Date: 12/14/1999
From: Jim Geary
Newsgroups: rec.gambling.poker
Subject: Re: HOLDEM: Lost a bet
For Abdul's problem[1],
Given that you're not laying it down under any circumstances,
the rudimentary equation is:
p(you win) ?>= 2*p(trouble)
for you to raise.
Expanding:
p(62s) + p(62o) + p(66) + p(K6o) + p(K6s) + p(A2s) +p(K2s) + p(23s) + p(2x) +p(other crap) ?>= 2*(p(AA) + p(22))
where other crap just means the distribution of hands not listed otherwise.
Crap could mean QQ.
Pre Bayesian consistency testing, this looks like[2]:
2*c(62s) + 4*c(62o) + 3*c(66) + *7c(K6o) + 2*c(K6s) + 1*c(A2s) +1/3*c(K2s) + 2*c(23s) + 76*c(2x) +z*c(crap) ?>=
2* (3*c(AA) + 1*c(22))
where the c stands for consistency shading function[3],
and z is an unknown, as it is difficult for me to quantify
the set of other possible holdings, so for now it is just z.
Now, one must expand the consistency function to read:
c(hand) = d*i(hand) +(1-d)*r(hand)
where d is the coefficient of donutosity
i is the likelihood that the hand was irrationally played thus and
r is the likelihood that the hand was rationally played thus.
This can be rewritten c = d*(i-r)+r. [4]
so,
c(62s) = d*(i(62s) - r(62s)) + r()
c(62o) = d*(i(62o) - r(62o)) + r()
c(66) = d*(i(66) - r(66) ) + r()
c(K6o) = d*(i(K6o) - r(K6o)) + r()
c(K6s) = d*(i(K6s) - r(K6s)) + r()
c(A2s) = d*(i(A2s) - r(A2s)) + r()
c(K2s) = d*(i(K2s) - r(K2s)) + r()
c(23s) = d*(i(23s) - r(23s)) + r()
c(2x) = d*(i(2x) - r(2x) ) + r()
c(crap)= d*(i(crp) - r(crp)) + r()
c(AA) = d*(i(AA) - r(AA) ) + r()
c(22) = d*(i(22) - r(22) ) + r()
Evaulating the terms the best I can[5],
c(62s) = d*((.4) - (.02) ) + .02 [6]
c(62o) = d*((.3) - (.01) ) + .01
c(66) = d*((.5) - (.02) ) + .02
c(K6o) = d*((.4) - (.01) ) + .01
c(K6s) = d*((.3) - (.01) ) + .01
c(A2s) = d*((.3) - (.5) ) + .5 [7]
c(K2s) = d*((.19) - (.01) ) + .01
c(23s) = d*((.1) - (.05) ) + .05
c(2x) = d*((.03) - (.01) ) + .01
c(crap)= d*((0) - (0)) + 0
c(AA) = d*((.4) - (.2) ) + .2
c(22) = d*((.8) - (.3) ) + .3
further reducing to:
c(62s) = .38d + .02
c(62o) = .29d + .01
c(66) = .48d + .02
c(K6o) = .39d + .01
c(K6s) = .29d + .01
c(A2s) = -.2d + .5
c(K2s) = .18d + .01
c(23s) = .05d + .05
c(2x) = .02d + .01
c(crap)= 0
c(AA) = .2d + .2
c(22) = .5d + .2
Substituting into our original inequality query:
{
2*(.38d + .02) +
4*(.29d + .01) +
3*(.48d + .02) +
7*(.39d + .01) +
2*(.29d + .01) +
1*(-.2d + .5) +
1/3*(.19d + .01)
+ 2*(.05d + .05)
+ 76*c(.02d + .01)
+z*(0) [8]
}
?>=
2* (3*(.2d + .2) + 1*(.5d + .2))
or
(.76 + 1.16 + 1.44 + 2.73 + .58 + (-.2) +.6 +.1 + 1.52)*d +
(.04 + .04 + .06 + .07 + .02 + .5 + .003 + .1 + .76) ?>=
2* ((.6d + .5d)+ .8)
or
8.69d + 1.593
>?=
2.2d + 1.6
or
6.49d ?>= .007
or as long as d > .001.
So, if there is a 1/1000 chance that your opponent was thinking of donuts, raise.
If d is zero, it's basically a push.
I haven't gone over the math with a fine tooth comb, but this kinda alines with
my intuition.
JG
[1] "30-60. A known fish, whom I sometimes chase into 100-200, open-limps a three
off the button. I raise with KK in the small blind, big blind reraises, fish cold
calls, I 4-bet (not the cap in Vegas), and they both call.
Flop comes AK2 rainbow. I bet, BB raises, fish cold calls, I reraise, and they
both call.
Turn is a 6. I bet, BB raises, fish cold calls. Now I pause,
partially to try to represent that I might have a hand less than a set of kings,
and partially because I'm trying to figure out how to extract the most money out
of the fish weighed against the possibility that the big blind has AA. I don't
like giving away my hand, but I 3-bet, and they both call.
The river is a 2. I bet, big blind thinks and thinks and thinks and folds, and
the fish raises me. I'm a bit taken aback, since I think I've made my hand
perfectly clear. Her raise says she can beat kings full. I suspect she has
22, though a very slow-played AA is a possibility. I just call. She has 62s.
She got really "lucky" to make a full house. The big blind apparently had K6s,
from his comments and what I thought I saw flash as it went into the muck."
[2] Not enough information given in Abdul's post to guess exactly the suit
distributions, so I arbitrarily assumed it went:
KA2 6 2
SHD C C
Tho with enough analysis, one could probably Bayesianize the identity of the
second deuce's suit, tho that is beyond the scope of the problem.
[3] As mentioned by Abdul in the latest issue of Intelligent Gambler, but
I seem to recall someone else using it in conversation with said author prior. :)
[4] Note that for most rational players, d is a very small number so we reduce
to c = r. This is just the reduced case we call normal poker.
[5] Getting into the mind of the irrational player is tricky business. This
would be the most likely introduction of error.
[6] Note that a lot of the r terms get very small because a rational player
either would not have gotten involved with the hand, or had, would not have
played it in the manner described once they did.
[7] Note that for any term where the right-hand parenthetic quantity is greater
than the left indicates that a rational player was more likely to be there than
an irrational one.
[8] Magically, the unknowable z quantity goes away. How many of you physics
guys saw that coming?
Last Modified 2/9/00