Crazy Sklansky Problem
Tom Weideman
regarding game theory applied to a three-handed poker game. I won't go into
the details of this, except to bring up one rather subtle point. I haven't
examined the proposed solution(s) posted there, but there is a pretty subtle
aspect of this problem that may have been missed by those involved. Maybe
someone reading this can find out for us if this is the case. Anyway, I'll
form it as a question...
Three players have "hands" that are real numbers ranging from 0 to 1 (1 is
the nuts). The first player to act has a calling cutoff of "a". That is,
he will call with all hands that have a value of "a" or higher, and fold all
other hands. The second player has a cutoff of "b". I'm not going to go
into the details of "b" depending on whether the first player called or not,
because I want to focus on something simpler and vital to all parts of the
calculation that Sklansky has set into motion.
Let's assume that b > a and that both players have already called (the
second player certainly would not enter the pot with a hand he knows to be
worse than the first player's worst possible hand). You now hold a hand
with a value of x, which happens to be greater than b. What is the
probability that you will win the hand if you call? This is not at all
trivial. I'll provide one hint... It is NOT equal to the product of the
probabilities of beating each player individually (each of which is rather
easy to determine).
Tom Weideman
Tom Weideman
For those of you beating your heads against the wall, I apologize. I wrote
the last paragraph too fast, and it didn't come out right. Here's what it
SHOULD read:
Let's assume that b > a and that both players have already called (the
second player certainly would not enter the pot with a hand he knows to be
worse than the first player's worst possible hand). You now have to decide
on your own cutoff hand (call it "x") which is your minimum calling hand.
Given that you choose cutoff "x", what is your probability of winning the
hand? This is not at all trivial. I'll provide one hint... It is NOT equal
to the product of the probabilities of beating each player individually
(each of which is rather easy to determine).
Again, I apologize if I made anyone question their understanding of Bayes'
Theorem, heh.
Tom Weideman
Jerrod Ankenman
Tom Weideman wrote:
>
> Tom Weideman wrote:
>
> > On a rare lurking trip over at 2+2, I saw a problem posed by Sklansky
> > regarding game theory applied to a three-handed poker game. I won't go into
> > the details of this, except to bring up one rather subtle point. I haven't
> > examined the proposed solution(s) posted there, but there is a pretty subtle
> > aspect of this problem that may have been missed by those involved. Maybe
> > someone reading this can find out for us if this is the case. Anyway, I'll
> > form it as a question...
> >
> > Three players have "hands" that are real numbers ranging from 0 to 1 (1 is
> > the nuts). The first player to act has a calling cutoff of "a". That is,
> > he will call with all hands that have a value of "a" or higher, and fold all
> > other hands. The second player has a cutoff of "b". I'm not going to go
> > into the details of "b" depending on whether the first player called or not,
> > because I want to focus on something simpler and vital to all parts of the
> > calculation that Sklansky has set into motion.
>
> For those of you beating your heads against the wall, I apologize. I wrote
> the last paragraph too fast, and it didn't come out right. Here's what it
> SHOULD read:
>
> Let's assume that b > a and that both players have already called (the
> second player certainly would not enter the pot with a hand he knows to be
> worse than the first player's worst possible hand). You now have to decide
> on your own cutoff hand (call it "x") which is your minimum calling hand.
> Given that you choose cutoff "x", what is your probability of winning the
> hand? This is not at all trivial. I'll provide one hint... It is NOT equal
> to the product of the probabilities of beating each player individually
> (each of which is rather easy to determine).
>
> Again, I apologize if I made anyone question their understanding of Bayes'
> Theorem, heh.
Right. OK. So now let's look at this problem as four different
situations:
Assuming that the hands are uniformly distributed across (0,1]:
Case 1:
Both P1 and P2 hold hands < x.
You win all the time.
Case 2:
P1 holds a hand on [x,1].
P2 holds a hand < x.
You win half the time.
Case 3:
P1 holds a hand < x.
P2 holds a hand on [x,1].
You win half the time again.
Case 4:
P1 and P2 both hold hands on [x,1].
You win a third of the time.
Probabilities of each case occurring:
P(1) = (x-a)/(1-a) * (x-b)/(1-b) <=> (x^2-ax-bx+ab)/(1-a)(1-b)
P(2) = (1-x)/(1-a) * (x-b)/(1-b) <=> (x-x^2+bx-b)/(1-a)(1-b)
P(3) = (x-a)/(1-a) * (1-x)/(1-b) <=> (x-a-x^2+ax)/(1-a-b+ab)
P(4) = (1-x)/(1-a) * (1-x)/(1-b) <=> (1-2x+x^2)^2/(1-a-b+ab)
(you can validate for yourself that these four add up to 1)
So then our overall chance of winning is:
6*(x^2-ax-bx+ab) + 3*(x-x^2+bx-b) + 3*(x-a-x^2+ax) + 2*(1-2x+x^2)
-----------------------------------------------------------------
6*(1-a)(1-b)
2x^2 - 3ax - 3bx + 6ab + 2x - 3b - 3a + 2
-----------------------------------------
6*(1-a)(1-b)
validating: let's say that a = .5, b.=.7, and x=.8
Running it through the above yields a win pct of 55.6%; i ran 10,000,000
trials of random #s, of which 300,000 met the .5/.7/.8 criteria, and got
55.57%.
