[1/3,5/9] [5/9,2/3] 1 2/81 2/81
[1/3,5/9] [2/3,1] 1 6/81 6/81
[5/9,2/3] [0,1/9] -1 1/81 -1/81
[5/9,2/3] [1/9,1/3] -1 2/81 -2/81
[5/9,2/3] [1/3,5/9] -1 2/81 -2/81
[5/9,2/3] [5/9,2/3] 0 1/81 0
[5/9,2/3] [2/3,1] 0 3/81 0
[2/3,1] [0,1/9] -1 3/81 -3/81
[2/3,1] [1/9,1/3] -1 6/81 -6/81
[2/3,1] [1/3,5/9] -1 6/81 -6/81
[2/3,1] [5/9,2/3] -1 3/81 -3/81
[2/3,1] [2/3,1] 0 9/81 0
Adding up that column on the right yields -9/81, or -1/9.
That's the value of this game for Y: 1/9.
Compare that to the value of the two bet game without check-raise we
solved in part 2: 1/8. Think about that for a second. This game is just
about the perfect conditions for X to check-raise - Y can never fold, Y
can't reraise. In fact, the game seems set up for X to do much better by
check-raising Y.
And yet the difference in value between the game with check-raise and
the game without it is a paltry 1/72.
Recap of game values (for player Y):
1 bet, no folding: 0
2 bets, no folding: 1/8 (.125)
inf bets, no folding: .11327
2 bets, check-raise: 1/9 (.1111)
Next time, we'll extend our math skills and solve the infinite bet game
again, but this time allow check-raise.
Next: The full infinite pot game, Game #5
T. Pascal
> The [0,1] game and Derivatives: A Study of Some Poker-Like Games
> Bill Chen/Jerrod Ankenman
>
> Part 4: Check-raising and Game #4
>
>
This is the most fascinating and gripping part of the series yet (to
me). In fact, Sklansky notes three criteria to meet before
check-raising, and the first is that your hand must be very, very
good.
In the human poker (aside from these mathematical constructs), the
theory is that you can check your best hands with others to act after
you because you are not in danger of free cards. Also you can check
your worst hands so your opponent will fear a check-raise.
In summary, check-raise some (most? all?) of your very best hands is
the sound way to go.
NWBurbsCouple
sound way to go.
Gee, who knew?
Jerrod Ankenman
"T. Pascal" wrote:
>
> This is the most fascinating and gripping part of the series yet (to
> me). In fact, Sklansky notes three criteria to meet before
> check-raising, and the first is that your hand must be very, very
> good.
>
> In the human poker (aside from these mathematical constructs), the
> theory is that you can check your best hands with others to act after
> you because you are not in danger of free cards. Also you can check
> your worst hands so your opponent will fear a check-raise.
>
> In summary, check-raise some (most? all?) of your very best hands is
> the sound way to go.
When we worked out this result, we were quite shocked at how little
value there was in check-raising. However, in investigations that will
show up later, we've found that the value of the check-raise becomes far
more prominent in at least two cases:
1) When the players have different distributions of hands. Say X gets a
number from [0,10] while Y gets a card from [0,1]; it's correct for X to
check ALL the time and check-raises his good hands.
2) When the distribution of hands for the first player is non-uniform,
particularly when it is more bimodal - that is, let's say that X either
has the nuts (50%) or a hand that can't win. We can show that under some
typical circumstances, practically all the value of the game comes from
X's check-raising.
Part 5 coming soon!
Jerrod Ankenman
Michael Maurer
> hands, instead of betting his best hands and check-raising some lesser
> hands?"
This question anticipates a problem I have seen coming later in your
series. In the no-foldem case, it is reasonable to assume -- and you
also demonstrate mathematically -- that the stronger your hand, the
more bets you will put into the pot. But once you allow folding that
relationship need not hold.
So my confusion is... won't that pull the rug out from under your
approach? You are able to set up N equations in N unknowns in the
no-foldem case because you have *parametrized* the solution. It
happens for the no-foldem case that there is an easy parametrization
having N degrees of freedom that can represent the optimal solution.
You know that somewhere on the interval [0,1] there will be a boundary
between going K and K+1 bets. But why should we expect this to hold
in the general case?
As an example, consider one of the simplest "real number poker" games
that allows folding. This is one of Von Neumann's poker examples.
There are two players, an initial (finite) pot, with one bet max. The
first player is required to check. Now the second player can either
check also (and end the game) or bet; if the second player bets, then
the first player can call or fold (and end the game). The solution is
for the second player to bet his weakest and strongest hands while
checking the medium ones.
So if you are clever enough to guess a parametrization that includes
this possibility, then your N simultaneous equations approach will
work for this problem too. But when you allow more raises, or make
the players' hand distributions non-uniform, it isn't obvious to me
that you will be able to guess the parametrization. Indeed, in the
worst case you would need to try ALL of the possible mappings from the
set of player actions to the set of subintervals. This combinatorial
search is made more difficult by the possibility that some actions may
never be taken.
