The [0,1] Game: Part 6
Jerrod Ankenman
Bill Chen/Jerrod Ankenman
Part 6: The value of the full infinite pot game
Recap to this point: In parts 1-5 we have solved a number of different
games with infinite pots, finding optimal strategies and the values of
games. In Part 6, the last part dealing with one-street infinite pot
games, we find the value of the infinite pot game with check-raises and
infinite raises allowed, and provide a look at what's coming.
Calculating the value of this game:
Once again we're going to deal with the ex-showdown value of this game.
Calculating this value straightforwardly (as we did when no check-raise
was allowed) is rather difficult; what we're going to do instead is take
advantage of a property of optimal strategies that makes the
calculations much simpler.
Remember, our game looks like this:
x2 x1
|----|----|----|----|---------|---------|
0 y3 y2 y1 1
Both X's strategy and Y's strategy are optimal and designed to make the
opponent indifferent at threshold points. Neither X nor Y can improve
their EV by changing their strategies.
But let's investigate what happens if one of them changes their strategy
unilaterally in terms of EV, while holding the other strategy constant.
Holding Y's strategy constant:
Let's move x1 around between y2 and y1. We only need to be concerned
with what the difference in expectation is. It's true that if X moves
his strategy, Y will be able to exploit the resulting strategy. But Y
isn't going to do that here, since Y's strategy is fixed.
What's going to happen if x1 moves between y2 and y1?
X's hand Y's hand What happens
[y2,x1] [0,y2] X loses a bet
[y2,x1] [y2,y1] same value/1 bet in the pot either way
[y2,x1] [y1,1] X wins a bet
[x1,y1] [0,y2] X saves a bet
[x1,y1] [y2,y1] same value/1 bet in the pot
[x1,y1] [y1,1] X wins a bet
Let's sum these up to get a formula for X's EV:
-1*(y2-x1)*(y2) + (y2-x1)*(1-y1) + (y1-x1)*(y2) - (1-y1)*(y1-x1)
-y2^2 + x1y2 + y2 - x1 - y1y2 + y1x1 + y1y2 - y2x1 - y1 + y1^2 + x1 -
y1x1
-y2^2 + y2 - y1 + y1^2
y1^2 - y1 - (y2^2 - y2)
There are no x1 terms left in this formula. This means that X's EV is
actually independent of where x1 is located, as long as y1 and y2 are
constant and x1 is between them.
This is actually true for all the intervals [y_n, y_(n+1)]. We can use
this fact to help us solve the ex-showdown value of the game.
Let's do the following: For odd n, move x_n to y_n. For even n, move x_n
to y_(n+1). So now our graph looks like this:
x2/x3 x1
|----|----|----|----|---------|---------|
0 y3 y2 y1 1
Notice that X never check-raises in this transformed game. This makes it
much easier to calculate the value of the game. It will be the same
value as the optimal strategies, even though X is not employing an
optimal strategy.
As an aside, this is a very intriguing feature of these optimal
strategies - what happens often when using optimal strategies is that if
the opponent plays close to optimally, often the EV of their slightly
suboptimal strategies is the same as it would be if they played strictly
optimally. However, as they stray further and further from optimal, the
amount of mistakes that generate EV grow significantly.
The following 4 scenarios exist:
X's hand Y's hand X's result Probability
[y1,1] [y1,1] no bets (1-y1)^2
[0,y1] [y1,1] +1 (1-y1)*y1
[y1,1] [0,y1] -1 (1-y1)*y1
[0,y1] [0,y1] special y1^2
Within the region marked "special," we can look at the region [0,y1].
But we'll renormalize the interval to [0,1]. So now all of our threshold
points (y2,y3) are multiples of r.
X's hand Y's hand X's result Probability
[0,r] [0,r] special* r^2
[0,r] [r,1] 0 r(1-r)
[r,1] [0,r] -1 r(1-r)
[r,1] [r,1] 0 (1-r)^2
* two bets go into the pot, and we play the game again. Since both
players have the same range of hands, their equity in the two bets is
equal.
If g is X's EV for this game, then in the second special* region, he'll
have -g expectation.
So:
g = -gr^2 - r(1-r)
g + gr^2 = -r(1-r)
g(1+r^2) = -r(1-r)
remember that r^2 = 1 - 2r, so:
g(2-2r) = -r(1-r)
g = -r/2
So X loses r/2 bets in the "special" scenario. Going back to the main
game, we find that the special scenario is the only one that generates
EV for the game, as the middle regions cancel out. So we have:
EV(X) = (-r/2) * y1^2
From the solution to the game, we have y1 = r/(1-r) = 1/sqrt(2).