I have a nagging suspicion that some algebra will reduce that expression
further, but I keep running into a problem involving the lack of ax^2
and bx^2 terms.
Anyway...
Jerrod Ankenman
eleaticus
problem, numerically or algebraicly.
Consider a five card game, standard hand values.
A. 9876 rainbow
B. kkqq
C. 4442
With one card to come those hands fit your definition.
Consider another set that fits your definition:
A. 9876 rainbow
B. 9988
C. 7778
I fail to see how you have sufficiently defined the problem.
To the extent you have defined a problem - non-poker - you have two implied
probabilities, that for b winning over a and c winning over b, but a third
is required and is that of c winning over a. And even that is an
insufficient definition, because b beating a - or vice versa - may condition
the likelihood that c's winner over a is also a winner over b, etc.
Eleaticus
"Tom Weideman" <zwi...@attbi.com> wrote in message
news:BA32574F.228EE%zwi...@attbi.com...
Tom Weideman
> 2x^2 - 3ax - 3bx + 6ab + 2x - 3b - 3a + 2
> -----------------------------------------
> 6*(1-a)(1-b)
>
> validating: let's say that a = .5, b.=.7, and x=.8
>
> Running it through the above yields a win pct of 55.6%; i ran 10,000,000
> trials of random #s, of which 300,000 met the .5/.7/.8 criteria, and got
> 55.57%.
I didn't check the details, but it looks right at a glance. My hint becomes
apparent when you compare with just the parlay of the two heads-up
match-ups:
probability of winning with a hand in region with min cutoff x=0.8 vs. a
hand in region with min cutoff a=0.5 is 4/5
probability of winning with a hand in region with min cutoff x=0.8 vs. a
hand in region with min cutoff b=0.7 is 2/3
The parlay of beating both is: 4/5 * 2/3 = 8/15 = 53.3%.
Bayes comes into play because if your hand happens to beat one opponent's
hand it is now slightly more likely that it will also beat the other
opponent's hand. So the probability of beating both simultaneously is
higher than the parlay probability of beating each individually.
Anyway, I have my doubts that any of the 2+2 solutions incorporated the
messy expression Jerrod derived above into the calculations. I could be
wrong, and I'd be interested if anyone could let me know otherwise. There
are other "complications" with the problem (those that prompt me to call it
"crazy"), but I thought this was as good of a starting place as any.
Tom Weideman
eleaticus
Not only was there nothing in yur definition to specify - for example - that
if the apparent magnitude of c,b,a fit 2*b=4*a that such is the ratio of
probabilities.
And, as I pointed out in the previous (in less direct wording), the idea
that the ratio of hand C's likelihood of beating hand B versus the
likelihood of hand B's beating hand A has anything much/necessary to do with
the likelihood of hand C beating hand A is ludicrous.
Your pedantry is equaled only by your equinity, so to speak, and we're not
talking about the noble horse.
Eleaticus
"Tom Weideman" <zwi...@attbi.com> wrote in message
news:BA3294FB.2291A%zwi...@attbi.com...
Renaud Desferet
integral(a,1)integral(b,1)integral(x,1)(1{w>u}1{w>v})dudvdw
=integral(a,x)integral(b,x)integral(x,1)(1{w>u}1{w>v})dudvdw
+integral(a,x)integral(x,1)integral(x,1)(1{w>u}1{w>v})dudvdw
+integral(x,1)integral(b,x)integral(x,1)(1{w>u}1{w>v})dudvdw
+integral(x,1)integral(x,1)integral(x,1)(1{w>u}1{w>v})dudvdw
=(1-x)*(x-a)*(x-b)+(x-a)*(1-x)^2/2
+(x-b)*(1-x)^2/2+(1-x)^3/3
eleaticus
Eleaticus
"Tom Weideman" <zwi...@attbi.com> wrote in message
news:BA3294FB.2291A%zwi...@attbi.com...
Darryl Parsons
[(u-a)*(u-b)]/[(1-a)*(1-b)]du
I can't be bothered to simplify it, but is this right? It's been
about 12 years since I've solved such problems and I'm wondering if
I've still got it. If so, it'll give me extra confidence when I try
to pick up young chicks at the upcoming New Year's Eve party.
Darryl Parsons
Renaud Desferet
9*integral(a,1)integral(b,1)integral(x,1)(1{w>u}1{w>v})dudvdw
-6*integral(a,1)integral(b,1)integral(x,1)(1{w<max(u,v)})dudvdw
=9*(integral(a,x)integral(b,x)integral(x,1)(1{w>u}1{w>v})dudvdw
+integral(a,x)integral(x,1)integral(x,1)(1{w>u}1{w>v})dudvdw
+integral(x,1)integral(b,x)integral(x,1)(1{w>u}1{w>v})dudvdw
-6*(
integral(a,x)integral(b,x)integral(x,1)(1{w<max(u,v)})dudvdw
+integral(a,x)integral(x,1)integral(x,1)(1{w<max(u,v)})dudvdw
+integral(x,1)integral(b,x)integral(x,1)(1{w<max(u,v)})dudvdw
+integral(x,1)integral(x,1)integral(x,1)(1{w<max(u,v)})dudvdw)
=9*((1-x)*(x-a)*(x-b)+(x-a)*(1-x)^2/2
+(x-b)*(1-x)^2/2+(1-x)^3/3))
-6*((x-a)*(1-x)^2/2+(x-b)*(1-x)^2/2+(1-x)^3-(1-x)^3/3)
When I have rewritten it, it looks a bit different from what I
remember, so I will have to review my calculations once again...my
last results looked so nice though...
