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Old 01-19-2008, 10:17 PM   #1
The Bryce
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Maximizing EV to Find a GTO Betsize

Half-street Game:

- Our opponent can only check and call or check and fold
- Villains calls always lose to our non-bluffs, and always beat our bluffs
- Out opponent's entire range is bluff-catchers
- The bet size and pot size are known

In this game our GTO bluffing frequency is determined by looking at the price we are laying when we bet. For example, if we are laying 2:1 we would bluff 33% of the time, as any deviation of that would be exploitable.

a=s/(p+2s)

a: the amount our opponent needs to win the pot based on the price we are laying in order for our distribution to be GTOl*
s: bet size
p: pot size

Note that this isn't the same "a" that Chen uses in the AKQ betsizing game, though the equation looks similar.

In any case, I wanted to jazz this up a bit and say that n% of my opponent's range beats my non-bluff range, but my bluff range beats his whole range o% of the time (this is a clunky attempt to simulate the equity of a semi-bluff). We know that the amount we're bluffing and lose plus the amount we're not bluffing and lose must equal a%, so:

o:The % our opponent wins when he calls and we're bluffing.
n: The % of our opponent's range that beats our non-bluffing range.

b*o+(1-b)n=a
b*o+(1-b)n=s/(p+2s)

My lack of a formal math education catches up with me here, as I do know how to simplify that to b= (or perhaps it can't be done).

The other thing I was curious about, was that if the bet size was unknown and you could simplify that expression to b= whether you would be able to write up an EV calc that took into account all the permutations of what you would win or lose with bet size "s" and then maximize "s" for the highest possible EV output (which you normally can't do, I think, since the amount we bluff (b) and the amount we bet (s) are co-dependent, but if we can substitute the other side of b= for "b" then it should work).

Anyways, to those who actually understand how to do this stuff: thanks for humoring me
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Old 01-20-2008, 03:09 AM   #2
schubes
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Re: Maximizing EV to Find a GTO Betsize

I can't make any sense of your equation or definition of a, but I'll show you how to find the optimal bet size (ignoring the semibluff aspect for now).

For simplicity I'll say p=1, we can always change the scale of our answers, so there's no loss of generality.

Villain wants to call enough to make Hero indifferent to bluffing and checking his bluffing hands. If you solve the indifference equation you'll find he'll call with 1/(s+1) of hands, including his nut hands.

So hero gets no EV from his bluffing hands and his (ex-showdown) EV for his value bets is:

EV = n*-s + (1/(s+1) - n)*s = s/(s+1) -2ns

To find the max EV we set the derivative equal to zero...

dEV/ds = 1/(s+1) - s/(s+1)^2 -2n = 0 --> (s+1)^2 - 1/2n = 0

So the optimal bet size is s = -1 + (2n)^-.5

For n >= 1/2, the optimal bet size is zero as we'd expect
if n=2/9, it's half the pot
if n=1/8, it's pot size etc.
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Old 01-20-2008, 04:07 PM   #3
The Bryce
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Re: Maximizing EV to Find a GTO Betsize

Ah, very good. I'm kind of doing the tour de force of mucking everything up in trying to get familiar with the higher end math stuff, but that's good. Let me just talk through the unsimplified equation first so that I can make sure I'm understanding it correctly:

EV = n*-s

The amount we lose when our opponent calls us with the nuts...

+ (1/(s+1) - n)*s

... plus the amount we win a bet when our opponent calls us without the nuts equals our EV.

I was a little stuck in trying to figure out why were weren't taking into account how often we won the pot by bluffing, but then I realized that if we're indifferent to bluffing the EV on our bluffs is 0. Am I correct in my line of thinking there?

So if were going to add equity into this what that does is change our point of indifference when bluffing. Say we are betting 1 unit into a pot of 2 units, in a nuts or bluff scenario our point of indifference when bluffing is if our opponent calls 33% of the time. If we only lose the pot 85% of the time when bluffing (15% equity), however, our point of indifference becomes 0.33/.85=0.388. So we could represent this in the equation:

1/(s+1)/o

Where o is the % our opponent wins when we're bluffing.

Now, I know that this is a pretty clunky solution, since if our opponent's range is split into something like the nuts and bluff catchers the amount we're bluffing affects our average equity with our bluffs when called and vice versa, but for the sake of functionality it actually would be fairly reasonable, since you would only have to tweak the numbers back and forth a few times until you were fairly close to on the money.

What we could further do, if the above is true, is change n in the equity equation so that it represents our opponent's average equity when calling us.

Am I on track so far?
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Old 01-20-2008, 05:07 PM   #4
schubes
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Re: Maximizing EV to Find a GTO Betsize

Quote:
Originally Posted by The Bryce View Post
I was a little stuck in trying to figure out why were weren't taking into account how often we won the pot by bluffing, but then I realized that if we're indifferent to bluffing the EV on our bluffs is 0. Am I correct in my line of thinking there?
Yes.