So EV(X) = -r/4.
Y's advantage in the infinite-bet, check-raise allowed game is r/4.
(about .1035)
Quite a startlingly simple and elegant result.
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)
inf bets, check-raise: r/4 (.1035)
This seems like as good a time as any to recap briefly. We began with
the simplest example of any interest at all and then defined and refined
some methodologies that we used to study a number of [0,1] games - from
one bet remaining to a full headsup game with check-raising and
unlimited raises allowed. In doing so, we ran into "r", the raise ratio,
a constant that we see again and again in this work. We saw the
relatively low value of check-raising when faced with an optimal
opponent.
These infinite-pot strategies are not in fact useless or confined to
theory. Frequently in real-life poker we are faced with pots that are
sufficiently large, for example, that we simply make up our mind that we
will never fold on the river; in these circumstances playing
infinite-pot strategies is an approximation to optimal play, given our
resolution not to fold. Of course, we would caution the reader about
trying to draw too easy a parallel between these [0,1] games and the
discrete, noisy games of poker, particularly holdem.
But for now, this will conclude our studies of infinite pot, one-street
[0,1] games.
Just as a teaser about things to come, future installments in this
series will contain information on the following topics:
- Multi-street games. (Surprisingly, some multi-street games are far
more trivial than we thought at first; but moving beyond simple examples
leads to the complexity we expected)
- High/low games. (Cartesian coordinates are so much more fun to draw
than number lines.)
- Pot-limit and no-limit, including bet sizing. (There are situations
where it's optimal to bet amounts that GIVE YOUR HAND AWAY.)
But in our next series, we will take the results we have generated in
parts 1-6 and generalize them to finite pots. At last, games where our
opponents can fold. (Those of you who have played with me know how
relieved I am at that fact...)
Next: The finite pot, bluffing, and Game #7.
jacksup
But this time I'm confused.
Here are my questions...
> The following 4 scenarios exist:
>
> X's hand Y's hand X's result Probability
> [y1,1] [y1,1] no bets (1-y1)^2
> [0,y1] [y1,1] +1 (1-y1)*y1
> [y1,1] [0,y1] -1 (1-y1)*y1
> [0,y1] [0,y1] special y1^2
Fine.
> Within the region marked "special," we can look at the region [0,y1].
> But we'll renormalize the interval to [0,1]. So now all of our threshold
> points (y2,y3) are multiples of r.
>
> X's hand Y's hand X's result Probability
> [0,r] [0,r] special* r^2
> [0,r] [r,1] 0 r(1-r)
Shouldn't X's result here be 1? (X bets and Y calls with a weaker hand.)
> [r,1] [0,r] -1 r(1-r)
> [r,1] [r,1] 0 (1-r)^2
>
> * two bets go into the pot, and we play the game again. Since both
> players have the same range of hands, their equity in the two bets is
> equal.
>
> If g is X's EV for this game, then in the second special* region, he'll
> have -g expectation.
Because in that special region, it's just the same game with X acting second, right?
> So:
>
> g = -gr^2 - r(1-r)
> g + gr^2 = -r(1-r)
> g(1+r^2) = -r(1-r)
This will make sense to me after my first question above is answered.
Thanks in advance,
Matt
Jerrod Ankenman
>
> > Within the region marked "special," we can look at the region [0,y1].
> > But we'll renormalize the interval to [0,1]. So now all of our threshold
> > points (y2,y3) are multiples of r.
> >
> > X's hand Y's hand X's result Probability
> > [0,r] [0,r] special* r^2
> > [0,r] [r,1] 0 r(1-r)
>
> Shouldn't X's result here be 1? (X bets and Y calls with a weaker hand.)
Sorry, I see why this isn't clear.
What we're looking at here is the additional bets (past the first one).
We know that since both players are to the left of y1, 1 bet will get
into the pot, and since they have the same distribution of hands, they
have equal equity in that one bet. When X is on [0,r] and Y is on [r,1],
no MORE bets get into the pot, because Y won't raise. But when those
ranges are reversed, Y will raise and win 1 extra bet.
>
> > [r,1] [0,r] -1 r(1-r)
> > [r,1] [r,1] 0 (1-r)^2
> >
> > * two bets go into the pot, and we play the game again. Since both
> > players have the same range of hands, their equity in the two bets is
> > equal.
This should say one bet, and the game should be reversed.
> >
> > If g is X's EV for this game, then in the second special* region, he'll
> > have -g expectation.
>
> Because in that special region, it's just the same game with X acting second, right?
Right.