Darryl Parsons
know because he is superior to me and I am superior to you. Trying to
pretend otherwise makes you look like a complete fool.
Darryl Parsons
Renaud Desferet
one term in the expectation that I have rewritten for you.
My new solution won't be posted before monday as my Powell
n-dimensional minimization algorithm in order to cancel the
derivatives is at work and my ev calculation details are there also.
I don't see what is crazy with this problem.
Renaud Desferet
Renaud Desferet
Divide by (1-x)(1-a)(1-b) if you want the conditionnal probability you
ask about, but of course when writing the expectation absolute
probabilities are the ones to consider.
Robert Ladd
Robert Ladd
"Renaud Desferet" <renaud_...@yahoo.com> wrote in message
news:925658d.02122...@posting.google.com...
Barbara Yoon
>> Three players have "hands" that are real numbers ranging from 0 to 1
>> (1 is the nuts). The first player to act has a calling cutoff of "a".
>> That is, he will call with all hands that have a value of "a" or higher,
>> and fold all other hands. The second player has a cutoff of "b".
>> (the second player certainly would not enter the pot with a hand he
>> knows to be worse than the first player's worst possible hand).
>> You now have to decide on your own cutoff hand (call it "x") which
>> is your minimum calling hand. Given that you choose cutoff "x",
>> what is your probability of winning the hand? This is not at all trivial.
>> each player individually (each of which is rather easy to determine).
Jerrod Ankenman:
> Right. OK. So now let's look at this problem as four different situations:
> [...calculation details...]
> So then our overall chance of winning is:
>
> 2x^2 - 3ax - 3bx + 6ab + 2x - 3b - 3a + 2
> ---------------------------------------------------
> 6*(1-a)(1-b)
>
> validating: let's say that a = .5, b.=.7, and x=.8
>
> Running it through the above yields a win pct of 55.6%; i ran 10,000,000
> trials of random #s, of which 300,000 met the .5/.7/.8 criteria, and got
> 55.57%. I have a nagging suspicion that some algebra will reduce that
> expression further, but I keep running into a problem involving the lack
> of ax^2 and bx^2 terms. Anyway...
Guys.....I did it by integration, resulting...
1/3 - a/2 - b/2 + ab - x^3/3 + ax^2/2 + bx^2/2 - abx
---------------------------------------------------------------
(1-a)(1-b)(1-x)
...which I've checked no further than to see that it yields the same
5/9 = 55.6% for Jerrod's "a = .5, b.=.7, and x=.8" test case.....OK?!
Bennett Niizawa
Parsons) wrote:
DON'T DRINK AND DERIVE!!!
eleaticus
"Darryl Parsons" <dpar...@mailandnews.com> wrote in message
news:574a08c4.02122...@posting.google.com...
> Oh, be quiet. Tom Weideman is a superior mathematician to you. I
> know because he is superior to me and I am superior to you. Trying to
> pretend otherwise makes you look like a complete fool.
If you mean: "AND you, Darryl, are a pedant-sucking dope" you are right.
Superior math is orthogonal to moronic problem-posing, where you are
incapable of realizing that your problem is insufficiently posed and your
solution doesn't work.
In general not only is one hand's probability of improving dependent on the
particulars of other hands, but also on whether they improve, and neither is
in the equations produced?
This is highlighted by the example pair I provided.
A hell of a lot more than three bare probabilities are required.
The problem as posed (actually, answered; the posing did not specify enough
to justify any kind of 'solution') says a hand has a particular probabilty
of winning.
But more is required. and not just, for instance, (my first example):
the p(a) that a 4-card str8 will improve to a full str8 with one card to
come
the p(b) that two pair will improve to a full boat with one card to come
the p(c) that trips and a kicker will improve to quads or a boat with one
card to come,
but down to more particulars.
These particulars are not solved by interaction terms between the supposed
three basic probabilities.
> One subtle point you have missed is that you have not defined a calculable
> problem, numerically or algebraicly.
>
> Consider a five card game, standard hand values.
>
> A. 9876 rainbow
> B. kkqq
> C. 4442
>
> With one card to come those hands fit your definition.
>
> Consider another set that fits your definition:
>
> A. 9876 rainbow
> B. 9988
> C. 7778
P.S. And then consider the fact that the game still hasn't been fully
definied. The dependencies vary greatly just according to whether it is a
four-card pocket and 1-card board game, or five-cards per player game.