But I sense that you have already applied your method to such
problems. So I am looking forward to see how you have approached it.
-Michael M
Bill chen
at this point for later in the series. What you have correctly
pointed out actually applies or x_n:
> As an example, consider one of the simplest "real number poker" games
> that allows folding. This is one of Von Neumann's poker examples.
> There are two players, an initial (finite) pot, with one bet max. The
> first player is required to check. Now the second player can either
> check also (and end the game) or bet; if the second player bets, then
> the first player can call or fold (and end the game). The solution is
> for the second player to bet his weakest and strongest hands while
> checking the medium ones.
I claim there are two ranges of hands you want to put in the n-th bet
with, one is with hands that cannot win (i.e. have expectation -1) if
your opponent calls optimally and ones that have positive expectation
if your opponent calls. There is a gap between these two ranges,
which we call the "value bet/raise" interval and the "bluff" interval.
The x_n, y_n in the infinite pot case correspond roughly to the
"value bet/raise" thresholds.
For the finite pot case each x_n will have a corresponding x_n~ for
bluffing and an x_n* as the cutoff for calling/folding after putting
in the nth raise. The number of equations triple but in the end you
still get 3n equations and 3n unknowns and when you get the solution
you can still go through the process to verify your solution is
correct by showing neither side can change its strategy to improve its
expectation. This is a method we use later to prove "persistence"
theorems which show that a seemingly more complex game with many
streets actually maps to a simpler one street game.
Bill
mjma...@yahoo.com (Michael Maurer) wrote in message news:<bb812c4e.0301...@posting.google.com>...
Barbara Yoon
conclusions here, but to make this interesting, but quite complicated
material easier for me to follow, would you kindly re-check a few of
your things below that I've marked by "[*...*]"...?! Thanks...
> Game #4: infinite pot, two bets allowed. Check-raise is allowed.
> ...our graph. ...from 0 to x2, X will checkraise, from x2 to x1,
> X will bet, and from x1 to 1 X will check-call.
>
> x2 x1
> [*0 = best*] |-----|----|---------|----------|----------| [*1 = worst*]
> 0 y2 y1 1
>
> ...indifference equations. At y1, player X wants to make player Y
> indifferent to betting or checking, just like last time.
>
> The values [*to Player Y*] of actions at y1:
> X's hand Bet Chk Diff
> [0,x2] -2 0 -2 [*(-2) - (0) = (-2)*]
> [x1,y1] -1 0 -1 [*(-1) - (0) = (-1)*]
> [y1,1] 1 0 1 [*(1) - (0) = (1)*]
>
> So our indifference equation is:
> 2x2 + y1 - x1 = 1 - y1
> 2y1 = x1 - 2x2 + 1 [1]
>
> Player Y wants to make player X indifferent to betting or checking at x1.
>
> The values [*to Player X*] of actions at x1:
> Y's hand Bet Chk Diff
> [0,y2] -1 -2 -1 [*(-1) - (-2) = (-1)??!!*] [*or*] [(-2) - (-1) = (-1)*]
> [y1,1] 1 0 1 [*(1) - (0) = (1)*]
Bill chen
> Bill Chen / Jerrod Ankenman.....I'm not challenging your bottom-line
> conclusions here, but to make this interesting, but quite complicated
> material easier for me to follow, would you kindly re-check a few of
> your things below that I've marked by "[*...*]"...?! Thanks...
>
>
> > Game #4: infinite pot, two bets allowed. Check-raise is allowed.
> > ...our graph. ...from 0 to x2, X will checkraise, from x2 to x1,
> > X will bet, and from x1 to 1 X will check-call.
> >
> > x2 x1
> > [*0 = best*] |-----|----|---------|----------|----------| [*1 = worst*]
> > 0 y2 y1 1
Yeah 0 is the best hand.
> > ...indifference equations. At y1, player X wants to make player Y
> > indifferent to betting or checking, just like last time.
> >
> > The values [*to Player Y*] of actions at y1:
Right these are player Y values.
> > The values [*to Player X*] of actions at x1:
> > Y's hand Bet Chk Diff
> > [0,y2] -1 -2 -1 [*(-1) - (-2) = (-1)??!!*] [*or*] [(-2) - (-1) = (-1)*]
> > [y1,1] 1 0 1 [*(1) - (0) = (1)*]
> > (note: there are other ranges with values, but they're the same, check or bet)
Yeah this was atypo. Here it should read, the value to Player X for y being in:
Bet Check Diff
[0, y2] -2 -1 -1
[y1, 1] 1 0 1
OK?! :-)
> >
> > So the indifference equation is:
> > y2 - 0 = 1 - y1 or
> > y2 = 1 - y1 [2]
This part is still right.
Bill