Quote:
Originally Posted by The Bryce View Post
So if were going to add equity into this what that does is change our point of indifference when bluffing. Say we are betting 1 unit into a pot of 2 units, in a nuts or bluff scenario our point of indifference when bluffing is if our opponent calls 33% of the time. If we only lose the pot 85% of the time when bluffing (15% equity), however, our point of indifference becomes 0.33/.85=0.388. So we could represent this in the equation:

1/(s+1)/o

Where o is the % our opponent wins when we're bluffing.
Well, it's an easy mistake to make but the opponent would call 2/3 of the time not 1/3. More importantly, I think you're jumping ahead too fast here. 1/(s+1) came from making us indifferent to betting with our bluffing hands. The semibluff aspect means that our bluffs lose less when they are called, but it also means that we win less when our opponent folds compared to checking, since we can check and still have some equity in the pot. You really should start with the indifference equation.

I'll say o is the chance our semibluff draw misses, c is chance opponent calls.

ex-showdown EV of bluff = c*o*-s + c*(1-o)*s + (1-c)*o*1

The last term is the probability opponent does not call, times chance we wouldn't win a showdown, times the pot size 1.

Setting this equal to zero and solving for c we get:

c = 1/((1-w)s + 1)

where w is the fraction of hits to misses with our semibluff. (w = (1-o)/o)

You can also set up an indifference equation for villain calling and folding with his bluff-catchers to find out how often hero bluffs. You'll find he'll bluff more often when his bluffs have equity, as you'd expect.
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Old 03-03-2008, 06:55 AM   #5
Alix
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Re: Maximizing EV to Find a GTO Betsize

This reminded me of something. I searched into the news archives and found this :

http://groups.google.fr/group/rec.ga...97e1d30730d623

12 years ago, but I do find the same result for optimal pot size when there is a risk of running into the nuts.
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Old 03-05-2008, 02:44 AM   #6
stealthcow
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Re: Maximizing EV to Find a GTO Betsize

Hey guys,

A few questions. First I ordered the mathematics of poker but it hasn't arrived yet. Can someone explain to me the "indifference equation and how it goes to 1/(1+S) ? Shouldn't it be 1/(1+2s), where 1 = the pot size?

2nd, I tried to break both the c/cer (player 1) and the bettor (player 2) into their playing ranges. We assume that at this point of time player 2 has already bet.

P = current pot size on the river
S = Bet size

For the check caller:
A - range where he beats everything
C - He beats our bluffs, loses to our legit hands
E - Loses to our bluffs/ everything

For the range of the Bettor
B - Loses to A, beats C, beats E
D - Loses to C, beats E

So of course, this notation is such that A > B > C > D > E

Also note B + D = 1.

So the EV of the guy calling is...

B(A(P+S) - CS - ES) + D(A(P+S) + C(P+S) - E(S))
now sub in D = 1-b and begin to simplify
B(AP + AS - CS - ES) + (1-B)(AP + AS + CP + CS - ES)
BAP + BAS - BCS - BES + (AP + AS + CP + CS - ES) - (BAP + BAS + BCP +BCS - BES)
BAP + BAS - BCS - BES + AP + AS + CP + CS - ES - BAP - BAS - BCP - BCS + BES
we can eliminate some terms
AP + AS + CP + CS - ES - BCP - 2BCS
now combine terms with P and those with S in it
P(A+C-BC) + S(A+C-E-2BC)

I think this is all sound, although i'm a bit confused on what A+C+E should add up to. If its supposed to add up to less than 1, then I think this is okay, if it should add up to 1, then i'm confused. Because from here i'd say that when we bet our ev is

(% vill folds)*pot size - (1- % vill folds)*(Caller's EV)

If villain folding is just (1-A-C-E), then to solve this we just need to say that
P(1-A-C-E) - (A+C+E)(P(A+C+BC) + S(A+C-E-2BC)) = 0.

From here you can get the value of S/P, given you know A,B,C and E

i'm pretty 50/50 on if i made a mistake somewhere in here. Please let me know if i did, as i'd like to learn how to do these calculations and improve on them.

thanks,

stealthcow-
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Old 03-05-2008, 10:00 PM   #7
ImsaKidd
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Re: Maximizing EV to Find a GTO Betsize

Quote:
Originally Posted by The Bryce View Post

b*o+(1-b)n=a


My lack of a formal math education catches up with me here, as I do know how to simplify that to b= (or perhaps it can't be done).
b*o+(1-b)n=a

b*o+1*n-b*n=a

b*o-b*a=a-n

b(o-a)=a-n

b= (a-n)/(o-a)

Is that what you were looking for?
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