Eleaticus
Simon White
<elea...@bellsouth.net> wrote:
(snip)
>To the extent you have defined a problem - non-poker - you have two implied
>probabilities, that for b winning over a and c winning over b, but a third
>is required and is that of c winning over a. And even that is an
>insufficient definition, because b beating a - or vice versa - may condition
>the likelihood that c's winner over a is also a winner over b, etc.
>
>Eleaticus
My reading of Tom's post was that a hand had a rating x {x in R such
that 0<x<1 } and that a hand with a higher rating beats a hand with a
lower rating. Given [b beats a], [c beats b] implies [c beats a]
surley?
If you want to consider a "real" application, you could consider x to
be the probability that a given hand will win the pot under the
current board, ignoring cases where the pot is split.
--
Simon White
Department of Electronics and Computer Science
Faculty of Engineering
University of Southampton
Mike McClain
"eleaticus" <elea...@bellsouth.net> wrote :
>
> In general not only is one hand's probability of improving dependent on the
> particulars of other hands, but also on whether they improve, and neither is
> in the equations produced?
>
Are you not capable of reading? What is wrong with you?
Tom stated that the 'hands' are real numbers from 0 to 1, and
the highest-valued hand at showdown is the winning hand. There
are no cards, there is no board, there are no draws, there is
no way for a hand to 'improve'.
Just read the damn thing. None of your concerns are at all
applicable.
Mike.
eleaticus
"Simon White" <smw...@ecs.soton.nospam.ac.uk> wrote in message
news:3e0df9c7...@news.zen.co.uk...
> On Sat, 28 Dec 2002 01:00:48 -0600, "eleaticus"
> that 0<x<1 } and that a hand with a higher rating beats a hand with a
> lower rating. Given [b beats a], [c beats b] implies [c beats a]
> surley?
Not only was the problem 'solved' as if the x were independent of the
particulars of the other hands but 'ratio' scale - not a fact in most poker
games with finite decks .
> If you want to consider a "real" application, you could consider x to
> be the probability that a given hand will win the pot under the
> current board, ignoring cases where the pot is split.
Consider? If I have to 'consider' the problem wasn't posed sufficiently.
There is nothing about the concept of 'rating' that allows me to 'consider'
a rating as a probability, even when 0<=x<=1. Consider the USCF/FIDE
ratings; there is nothing about them that doesn't allow us to normalize to
that range in practice.
Current board or future draw? Makes a big difference. In the case of a
one-card board, for instance, one hand's improvement could mean the others
don't or must.
The 'solutions' supplied might be reasonable only where there is an infinite
deck and a non-board game (or an infinite board?) and then only with further
specification?
If you don't know your own assumptions, you don't know what you are doing.
Barbara Yoon
>>
>> 1/3 - a/2 - b/2 + ab - x^3/3 + ax^2/2 + bx^2/2 - abx
>> ---------------------------------------------------------------
>> (1-a)(1-b)(1-x)
>>
>> ...which I've checked no further than to see that it yields the same
>> 5/9 = 55.6% for Jerrod's "a = .5, b.=.7, and x=.8" test case.....OK?!
..."by integration".....I mean the integral from c=x to c=1 of...
(c-a) (c-b) dc
------ x ------ x ------
(1-a) (1-b) (1-x)
.....OK?!
Tom Weideman
Excellent! Part of my belief that you probably hadn't was based on the
amount of work you would be willing to do for this problem. Best of luck
getting to the point where you're certain you have it right. I look forward
to seeing the solution you are happy with.
> I don't see what is crazy with this problem.
I see a couple of things crazy about it. First, it's just a hell of a lot
of work to do at the whim of someone who very likely doesn't have the tools
to do it himself, especially when the solution is not going to be
particularly enlightening for real poker. My hat's off to you for your
generosity with your time. I've been known to work on poker/game theory
problems that are quite complicated, but I've always reserved this hard work
for ideas I have myself and which are likely to help my development of the
theory (which I then rarely, if ever, share with anyone). I'm not inclined
to put in this much work for someone who wants the work named after himself,
is not inclined to do any of the work himself (but implies he could if he
wanted to), and who apparently feels like he's some sort of thesis advisor
doling out grunt work to his lowly graduate students.
As if all of that was not enough, nowhere in Sklansky's description of the
problem does he address the single most troubling aspect of multiplayer
games (and one which is very important to real poker games): The absence of
an equilibrium solution when coalitions could be present. Sure, he asks
questions about collusion where partners know each other's hands, but that's
not what I am talking about. Poker frequently contains situations where
players form implicit (non-cheating) partnerships against a third player,
and this wreaks havoc with game-theoretical treatment of multiplayer poker.
I would expect at least some mention of this in such a problem, if for no
other reason but that those who he sets to work on it would know what
assumptions to make when he says "solve this game".
Tom Weideman
eleaticus
"Mike McClain" <mmcclain_...@omsoft.com> wrote in message
news:cXmP9.26152$BP.13495@fe01...
He also said that b is greater than a and x is greater than b, which makes
your comments as nonsensical as his hemi-posed problem.
Your interpretation - which is reasonable in context of the weevil's
idiocy - gives us a 100% probability that x wins under the circumstances
named.
He stated that x>b>a so obviuosly x wins under your reading of what he said.
You know, that the higher number wins?
Not to mention that in his setup and your interpretation, where b>a, the
introduction of a at all was complete nonsense, was completely immaterial
unless you were playing hi-low!
And to back off his own statement that x>b - thus you don't know x is
greater than b and a - you must then posit the probability distribution of
the range of hand values. There was nothing stated to allow any kind of
inference in that regard from what he said.
For instance a hand could be physically represented as a roll of a pair of
dice, one pair per hand.
One pair has mostly zero faces, one mainly .1, one mainly .2, all other
faces being equal in size and number from one pair to the other (.5 max face
size).
The probability distribution would not be uniform.
Don't you pedantic types ever stop to think? It is one thing to overlook
something to start with, but to be incapable upon repeated prompting is ...
well, proof the world is all a dream. A nightmare!
lol
"Anything and everything that requires or encourages systematic examination
of premises and logic."
Hell, your mottos seem to be "Anything ... to ALWAYS avoid thinking about
premises and logic".
Eleaticus
>
>
> Mike.
>
>
>
eleaticus
> As if all of that was not enough, nowhere in Sklansky's description of the
> problem does he address the single most troubling aspect of multiplayer
> games (and one which is very important to real poker games): The absence
of
> an equilibrium solution when coalitions could be present. Sure, he asks
> questions about collusion where partners know each other's hands, but
that's
> not what I am talking about. Poker frequently contains situations where
> players form implicit (non-cheating) partnerships against a third player,
> and this wreaks havoc with game-theoretical treatment of multiplayer
poker.
> I would expect at least some mention of this in such a problem, if for no
> other reason but that those who he sets to work on it would know what
> assumptions to make when he says "solve this game".
Such situations having been the subject of a number of threads here this
last month.
Mainly the allin/short stack tourney player situation. How small does
another stack have to be where it becomes reasonable to try getting the
other non-allin out of the pot when you aren't sure to win if he folds?
Eleaticus
>
>
> Tom Weideman
>
GEps222
The probability of beating both is the product of the individual probabilities.
So if one is 80% and the other is 60%, the the probability of winning the hand
is 80% x 60% = 48%. That's why my book (The Greatest Book of Poker for
Winners!) has the strategy of raising to reduce the size of the platying field
(RSPF) when you are dealt a high pair. The more opponents staying in, the less
likely your high pair will hold up -- or improve -- to win the pot. Ideally,
you would like to be heads up in such a case.
george epstein
eleaticus
Eleaticus
"GEps222" <gep...@aol.com> wrote in message
news:20021228190616...@mb-mu.aol.com...
Tom Weideman
Heh, I had to actually check at Amazon.com to see if this was for real or if
it was a joke. Amazingly, it is for real. Good luck with the book sales,
George. If the exclamation point in the title and the snappy little acronym
"RSPF" aren't enough to drive up sales by themselves, I'm sure this post of
yours will help. It pretty much says it all.
Tom Weideman
Dsklansky
example where the first two players will call with the top half of their hands
and the third player needs a hand in the top 25% to overcall. If they all call,
the third player has a 3/4 chance of beating the first player (because the
first player will be in the same top 25% and have an even chance half the
time). He also has a 3/4 chance of beating the sceond player. But his chances
of beating them both are not 9/16 because if he beats one, he is more than 3/4
to beat the other.
His actual chances are the sum of four possibilites:
1. Neither of the first two players have a hand in the top 25%. That is 25/50 x
25/50 or 1/4.
2. Both players, like him are in the top 25% but he wins anyway. Thats 25/50 x
25/50 x1/3 or 1/12
3. The first player is in the top 25%, the second player isn't, and the third
player beats the first. That's 25/50 x 25/50 x 1/2 or 1/8.
4. The first player isn't in the top 25%, the second player is and the third
player beats the second. Same as above or 1/8
Altogether that is 7/12 which is a bit more than the 9/16 you get by squaring
3/4.
Using the same method for the case where the first player is in the top .5 the
second in the top .3 and the third in the top .2. The four cases are:
1. Neither of the first two player reaches the top 20% is 30/50 x10/30 or 1/5
2. They all reach the top 20% but Player 3 wins anyway is 20/50 x 20/30 x1/3 or
4/45
3. The first guy reaches the top 20%, the second guy doesn't, and the third guy
wins is 20/50 x 10/30 x 1/2 or 1/15
4. The second guy gets into the top 20% ,the first guy didn't, and the last guy
wins. That's 20/30 x 30/50 x1/2 or 1/5
1/5 + 4/45 +1/15 + 1/5 is 25/45 or the same 5/9 that the calculus guys got. It
is simple to do the same technique for the general case.
eleaticus
point, he properly directs one at me in regard to the basic definition of
a,b,x.
Mea culpa.
Thank you, McCLain
Eleaticus
"Mike McClain" <mmcclain_...@omsoft.com> wrote in message
news:cXmP9.26152$BP.13495@fe01...
>
Bennett Niizawa
I dunno - if I get a high pair, I prefer Raising But Giving The
Illusion Of A Level Platying Field, or RBGTIOALPF. It doesn't quite
roll off the tongue as acronyms go, but I think the RBGTIOALPF method
is a little more profitable. I mean, do you really want to win the
blinds with aces? Then, you would be suffering from Where Did All My
Callers Go? (WDAMCG?)
Tom Weideman
<snip>
> 1/5 + 4/45 +1/15 + 1/5 is 25/45 or the same 5/9 that the calculus guys got.
> It is simple to do the same technique for the general case.
This is how I have always done it, and I'm virtually positive it's what
Jerrod did (only we did it for the general case). Of course, for the REALLY
general case of non-uniform distributions, well, then being able to do
calculus is kind of vital.
Tom Weideman
JonCooke
Yes, that's right.
That's how I did it too; B Yoon has verified it yields the same result
as Jerrod's, non-calculus approach.
Been quite a lot of sums on here lately:)
Good thoughts, Tom...
JonCooke
> One subtle point you have missed is that you have not defined a calculable
> problem, numerically or algebraicly.
Then how come we all managed to solve it?
A. Prock
Yes and no. Doing calculus becomes "kind of" vital if you are
going to make the simplifying assumption that the discrete game
in question (poker) can be modeled using continuous variables.
If you stay in the discrete realm, then calculus won't avail you
of much. I'm not really sure that modeling poker-like problems
as continuous ones is very useful in the long run. There is a
fundamental independence assumption that you make in these
problems which is completly blown in practical poker games.
This is why Morton's Theorem is so important, it illustrates
how these specific kind of "simple" games can lead you to a
very wrong conclusion.
An interesting related question which I've been long pondering,
and have done considerable work on -- and which is almost actually
usefull enough to merrit your public inattention -- is the problem
of approximating the hand distribution of a player in a given hand,
based on whatever kind of prior knowledge you care to use.
The method suggested by your (not so) simple game examples seems
to indicate you like linearizing hands based on some metric, and then
treating them as a probability distribution. But this is not a
strictly clear task, as players who play the game tend to use
different metrics to gauge their hands.
Well, just some grist for the mill. All of these baby games are
interesting for what they are, but they seem to step too far away
from actual poker to be of much practical interest.
- Andrew
Darryl Parsons
You also don't need the sum function when working with Excel
spreadsheets; after all, you can always add up the numbers by hand.
The sheets are always finite, so that proves it is always possible. I
hope you are not trying to send any sort of message that says it's ok
to be ignorant of powerful mathematical tools because they are hardly
ever needed. That would be inconsistent with your other teachings as
well as a shame in itself.
Darryl Parsons
Dsklansky
>hope you are not trying to send any sort of message that says it's ok
>to be ignorant of powerful mathematical tools because they are hardly
>ever needed. That would be inconsistent with your other teachings as
>well as a shame in itself.
My message has always been that it is very un OK to dismiss math when
analyzing decisions. However powerful mathematical tools can make you
intellectually lazy. They also scare off people who don't have these tools. I
like to show them they often don't need them.
eleaticus
actually knowing the premises of what you are solving?
Did you or did you not notice there was no indication that the 'solutions'
were only valid if
a. the density function of x was uniform
b. x was continuous or the 'sampling' a hand represented was essentially
with replacement or from an infinite 'deck'
c. the order of calling is important.
d. the opening/calling standards were unknown to players A and B.
and others I forget for the moment.
For instance, let x represent probabilities from a deck of two suits, 26
cards and let the standards be known to all players.
Let A open/call only w/QQ or better, B w/KK or better, and C, the final
hand, only AA or better.
If A enters the pot and B has KK he knows A has either QQ or AA, and if he
has AA, he knows he has a lock.
But if C has AA he knows he has a lock, and this was a condition of the
problem: that C's hand be sufficient to meet his standards, and A and B
already be in the hand.
But what does your 'solution' say about this?
The example covers part of the unstated, unconsidered, and contrary
conditions that fit the problem but not the 'solutions'.
Eleaticus
"JonCooke" <jco...@pmsi-consulting.com> wrote in message
news:53ebc436.02122...@posting.google.com...
Jerrod Ankenman
eleaticus wrote:
>
> Because you are long adapted to a schizo 'solving' of things without
> actually knowing the premises of what you are solving?
>
> Did you or did you not notice there was no indication that the 'solutions'
> were only valid if
>
> a. the density function of x was uniform
> b. x was continuous or the 'sampling' a hand represented was essentially
> with replacement or from an infinite 'deck'
> c. the order of calling is important.
> d. the opening/calling standards were unknown to players A and B.
Well, there's also this part of this problem that pertains to those of
us who have worked on this kind of thing before, which is that we
already know that a) is true, even though Tom didn't state it, because
that's the way these problems are always posed; it's like when a physics
textbook says "Earth" it means our Earth and not some other planet
called "Earth" by its inhabitants. If we wanted a non-uniform
distribution, it would be specified. Same thing with b - it's basically
irrelevant to the gist of the solution whether it's with replacement or
not - these are real numbers after all. c. was specified (the order of
calling was), but it doesn't matter anyway. If player B foolishly wishes
to call with a when the first player only calls with B, it's the same
result. d. doesn't matter, because we're assuming that A and B have
already acted and have specified the strategies they used to act. It
doesn't matter whether they know them.
