Sklansky asked me this question
Howard Lederer
purely theoretical poker problem. The answer to the problem has
implications that can be applied in many different poker situations.
Assume that player A has the best hand with one card to come. Player B will
make a hand that beats player A's current hand exactly 50% of the time.
But, player A will also improve his hand 50% of the time, and when he does
improve, player B can't win. This means that both players will miss 25% of
the time and player A will win, both players will improve 25% of the time
and player A will win, player A will improve and player B will miss 25% of
the time and player A will win, and player A will miss and player B will
improve 25% of the time and player B will win.
Let's further assume that the cards have been turned face up going into 7th
street, so the players know exactly where they stand. Lets also assume that
the pot size going into 7th street approaches infinity, and the limit on 7th
street is $100. This means that no one will ever fold if they have any
chance to win. Check and raise is permitted.
If player A must announce his final round betting strategy before looking at
his final card, what should it be? And what will his EV be using this
strategy?
Howard Lederer
Hamlet
always bet. B should never raise because he will gain 1 extra bet 50% of
the time and lose 2 (since A can re-raise) 50% of the time. The EV of the
bet is $75, since A wins 3 out of 4 times.
Am I missing some weird game theory optimum strategy?
"Howard Lederer" <how...@lvcm.com> wrote in message
news:H_mH9.8019$Xb3.4...@news1.west.cox.net...
Hamlet
answer.
"Hamlet" <ham...@earthlink.net> wrote in message
news:BbpH9.3888$hM3.4...@newsread2.prod.itd.earthlink.net...
CapnD
propostion.
_________________________________________________________________
Posted using RecPoker.com - http://www.recpoker.com
CapnD
improve, in other words if B is a complete moron, then check-call is the
best move.
JM
If Player A improves and Player B bets, Player A should raise.
If Player A does not improve and Player B bets, Player A should call.
With this strategy Player A can yield an EV up to +$25. (Depending on
Player B betting when he improves.)
---
The instinctive answer would be for Player A to always bet. And, if
Player B raises to re-raise if Player A improves. With this strategy
Player A has an EV = 0.
If Player B misses, he folds, so 2 times Player makes nothing.
If Player B calls when he improves, 1 time Player A makes $100 and 1 time
Player A loses $100. Therefore, over the 4 possible outcomes Player
breaks even.
(You can try the math yourself for cases where Player B raises if he
improves, but this is a losing proposition for Player B. Player A would
have an EV = +25 if Player B decided to raise when he improves.)
---
However, if Player A always checks. Even if Player B never bets, Player
A's EV is still EV = 0. Player A has nothing to lose if he always checks.
BUT IF PLAYER B BETS WHEN PLAYER B IMPROVES, PLAYER A CAN GAIN SOME EV.
If Player B misses, he will still check. So, 2 times Player A gains
nothing. But if Player B improves and bets, Player A can check-raise when
Player A also improves. So, one time Player A gains $200 and the other
time when Player does not improve and calls, he loses $100.
Therefore +$100 over 4 outcomes yields an EV of +$25.
This strategy depends on Player B betting, though. All other strategies
yield a neutral EV and there is no chance for a positive EV, unless Player
B will raise Player A when Player B improves. Even an idiot could reason
that raising in Player B's shoes is not wise. However, if Player A always
checks it is more likely Player B will bet when he improves.
---
There is one more case where Player A's strategy would be to check and
call. Obviously, this yields an EV = 0 as well.
Later,
JM
_________________________________________________________________
Octo the Genarian
never raise, because even if he hits there is a 50% chance A hit too, in
which case A would reraise, making raising a losing proposition for him.
So betting has an EV of 0 for A. If B calls, he will have hit, and half the
time A will have hit and half the time A will have missed.
If A checks, B will check behind him half the time when he misses. He will
also check behind him when he hits, because he has to fear a raise the 50%
of the time that A has hit too.
So the way I see it it doesn't matter what he does, he will break even on
the last round of betting.
I have a feeling I missed something because that was too easy.
"Howard Lederer" <how...@lvcm.com> wrote in message
news:H_mH9.8019$Xb3.4...@news1.west.cox.net...
Matu
if B calls everytime then the EV is +50,
if B raises when he imroves and A reraises when also the EV goes to +75.
If A checks,
B checks, it's break even,
B bets when improves, A raises when improves, it's +25.
If A bets only when improves,
B calls and check otherwise, it's +25
if B bets when improves even if A bets, it's +75,
and there's a whole lot of ways for mixing the betting,
but no time to count...
_________________________________________________________________
MSA1213
>Date: 12/4/02 7:59 AM
First, the question is really MOOT. Any additional $100 bet is clearly
meaningless when the current pot "approaches infinity". It would now just
become "infinity +1" (or plus $100). :)
You failed to specify what Player B can do, when he can do it, and whether he
gets to see Player A's final card before he does it (as he's seen the rest).
You also omit how "bright" Player A knows Player B to be. (I'll assume that
Player B is known to be sharp.)
A.Player B would clearly just call if forced to decide before seeing either
last card. (He will also fold if he can see his last card and does not
improve).
B. If Player B could see both cards his bet:
Player B would raise any bet if he improved and Player A did not.
He would fold if he did not improve or if they both improved.
So in this circumstance, Player A should not bet as he can only lose from his
bet or break even, but never win.
C. If Player B can see only his own card first:
If Player B fails to improve, he will fold and not call any bet. (50%)
If Player B improves, he should only call.
a. 50% (conditional or 25% overall) Player A has improved and will beat him
so would reraise. Because of the infinity bit, Player B would have to call any
reraise, so does not want to raise..
b. 50% Player A has not improved and so would just call any raise, unless
bluffing (which gets into the near infinity problem as he would certainly be
called).
So in this case a bet by Player A does not help as Player B calls only when he
has improved and then it is 50/50 who wins.
The interesting case is the last one and there a bet does Player A no good. So
Player A may as well adopt a no bet strategy if Player B is sharp. If he isn't,
then under Case C, Player A can bet and then call any bet if unimproved/reraise
if improved (which latter case hopes that Player B mistakingly raises when both
improve and will call the reraise or reraise further).
The EV of no bet is zero unless Player B is foolish and makes the mistake of
betting when he has not improved (and then has to call a raise if Player A has
improved).
marc
MSA1213
>Date: 12/4/02 2:09 PM Eastern Standard Time
Curious how at 1:25 PM EST on 12/4/02, I can read your post of 2:09PM EST on
12/4/02?
How do you post an hour EARLY?
(Or is it late?)
marc
Matu
Perry Friedman
I am not sure what you mean by your statement:
>If player A must announce his final round betting strategy before looking at
>his final card, what should it be? And what will his EV be using this
>strategy?
Can his strategy incorporate what his last card is? Meaning, can he
say "I will bet 100% of the time when I hit and X% of the time when I miss.
If I am raised, I will reraise every time I hit, and call Y% and reraise Z%
when I have missed..." Or does he have to say ahead of time, "I will
bet the river X% of the time and check Y%"?
I assume you mean the former.
We can assume he folds every time he misses, so we can just examine
the case where he hits.
Let's assume you say "I will bet every time I hit, and I bluff X% when I miss."
If he ever folds when you bluff, you gain something (he loses something)
so he must always call if you bet (if there is any non-zero chance
you are bluffing). For now, I will ignore reraise scenarious.
Now, he should call if you bet, and bet if you check. And you must call
if he bets (again, otherwise he can bluff with any non-zero probability).
Since this is the case, it looks like everyone is correct about it being
EV neutral, since the 50% of the time you hit you make an additional 100,
and the 50% of the time you miss, you lose $100 (either because your bluff
was called or bet when you checked).
Hmm... let's suppose you can check some of the time you hit as well, and
may check raise! Now it becomes interesting, perhaps.
Now he can't necessarily bet every time you check.
I think if I try to work this out right now, I will end up with more
errors than usual, so I will leave that thought out there for
others and get back to it when I am not also playing a tournament on
Poker Stars at the same time.
Perry
Bing
wrote:
> > Lets also assume that
> > the pot size going into 7th street approaches infinity, and the limit on 7th
> > street is $100.
> Player A should Bet blind,
> if B calls everytime then the EV is +50,
The trick to this problem seems to be that the pot size is "near
infinity". So, your EV is already +(infinity). If you bet and he
calls, your EV goes up by 50, that's true. But, infinity plus 50
is still infinity.
In the 75% of the time you win, you gain (infinity). In the
25% of the time you lose, you lose whatever new money you
bet. So, by adding more to the pot, you can't be gaining
anything, you can only lose.
The answer is to check and call if he bets.
To see the point a little better, assume that "near infinity"
is a million dollars--some rich guy came along and tossed
a million into the pot.
If you both check, your EV is 75% of $1,000,000. That's
750,000. If you bet $100 and he calls, your EV is now
750,050. That's so small as to be inconsequential. If
you rerun the example with a billion, the result is
1,000 times smaller. A trillion is even smaller and,
finally, out at infinity, the difference is zero.
-- Bing Monopoly Expansion Set
Visit us at http://www.paxentertainment.com
Barbara Yoon
> The answer to the problem has implications that can be applied in many
> different poker situations. Assume that player A has the best hand with
> one card to come. Player B will make a hand that beats player A's current
> hand exactly 50% of the time. But, player A will also improve his hand
> 50% of the time, and when he does improve, player B can't win. This
> means that both players will miss 25% of the time and player A will win,
> both players will improve 25% of the time and player A will win, player A
> will improve and player B will miss 25% of the time and player A will win,
> and player A will miss and player B will improve 25% of the time and
> player B will win. Let's further assume that the cards have been turned
> into 7th street approaches infinity, and the limit on 7th street is $100.
> This means that no one will ever fold if they have any chance to win.
> Check and raise is permitted. If player A must announce his final round
> betting strategy before looking at his final card, what should it be?
> And what will his EV be using this strategy?
Howard.....just to see if I'm understanding your description here...
...is your "7th street" the equivalent of each player tossing a coin that
only he can see before the "final round betting," with 'heads' beating
'tails,' and player A winning all tie coin tosses...?! If so, then how
about a 'game theory' starting point of player A "announcing,"
"I will bet every time I've tossed a 'head,' plus, but only very rarely,
when I've tossed a 'tail' too, so I'll be bluffing just enough to force you
to call any time you yourself have tossed a 'head'"...?!
Dsklansky
that it took about two minutes for Howard to answer it. Hint: Keep in mind
that checkraise is allowed.
Matu
A checking 33% (11% hit 22% miss),
would result an positive EV, takes time to find optimal numbers,
but i'd guess the checks have to be in a ratio of 1-2 on the hit/miss scale
On Dec 4 2002 11:38PM, Barbara Yoon wrote:
> Howard Lederer:
> > The answer to the problem has implications that can be applied in many
> > different poker situations. Assume that player A has the best hand with
> > one card to come. Player B will make a hand that beats player A's current
> > hand exactly 50% of the time. But, player A will also improve his hand
> > 50% of the time, and when he does improve, player B can't win. This
> > means that both players will miss 25% of the time and player A will win,
> > both players will improve 25% of the time and player A will win, player A
> > will improve and player B will miss 25% of the time and player A will win,
> > and player A will miss and player B will improve 25% of the time and
> > player B will win. Let's further assume that the cards have been turned
> > face up going into 7th street... Lets also assume that the pot size going
> > into 7th street approaches infinity, and the limit on 7th street is $100.
> > This means that no one will ever fold if they have any chance to win.
> > Check and raise is permitted. If player A must announce his final round
> > betting strategy before looking at his final card, what should it be?
> > And what will his EV be using this strategy?
>
>
>
> Howard.....just to see if I'm understanding your description here...
> ....is your "7th street" the equivalent of each player tossing a coin that
> only he can see before the "final round betting," with 'heads' beating
> 'tails,' and player A winning all tie coin tosses...?! If so, then how
> about a 'game theory' starting point of player A "announcing,"
> "I will bet every time I've tossed a 'head,' plus, but only very rarely,
> when I've tossed a 'tail' too, so I'll be bluffing just enough to force you
> to call any time you yourself have tossed a 'head'"...?!
_________________________________________________________________
Perry Friedman
Well, if it only took him 2 minutes, I will give my off-the-cufff answer
and say:
bet 1/2 of the time I make my hand, and some infintessimal part of the
time that I miss (just enough to insure he must call me when I bet).
If he bets, I check-raise every time I have made my hand and some
infintessimal number when I miss (again, to insure he must call me
when I check raise).
We only have to consider the case where B improved, since he folds otherwise.
Now, if he just always bets if I check, out of 100 hands,
50 of them I have nothing and he wins 50 bets. But 25 of them (1/2 of the
50 hands I improved, I checked), I make 2 more bets, so he gains
nothing by betting if I check.
So, overall, with this strategy, I have an EV of $12.50 more than what
I have if there is no action at all on the river.
1/2 the time, he misses and folds and there is no +EV.
1/2 the time, he improves.
In that case, 1/2 of those times, I improved, and 1/2 of those, I extract 1
$100 bet. When I check, it is EV neutral with the above strategy.
So, I get 1/2 (he improved) * 1/2 (I improved) * 1/2 (I bet) * 100.
As for the bluffing strategy, since he must always call when I bluff, because
of the pot size, I believe it is clearly correct for me to minimize my
bluffing. However, there still must be SOME chance I will bluff or he can
indeed fold every time I bet. But you can make this number arbitrarily
close to 0.
Perry
Matu
A checking 50 (12,5-37,5),
makes EV +25 and B must call all bets and can't bet or raise
Jerrod Ankenman
Because of the stipulation that no player will ever fold if they have
a chance to win, we are only dealing with money won on the last
betting rounds.
Player A will never fold. Because of this, Player B should never bet
when he misses. On the 50% of hands where Player B misses, A's EV is
always zero using any strategy. Therefore, A's task is to maximize his
win when both improve while minimizing his loss when B improves but A
doesn't.
A has two possible strategies when he has improved that are not
dominated: to bet and to check-raise. Let's call the fraction of the
time that A bets the nuts when he has them X.
A also has two possible strategies when he has not improved that are
not dominated: to bet and to check-call. Let's call the fraction of
the time A bets when unimproved Y.
So A is going to bet, in total, .5*(X+Y)% of the time.
If A bets, B should fold when he is unimproved.
If B improved:
EV(B,call) = (Y-X)* .5 *(X+Y)
EV(B,raise) = (2Y - 3X) * .5*(X+Y)
2Y - 3X > Y - X, or Y > 2X
So B should raise if A will bet twice as often unimproved as with the
nuts.
If A checks, B should check when he is unimproved.
EV(B, check) = 0
EV(B, bet) = .5*(1-Y)*1 - .5*(1-X)*2 = .5 - .5Y - 1 + X = X - .5Y - .5
X - .5Y - .5 > 0, or Y < 1 - 2X
So B will bet when A checks if A's strategy is to bet unimproved less
than (1-2*pct bet the nuts).
So we end up with four sets of conditions:
Y > 2X and Y < 1 - 2X:
.25*3*X - .25*2*Y + 2*.25*(1-X) - .25*(1-Y) = .75X - .5Y +.5 - .5X
-.25 +.25Y = .25X-.25Y+.25
Total EV(A) = (X-Y+1)/4
Y < 2X and Y < 1 - 2X:
.25*3*X - .25*2*Y = .75X - .5Y = .75X-.5Y
Total EV(A) = (3x-2y)/4
Y > 2X and Y > 1 - 2X:
.25*1*X - .25*1*Y + 2*.25*(1-X) - .25*(1-Y) = .25X - .25Y +.5 - .5X
-.25 +.25Y = -.25X+.25
Total EV(A) = (1-X)/4
Y < 2X and Y > 1 - 2X:
.25*1*X - .25*1*Y = .25X - .25Y = .25X-.25Y
Total EV(A) = (X-Y)/4
Our opponent B will play the counterstrategy that yields the best EV
for him, so we can take the minimum value of these functions as our EV
and try to maximize our EV with different values of X and Y.
Turns out the answer is X=.5 and Y=0, so the strategy is:
If A improves, A bets 50% of the time and check-raises the rest of the
time. If A doesn't improve, A checks and calls.
This yields an EV of $12.50 per instance.
Jerrod Ankenman
Matu
there's a mistake on my part,
so betting 40 and checking 60 (20-40) would result +10 best i can do
JonCooke
are weak.
Since his bet threatens to win him 1 or lose him 2, we need only have
a hand half as often as not when we check.
So if improve bet half the time, check half.
Otherwise check.
Obviously bluff infinitesimally often when the pot is infinite.
B pays A off if B improves, B is indifferent whether be bets behind a
check.
So: 1/8 (A,B improve & A bets) x + 1
1/2 x 0 - B doesn't improve.
remainder: B improves and checks or bets: 0
EV of strategy = +1/8.
Give me a pot size Howard - then we have to consider bluffing
frequencies properly.
Avarice
player, since the pot is infinitely large, making an incorrect fold a
'catastrophic' mistake.
If A announces that he will check, but will check-raise when he
improves, B must check behind him. For B, a bet is a losing
proposition. If B does not improve, a bluff will fail due to the
'catastrophic' mistake clause above. If B bets, 50% of the time he
will gain one bet when A does not improve, and 50% of the time he will
lose two bets because he must pay off the check-raise.
If A announces that he will bet, B must call the bet if he improves.
A raise by B will lose money by the same argument as above.
However, if A announces that he will bet 2/3 of the time he improves
(and check-raise the other third) and bet 1/3 of the time that he does
not improve, then he does stand to gain some money on the last betting
round, regardless of whether or not B checks behind him or bets when A
checks and B improves.
Determining the correct percentages to maximize EV (or verifying the
optimality of this solution) is left to the reader as an exercise.
(Hint: when B's decisions between checking/betting results in the same
EV for him, the solution is optimal.)
(Note: A truly optimal solution would have to include a clause about
check-raise bluffing, so that B wouldn't be able to throw his hand
away to a check-raise, assuming he knew perfectly A's strategy, but
that's why the first assumption was made about 'catastrophic' folds.)
Barbara Yoon
> The answer to the problem has implications that can be applied in many
> different poker situations. Assume that player A has the best hand with
> one card to come. Player B will make a hand that beats player A's current
> hand exactly 50% of the time. But, player A will also improve his hand
> 50% of the time, and when he does improve, player B can't win. This
> means that both players will miss 25% of the time and player A will win,
> both players will improve 25% of the time and player A will win, player A
> will improve and player B will miss 25% of the time and player A will win,
> and player A will miss and player B will improve 25% of the time and
> player B will win. Let's further assume that the cards have been turned
> into 7th street approaches infinity, and the limit on 7th street is $100.
> This means that no one will ever fold if they have any chance to win.
> Check and raise is permitted. If player A must announce his final round
> betting strategy before looking at his final card, what should it be?
> And what will his EV be using this strategy?
I agree with JonCooke and Perry Friedman and Jerrod Ankenman...
JonCooke:
> ...we need to make it irrelevant whether B bets when we are weak
> [not improved]. Since his bet threatens to win him 1 or lose him 2,
> we need only have a hand half as often as not when we check.
> So if [we] improve bet half the time, check half. Otherwise check.
> ...bluff infinitesimally often when the pot is infinite. B pays A off
> if B improves, B is indifferent whether [he] bets behind a check.
> So: 1/8 (A,B improve & A bets)...EV of strategy = +1/8.
Perry Friedman:
> bet 1/2 of the time I make my hand, and some infintessimal part of
> the time that I miss (just enough to insure he must call me when I bet).
> If he bets, I check-raise every time I have made my hand and some
> infintessimal number when I miss (again, to insure he must call me when
> I check raise). We only have to consider the case where B improved,
> since he folds otherwise [except for necessary rare bluffs]. ...he gains
> nothing by betting if I check. ...overall, with this strategy, I have an EV
> of $12.50 more than what I have if there is no action at all on the river.
> So, I get 1/2 (he improved) * 1/2 (I improved) * 1/2 (I bet) * 100.
Jerrod Ankenman:
> Because...no player will ever fold if they have a chance to win, we are
> only dealing with money won on the last betting rounds. Player A will
> never fold. Because of this, Player B should never bet when he misses
> [except for necessary rare bluffs]. On the 50% of hands where Player B
> misses, A's EV is always zero... Therefore, A's task is to maximize his
> win when both improve while minimizing his loss when B improves but
> A doesn't. [*snip math details*] If A improves, A bets 50% of the time
Barbara Yoon
> Inspired by the razz hand I posted here last week, David came up with
> a purely theoretical poker problem. The answer to the problem has
> implications that can be applied in many different poker situations.
OK.....Howard.....so what was your answer to it...?!
Barbara Yoon
> When I posted this question on the General Theory Forum of 2+2,
> I mentioned that it took about two minutes for Howard to answer it.
> Hint: Keep in mind that checkraise is allowed.
OK.....so what was Howard's answer to it...?!
Tom Weideman
I don't like the way the question was posed (stating that the pot is
infinite is an ugly way to simplify the problem in an effort to make it seem
like its solution applies to real life). But if we simply come out and say
that player A cannot fold at all, and player B cannot fold if he improves,
then there is a simple short-cut answer that David no doubt intended:
Player A doesn't want to bet if he doesn't improve, because he will only be
called by a better hand. But if he checks these hands, and always bets when
he improves, he will be exploited by a value bet from player B every time B
improves. So player A needs to check raise just often enough so that player
B is indifferent to betting after a check. That is, A needs to check-raise
half as often as he check-calls, since B stands to lose twice as much on the
check-raise as he stands to gain on the check-call. Since the check-raise
leaves player B's strategy irrelevant, the we can just say that player B
always checks when checked to, leaving a contribution of 0 to the ev. Thus
all of the ev comes from A's betting for value, which he does half the time
that he improves (1/4 of the time overall), and since player B only calls
that bet when he improves (1/2 the time), player A wins 1/8 of a bet in ev.
The reason the infinite pot specification is so ugly is that you get
undefined "0*infinity" situations. For example, if player A told me (player
B) that he only bets when he improves, then my probability of winning by
calling drops to 0. The pot odds are infinite, so my ev is actually
undefined. If you instead specify that the pot is "very very large", or
"arbitrarily large but finite", or "essentially infinite", as you would do
when you try to claim that the problem has applications in real world poker,
then the solution crumbles.
Tom Weideman
Barbara Yoon
...and with Tom Weideman too...
Tom Weideman:
> I don't like the way the question was posed (stating that the pot is
> infinite is an ugly way to simplify the problem in an effort to make it
> seem like its solution applies to real life). ... ...all of the ev comes from
> A's betting for value, which he does half the time that he improves
> (1/4 of the time overall), and since player B only calls that bet when
> he improves (1/2 the time), player A wins 1/8 of a bet in ev. ...
Dsklansky
>that player A cannot fold at all, and player B cannot fold if he improves,
>then there is a simple short-cut answer that David no doubt intended:
>
Of course. By now you must realize that I am to lazy to do anything that
doesn't have a shortcut.
>The reason the infinite pot specification is so ugly is that you get
>undefined "0*infinity" situations. For example, if player A told me (player
>B) that he only bets when he improves, then my probability of winning by
>calling drops to 0. The pot odds are infinite, so my ev is actually
>undefined. If you instead specify that the pot is "very very large", or
>"arbitrarily large but finite", or "essentially infinite", as you would do
>when you try to claim that the problem has applications in real world poker,
>then the solution crumbles.
>
>Tom Weideman
It does not.
Howard Lederer
almost won as much as my wife did playing 15-30 online :-)
My answer was to bet half the time you improve and check the rest of the
time.
I spoke to Barry Greenstein about this hand and he did point out that there
must be a little bluffing by all parties in this hand. He felt that if the
pot size is N then player A should bluff every (1/2N) times he misses. And
player B should bet every (1/2N) times he misses and player A checks.
Likewise, I guess this would mean that player A should bluff check raise
every (1/2N) times he misses and player B bets.
Howard Lederer
"Barbara Yoon" <by...@erols.com> wrote in message
news:asmacl$ki2$1...@bob.news.rcn.net...
Howard Lederer
This problem highlights many issues that apply to a wide variety of poker
situations. I found the answer to this problem quickly because it is
answered by doing something I always try do when playing poker. I try to
think about what my opponent wants to do and then adopt a strategy to
neutralize his options. In all forms of poker (except lo-ball draw) there
are many situations where you must use the power of the check-raise to
maximize your expectation.
You have to be willing to give up a bet here or there to keep people off
your back. A common time to use this is in stud high on fifth street.
Let's say you opened with an ace up and bet fourth street. If you have hit
a non-threatening board, checking on fifth street is a great strategy.
Sometimes your opponent will bet and you can now check-raise your aces or
fold your nothing. If you opponents see you do this often enough, you will
start to get some crucial free cards when you can really use them. There is
nothing sweeter than hitting open aces on sixth street to win the pot when
you would have had to fold if your opponent had bet on fifth street. Of
course, you should compliment your opponent on saving that bet on fifth
street while you stack up your free money.
Howard Lederer
"Howard Lederer" <how...@lvcm.com> wrote in message
news:H_mH9.8019$Xb3.4...@news1.west.cox.net...
> Inspired by the razz hand I posted here last week, David came up with a
> purely theoretical poker problem. The answer to the problem has
> implications that can be applied in many different poker situations.
>
> Assume that player A has the best hand with one card to come. Player B
will
> make a hand that beats player A's current hand exactly 50% of the time.
> But, player A will also improve his hand 50% of the time, and when he does
> improve, player B can't win. This means that both players will miss 25%
of
> the time and player A will win, both players will improve 25% of the time
> and player A will win, player A will improve and player B will miss 25% of
> the time and player A will win, and player A will miss and player B will
> improve 25% of the time and player B will win.
>
> Let's further assume that the cards have been turned face up going into
7th
> street, so the players know exactly where they stand. Lets also assume
that
> the pot size going into 7th street approaches infinity, and the limit on
7th
> street is $100. This means that no one will ever fold if they have any
> chance to win. Check and raise is permitted.
>
> If player A must announce his final round betting strategy before looking
at
> his final card, what should it be? And what will his EV be using this
> strategy?
>
> Howard Lederer
>
>
Howard Lederer
Howard Lederer
"Howard Lederer" <how...@lvcm.com> wrote in message
news:owGH9.16462$Xb3.1...@news1.west.cox.net...
KrisppyKreme
B if he also checks. The reason is that the pot is so big (infinity) that
trying to do anything fancy is a waste of time. Win this pot and retire.
Now I have a question. What is the point of the inquiry?
1. Pots never reach infinity
2. Cards are never fully exposed
3. Exact percentages as in the example can never be realized.
Is this just a way to ask a math oriented question? As far as I can see this
has nothing to do with how real poker is played.
Krisppy Kreme
Tom Weideman
>> The reason the infinite pot specification is so ugly is that you get
>> undefined "0*infinity" situations. For example, if player A told me (player
>> B) that he only bets when he improves, then my probability of winning by
>> calling drops to 0. The pot odds are infinite, so my ev is actually
>> undefined. If you instead specify that the pot is "very very large", or
>> "arbitrarily large but finite", or "essentially infinite", as you would do
>> when you try to claim that the problem has applications in real world poker,
>> then the solution crumbles.
> It does not.
Yes it does.
It's true that if you solve the problem for pot size "P", the solution
converges to the one given as P approaches infinity. But that does NOT mean
that the P->inf solution is an approximation to the "very large P" solution.
The final EV's are close, but there is a dramatic difference in the
strategies required to attain these EV's. Namely, plays that you don't even
need to consider in the infinite pot case become essential in the large pot
case. If you ignore them because you are using the inifnite pot case as
your approximate strategy, then you can be exploited to the tune of 100% of
your ev.
For example, let's say the pot is $1 trillion, and you state that your
strategy is to bet half the hands you improve and check the rest, because
you are using the answer to the infinite pot case as your approximate
solution. I can exploit you and reduce your ev to zero by never calling
your bet and never betting when you check. If you worked out the more
general case involving a finite pot size, then you could include the other
plays necessary to regain your ev.
In your defense I will point out that in the real world a player does not
announce his river betting strategy, and the slightest doubt in the mind of
the opponent may be all it takes to make using the pot=infinity solution a
pretty good one (let's call it "quasi-optimal") for large pots. But if, in
the pursuit of a mathematical solution, we are suspending real world
conditions like both players knowing the situation with 100% certainty going
to the river and one player announcing his strategy, I am merely suggesting
that it is more precise to insist upon the "must call if not holding the
worst possible hand" rule by caveat than it is to insist upon an infinite
pot. After the solution has been found, then the advantages and drawbacks
of including real-world conditions like large-but-not-infinite pots and not
pre-announcing strategies can be examined.
Tom Weideman
Dsklansky
of the nineteenth century who claimed Newton and Leibniz (like Sklansky and
Lederer) were being unrigorous by using infintesimals in their invention of
calculus. I am not joking either. Howard and I are willing to ignore the fact
that player A needs to bluff without improving infintessimally often. Of course
in the real world Player B will never be able to fold with absolute impunity
when he improves if only because Player A might have overlooked his hand.
Newton and Leibniz were eventually exonerated by a guy named Abraham Robinson.
Who will come to our rescue?
Furthermore saying a pot approaches infinity is often a better way to tackle
these problems as it allows you to ignore very complex yet irrelevant options
in coming to your solution. As to why this question is relevant, Howard did a
very good job of explaining that earlier.
Mike McClain
"Dsklansky" <dskl...@aol.com> wrote :
> Tom Weideman and Barry Greenstein's quibbling is much like those mathmeticians
> of the nineteenth century who claimed Newton and Leibniz (like Sklansky and
> Lederer) were being unrigorous by using infintesimals in their invention of
> calculus.
LOL!! This might be the funniest thing you have ever written!!
> I am not joking either.
Doh... not sure if that makes it even funnier, or if it's just plain sad.
Look, David, you're no Newton.
If you wanted people to come up with the answer that you had in
mind, then you should have defined the question properly. Do you
really think it's fair to claim that Player B has to call because Player
A might have misread his hand? Why didn't you just say that
Player B was going to always call with his hand, even if Player A
has indicated in his strategy that he will only be betting when he
has the nuts? Instead you had to introduce the 'infinite pot',
without fully understanding the implications of such.
> Who will come to our rescue?
Paging Dr. Kevorkian...
Mike.
Tom Weideman
> Tom Weideman and Barry Greenstein's quibbling is much like those mathmeticians
> of the nineteenth century who claimed Newton and Leibniz (like Sklansky and
> Lederer) were being unrigorous by using infintesimals in their invention of
> calculus.
Heh. Newton and Leibniz would have solved the actual problem instead of
figuring out a way to dumb it down enough so they can "cleverly" state the
answer in a few lines. Then they could have explained how the strategy
really does behave for large-but-finite pot sizes instead of fumbling around
with feeble rationalizations.
Tom and Barry are more like the thousands of unnamed physicists that knew
almost instantly that the cold fusion results of chemists Pons and
Fleischman were bogus. [Sorry Barry, I just don't feel comfortable
comparing myself (and in this case by extension, you) to famous geniuses
like Sklansky does. It makes it too easy to be mocked when you can't live
up to the comparison, you know.]
Tom Weideman
Barbara Yoon
>> