19.3 Game Implications.

On this static model the defender starts with nearly a 85%
probability of having the best hand and still was only
able to call three bets less than 10% of the time. With
barely over 15% winning cards the clairvoyant wins over
40% of the pots.
In dynamic models of real poker hands the defender usually
has less than 50% chance unless she hits the flop big.
I believe even in the dynamic case pot/(pot+bet) calling
frequency is still correct. Provided calling is still
a viable option. Often a new board card is so detrimental
to defender, his linear hand strength(LHS) becomes too low
to consider calling.

His best response is to call if and only
if F(y) > B/(2B + 2).

Where B is the bet and 2 is the pot.
This is from 5.1 pp13-14. Case 1 on page 14.
This is a paper by Thomas Ferguson.
www.math.ucla.edu/~tom/papers/poker1.pdf
The raiser's distribution is often stronger than
the defender's. Often it's the raiser always bets and
the defender always folds.
For the defender it becomes fold when LHS is too low.
Certain hands are 'clear' calls. There's an inbetween
region where defender should sometimes call and other
times fold. Call with pot/(pot+bet) frequency with these
hands. Unlike the [0,1] toy games defender shouldn't
always choose the hands with the highest LHS. LHS only
measures probability of finishing with the best hand at
showdown. He favors drawing hands with potential to be
betting or raising hands. There may be some other vector
which measures chip winning potential better. Call with
the highest pot/(pot+bet) on that chip vector.
The defender is at such a great disadvantage that it may
be wrong to ever cold call an opening raise. He will be
facing a c-bet by opener raiser over 50% of the time and
will get a piece of the flop at most 35% of the time.
There just doesn't seem to be any percentage in calling.

MOP chapter 19

jerrod said,

"Yes, that's right. Here's what happens. (Ignore checking
hands - assume that checking on any street is tantamount
to giving up)

Suppose we have street 3. We know what the bet size is
going to be on street 3, so Y ends up with a distribution
of all his value bets and exactly enough bluffs to make
his bluffing ratio on the river correct. Call his value
bets V and his bluffs B = (alpha)V for the appropriate
alpha = 1/(p+1). If the bets on the first two streets
are both $60 and the initial pot is $10, then we have
a pot size of $250 and a bet size of $60, so alpha is
60/(250+60) = 6/31.

So if we have V value bets, we need 6/31 bluffs. So the
total number of hands we need in order to play the river
correctly is (37/31)V - 31/31 of which are value bets
and 6/31 of which are bluffs.

Now let's look at the 2nd street. The trick here is to
use as your value bets the (37/31)V number and not V
itself. So here we have a pot of $130, a bet of $60,
and 37/31V value bets. alpha is 60/190 = 6/19. So now
we need (6/19)(37/31)V hands to bluff with on the 2nd
street (these hands will be given up on the river). So
we have V(222/589) hands that are 2nd street bluffs only.

Now our total number of hands to play the 2nd street
properly is 222V/589 + 37/31V, or 925V/589.

Now on the first street, we have an alpha of 60/(60+10)
= 6/7. So 6/7 of the 2nd street hands will be bluffs
on the first street. So that's (6/7)(925/589)V that are
bluffs on the first street.

I attached a little Excel file to this post - it shows
the relative frequencies of each strategy for X, and the
value of each of Y's strategies. You can see that using
this method, Y is indifferent between calling any number
of streets and folding immediately, and the value of any
strategy for him is -$19.17.

Y1=(14.29%)($60)+(9.77%)($60)+(7.88%)($60)=$19.16
After weeks still can't see the -$19.17. What's the
payoff for the rest of the space?
Y is guaranteed -$3.46. Y only needs to bet AceKing
and give up all others. Any strategy which produces
less than -$3.46 is dominated.
Didn't understand ($10). Y antes $5. If he loses his
ante, that's only ($5).
X is guaranteed -$5. Just fold every hand on street one.
Therefore the value of this game with both playing optimal
strategy is bounded by -$3.46 for Y and -$5 for X.

To solve this game this matrix was constructed.
Values are net payoffs for Y.

.....\....................X
__Y__\___fold____foldS2___foldS3___call
__V___|____5_____65____125____185
_BBB__|____5_____65____125___-185
_BBCh_|____5_____65___-125___-125
_BCh__|____5____-65____-65____-65
__Ch__|___-5_____-5_____-5_____-5

For each street the ratio of AKs/others(f) were known.
The value betting line was combined with the others.

_VBBB_|____5_____65____125___185(2f3-1)
_VBBCh|____5_____65____125(2f2-1)
_VBCh
__VCh_


fixed bet in series
S1 Y gives 0.71429 ___-5 ___-3.57143
S1 X folds 0.24490 ____5 ____1.22449
S2 X folds 0.04082 ___65 ____2.65306
Y expected value ____________0.30612

$0.30612.
Wasn't able to produce a plus value by over-bluffing method.
Decided to solve geometric growth game.

2. Geometric growth. Y bets 11.66, 38.86, and 129.48.
X calls 10/(10+11.66) of the time on each street.
This was the solution for three one street games in
series.



Finally settled on this tree structure as representative
of the game.



Fewer area values to solve. Fewer roundoff errors.


geometric growth in series
S1 Y gives 0.76333 ___-5 _____-3.81666
S1 X folds 0.12741 ____5 ______0.63704
S2 X folds 0.05882 ___16.66 ___0.97998
S3 matrix 0.05044 ____55.52 ___2.80036
Y expected value ______________0.60073

Y's win frequency for each street.
street one: 2/13
street two: 7/13
street three: 65%

X could improve his results by folding to all bets on
street one. Therefore Y needs to bet over 50% of
his hands on street one.
5y - (1-y)5 > .60073
y > .560073
56.0073% hands to be precise. Y has given up on
44% of his hands.
Y also needs to bet over 50% on street two.
16.66y - (1-y)16.66 > .60073 + 2.19964 - 1.50755
y > .5388
53.88%(30.174% of the original) the remaining hands
are bet on street two. Y gives up on another 25.8%
of his hands.
Only street three was treated as a one street
matrix game.


geometric growth over-bluffing
S1 Y gives 0.43993 ___-5 _____-2.19964
S1 Xfolds 0.30151 _____5 ______1.50755
S2 Ygives 0.11925 ___-16.66 __-1.98682
S2 Xfolds 0.07500 ____16.66 ___1.24954
S3 matrix 0.06432 ____31.56 ___2.03009
Y expected value ______________0.60073

On each street Y gives up a portion of his hands.
Y's win frequency and remaining hands for each street.
street one: 2/13, 100%
street two: 27.5%, 56%
street three: 51%, 30.2%

Just like the two street games, both the in series
method and the over-bluffing method produced the
same results. The $0.60 for Y is within the
bounds of -$3.46 and $5.
Y overbluffed both on streets one and two. Nearly three
times as many bluffs as value bets on street one. Nearly
as many bluffs as value bets on street two.
Y started with 2/13 winning frequency. X knows Y is
overbluffing on streets one and two. Still X is not
able to hold Y under parity.

<>Only Y, the clairvoyant, is allowed to bet in this
toy game. Dealt only about 15% winners, Y was able to
reverse -$3.46 equity into +$0.60 equity. This shows
the advantage of the bettor over the caller. This game
demonstrates the power of the continuation bet. Even
after being called, it is correct to frequently fire
more continuation bets. Only on the final street of
betting does Y revert to bluffing at the low frequency
of the one street game.