You seem to really have a huge desire to prove how wrong everyone else
is, but i'm fairly sure that I've never seen you propose a useful toy
game to solve or provide a solution to anyone else's toy games. (I don't
read everything posted to rgp, so it could be that I missed some
things). Toy games are one of the useful ways we can study these topics
- and solving progressively more complicated toy games using results
obtained from previous similar games is a strong method for improving
our understanding of how poker might work. These are not formal results,
nor PhD dissertations, and in many cases an answer that is "very close"
is sufficient for these purposes.
Of course, you are free to continue to add zero value by simply
complaining about lack of rigor, just as we are free to continue
ignoring it.
Jerrod Ankenman
eleaticus
discrete distribution (a la poker) because ... ?
For that matter, the 'useful' toy game in question is useful because ... ?
Just what principle of play did it illuminate?
For what game that you can name?
And in just which post to the thread was any application mentioned?
Any application to applications?
Eleaticus
"Jerrod Ankenman" <jerroda...@yahoo.com> wrote in message
news:3E0F6F6B...@yahoo.com...
eleaticus
thread: the problem was not defined sufficiently.
This is a poker site and poker is not a game of continuous density
functions.
If you want to do something useful - and not belie by fact that the 'useful'
concept is misapplied to the weevil's toy, solve for discrete distributions.
Poker distributions.
Eleaticus
"Jerrod Ankenman" <jerroda...@yahoo.com> wrote in message
news:3E0F6F6B...@yahoo.com...
>
T. Pascal
question, "In theory you can program a statistically sound poker
playing program, but will that program be a winner?" I think not, as
bluffing is common with humans. In limit poker, the pot odds cannot
be manipulated easily, so statistical play is mostly correct. In
no-limit or pot-limit, it's less correct (or not?). So, I then
started asking, "Can you come up with an algorithm to place a
probability on a poker player's hand?" In this way, you could come up
with an approximation of bluffing and likely hands. I think so, and
based on some of my experience working with database experts. Please
bear with my "applied" "computer science" expertise (A.K.A. "database
administration").
It would be simple enough to create a relational database schema with
a few simple tables and index them according to the following fields
(add more as needed): Player ID, Relative position (8-10 of these, 1
for first to act or small blind pre-flop, etc.), Number of players
(8-10), Game turn (4 to 7 of these, 4 for holdem (pre-flop, flop,
turn, river, etc)), Pot size (before player's action), Action (check,
check-raise, check-call, check-fold, bet, raise, fold, call), Money
amount (of the bet, raise or call), Eventual hole cards if known (2-5
fields of 52 possibilities), etc.
So the table entries for my example schema would look a lot like:
Player 1, 3 players, 1 position, flop, $20 pot, bet, bet $10
Player 2, 3 players, 2 position, flop, $30 pot, fold, <empty field>
And you see how you could then query:
Select * from mytalbe where player=player1 and position=2 and potsize
< $100; etc. Then, simply count the frequency of each holding linked
to those actions. Or, alternatively, you query for pot sizes,
position, etc., and count the frequency of each action (bet, check,
raise, fold).
I have the database but only very limited hand samples (the record
keeping is hard unless you can get hand histories from the internet,
for example, and program a tranform into the database).
What do you think? Yes, please flame my reliance on Larry Ellison and
for using high-level SQL programming languages. Thanks in advance.
> The method suggested by your (not so) simple game examples seems
> to indicate you like linearizing hands based on some metric, and then
> treating them as a probability distribution. But this is not a
> strictly clear task, as players who play the game tend to use
> different metrics to gauge their hands.
>
> Well, just some grist for the mill. All of these baby games are
> interesting for what they are, but they seem to step too far away
> from actual poker to be of much practical interest.
>
> - Andrew
Any feedback on my thoughts is appreciated.
A. Prock
>It would be simple enough to create a relational database schema with
>a few simple tables and index them according to the following fields
>(add more as needed): Player ID, Relative position (8-10 of these, 1
>for first to act or small blind pre-flop, etc.), Number of players
>(8-10), Game turn (4 to 7 of these, 4 for holdem (pre-flop, flop,
>turn, river, etc)), Pot size (before player's action), Action (check,
>check-raise, check-call, check-fold, bet, raise, fold, call), Money
>amount (of the bet, raise or call), Eventual hole cards if known (2-5
>fields of 52 possibilities), etc.
>
>So the table entries for my example schema would look a lot like:
>
>Player 1, 3 players, 1 position, flop, $20 pot, bet, bet $10
>Player 2, 3 players, 2 position, flop, $30 pot, fold, <empty field>
>
>And you see how you could then query:
>
>Select * from mytalbe where player=player1 and position=2 and potsize
>< $100; etc. Then, simply count the frequency of each holding linked
>to those actions. Or, alternatively, you query for pot sizes,
>position, etc., and count the frequency of each action (bet, check,
>raise, fold).
The problem with an approach like this is that it would take
many thousands of hands to flush out any kind of discernable
pattern. This means that such information would only be
available for a limited subset of players. I think you really
need to approach the problem from a function approximation
standpoint, rather than from only known samples.
A. Prock
>And the 'useful' toy game in question is useful compared to one with a
>discrete distribution (a la poker) because ... ?
>
>For that matter, the 'useful' toy game in question is useful because ... ?
>
>Just what principle of play did it illuminate?
>
>For what game that you can name?
>
>And in just which post to the thread was any application mentioned?
>
>Any application to applications?
While I might argue that these kind of toy games aren't
of much use either, I think the purpose of this toy game
was fairly clear, and illustrated a specific property of
some multi-player games in general:
*) certain multi-player games are fundamentally different
from two-player games. In particular, you can't treat
a multi-player game as a "composition" of multiple
two-player games because certain independance assumptions
are violated.
Of course, poker is an example of this kind of multi-player
game.
eleaticus
what you say but with some additional necessary information, particularly
how many players have folded/limped/raised in the action as it comes to you.
(Also blnd size.) However much of the board that is dealt shows also.
How many have the pockets? Fewer, but I'm not sure how few. I only look
per-person.
Eleaticus
"T. Pascal" <t_pa...@my-deja.com> wrote in message
news:2611b663.02123...@posting.google.com...
eleaticus
"A. Prock" <jeffy...@yahoo.com> wrote in message
news:3e109fbd$0$1257$8026...@spool.cs.wisc.edu...
> According to eleaticus <elea...@bellsouth.net>:
> >And the 'useful' toy game in question is useful compared to one with a
> >discrete distribution (a la poker) because ... ?
> >
> >For that matter, the 'useful' toy game in question is useful because ...
?
> >
> >Just what principle of play did it illuminate?
> >
> >For what game that you can name?
> >
> >And in just which post to the thread was any application mentioned?
> >
> >Any application to applications?
>
> While I might argue that these kind of toy games aren't
> of much use either, I think the purpose of this toy game
> was fairly clear, and illustrated a specific property of
> some multi-player games in general:
>
> *) certain multi-player games are fundamentally different
> from two-player games. In particular, you can't treat
> a multi-player game as a "composition" of multiple
> two-player games because certain independance assumptions
> are violated.
And as I said a number of times, and I notice you pointed out in another
port, the continuous distribution is not appropriate and independence is not
appropriate.
Eleaticus
A. Prock
>> of much use either, I think the purpose of this toy game
>> was fairly clear, and illustrated a specific property of
>> some multi-player games in general:
>>
>> *) certain multi-player games are fundamentally different
>> from two-player games. In particular, you can't treat
>> a multi-player game as a "composition" of multiple
>> two-player games because certain independance assumptions
>> are violated.
>
>And as I said a number of times, and I notice you pointed out in another
>port, the continuous distribution is not appropriate and independence is not
>appropriate.
This is true, but it doesn't apply to the point being
made. The conclusion is valid for discrete distributions,
and for dependent distributions.
KrisppyKreme
(Dsklansky) writes:
> My message has always been that it is very un OK to dismiss math when
analyzing decisions. However powerful mathematical tools can make you
intellectually lazy. They also scare off people who don't have these tools. I
like to show them they often don't need them.
>
I use math while playing poker about 2% of the time. Mathematics in poker is
the least important concept.
Krisppy Kreme
T. Pascal
> According to T. Pascal <t_pa...@my-deja.com>:
> >It would be simple enough to create a relational database schema with
> >a few simple tables and index them according to the following fields
> >(add more as needed): Player ID, Relative position (8-10 of these, 1
> >for first to act or small blind pre-flop, etc.), Number of players
> >(8-10), Game turn (4 to 7 of these, 4 for holdem (pre-flop, flop,
> >turn, river, etc)), Pot size (before player's action), Action (check,
> >check-raise, check-call, check-fold, bet, raise, fold, call), Money
> >amount (of the bet, raise or call), Eventual hole cards if known (2-5
> >fields of 52 possibilities), etc.
> >
> >Select * from mytalbe where player=player1 and position=2 and potsize
> >< $100; etc. Then, simply count the frequency of each holding linked
> >to those actions. Or, alternatively, you query for pot sizes,
> >position, etc., and count the frequency of each action (bet, check,
> >raise, fold).
>
> The problem with an approach like this is that it would take
> many thousands of hands to flush out any kind of discernable
> pattern. This means that such information would only be
> available for a limited subset of players. I think you really
> need to approach the problem from a function approximation
> standpoint, rather than from only known samples.
>
hundreds of players (I don't know how, but if I had an online casino,
I'd have access to all those data) with thousands of hands each and
then fit a function to the observed statistics?
Again, I know my approach is probably laughable, but if the only tool
you know how to use is a hammer, everthing looks like a nail.
T. Pascal
>
My favourite proof is the proof that there is no largest prime number.
It is very simple and requires only basic arithmetic skills. This is
off-topic for Poker, obviously. But Sklansky is right, I can teach
anyone this proof (even a 6th grader) and their eyes will light up
with understanding. Making math and poker (there's the on-topic
discussion!) approachable is very worthwhile.
Why use a power-saw when scissors will work as well? When cutting
lumber, use the power saw. When cutting paper, use the scissors.