Game theoretic HE question.

This is a question that I've been thinking about for a while. If from a game theoretic point of view one would like to have optimum value betting frequency on the river, how is this problem solved. My initial approach was:


Given the stochastic variable C (Uniformly distributed in [c_l...1]) signifying call-hand strength and the variable B (Again, Uniformly distributed in [b_l...1]) signifying bet-hand strength. I don't consider bluffs in B to simplify things, they seem to get complicated enough as it is. The optimum calling frequency would then be determined by solving the differentiate of:


EV_call = P(C>B)*pot - P(B>C)


So,


d/dc_l EV_call = 0


would yeild the result. Problem 1. My knowledge of probability math is becoming faint so I have trouble solving the above equation. Anyway, let's assume whe have solved it. We'll then probably get a result that is a function of b_l. So let's introduce c_l_opt = f(b_l). Assuming that the opponent knows this function he'll call with hands C_opt in [f(b_l)...1].

The ev for a value bet on the river would then become (not considering the added value of him sometimes folding a better hand):


EV_valuebet = P(B>C_opt) - P(B>C_opt) = 2*P(B>C_opt) - 1


Again, solving:

d/db_l EV_valuebet = 0


should give us the optimum betting frequency for the river.


Has anyone solved this problem? I find when I start to grind it out on paper that the formulae get really big and really horrible really fast so I get stuck after a few calculations.


Another aproach would be to use normal distributions but the interpretation of the results would be kind of fuzzy to me. I mean, you should bet with my = 0.78 and stddev = 0.33 doesn't really make a whole lot of sense to me.


Any math wiz feeling up for the task of solving this? Or is there some better way to reason about it?


Sincerely, Andreas

Frankly, I kind of like the way Norman Zadeh and Mason Malmuth handle these situations in their (respective) books, "Winning Poker Systems", and Gambling Theory and Other Topics.

Ok, I'll give it a shot, but since I don't know much about game theory, I'll do it the other way around:


Hand values:


B in [b,1]

C in [c,1]


Bettor bluff bets for b0>B, checks for b1>B>b0 and value bets for B>b1.

Caller folds for c0>C and calls for C>c0.


EV_bet(B)=pot*p[c0>C]+(pot+1)*p[B>C|C>c0]-p[C>B|C>c0]

EV_check(B)=pot*p[B>C]


EV_call(C)=(pot+1)*p[C>B|b0>B or B>b1]-p[B>C|b0>B or B>b1]

EV_fold(C)=0


Conditions:


(1) EV_bet(b0)=EV_check(b0)

(2) EV_bet(b1)=EV_check(b1)

(3) EV_call(c0)=EV_fold(c0)


assumption: b1>c0>b1


ad (1):


pot*(c0-c)/(1-c)-(1-c0)/(1-c)=pot*(b0-c)/(1-c)


==> b0=c0-(1-c0)/pot


ad (2):


pot*(c0-c)/(1-c)+(pot+1)*(b1-c0)/(1-c)-(1-b1)/(1-c)=pot*(b1-c)/(1-c)


==> b1=(c0+1)/2


ad (3):


(pot+1)*(b0-b)/(b0-b+1-b1)-(1-b1)/(b0-b+1-b1)=0


Note that c0 doesn't appear in this equation: since b1>c0>b1 you either call a bluff and win or you run into a better hand, which got bet for value. c0 is only indirectly determined, since the bettor will adopt his bluffs and value bets to our calling standard.


using (1) and (2):


(pot+1)*(c0-(1-c0)/pot-b)=1-(c0+1)/2


==> c0=(2*b*pot^2+(3+2*b)*pot+2)/(2*pot^2+5*pot+2)


the proof that


[x-c] * EV_fold + integral[x,1] EV_call(C,b0(x),b1(x)) dC


does have a global maximum at x=c0 is left as an exercise to the reader. ;-)


Lets further investigate some special cases:


I. pot->0


we get c0=1 and b=1; for b0, we have to develope c0-(1-c0)/pot into a series and finally get b0=b.


So when there's no extra money in the pot, you only bet with the nuts and only call with the nuts, which is exactly what we would expect.


II. pot->infinity


we get c0=b, b1=(b+1)/2 and b0=b.


So we will value bet the better half of our cards and never bluff, as we know that we will get called by any hand with a non-zero chance to beat us.


III. pot->1


So there is only one big bet in the pot as is the case when the small blind calls the big blind and the hand get checked through until the river.


we get c0=(5+4*b)/9, b0=(1+8*b)/9 and b1=(7+2*b)/9.


So we will bluff with our worst 1/9 of hands, value bet with our top 2/9 of hands and check otherwise. The big blind will call with all hands that will beat at least 5/9 of the small blind's possible hands.


cu


Ignatius

Awesome! I have to read the post a couple of more times and mess around with the equations some.


Then there is always the question as to "top 2/9" really implies in terms of hold'em hands.


Sincerely, Andreas

> Then there is always the question as to "top 2/9"

> really implies in terms of hold'em hands.


On the river, you could do the following: Take your own hand plus all comb(45,2)=990 possible opponent's hands, sort them for hand strengh and map the postions on the interval [0..1]. It is clear that each player will use a slightly different mapping, since each player only knows his own holecards, but that shoudn't really matter, since your strategy is only determined by the value of your own holding and the above procedure should give you the best possible estimate for it.


A much more serious problem with the model is, that it doesn't include the possibility of the caller betting or reraising, so despite having position, being the caller is -EV. Creating an extended model which allows for the caller to play back, should be straightforward, however.


As for real life, the most important conclusion from the model is the correct ratio between value bets and bluffs, to keep your opponent guessing. This ratio is pot+1 : 1.


The above can be applied e.g. when you are on a non-obvious draw. When you have n outs on turn, then the correct frequency of bluffs on the river is n/(46*pot'), where pot' is the pot size after your bet. This is most easily done by selecting n/pot' cards and treat them as bluff-outs, i.e. you bluff when one of these cards comes.


cu


Ignatius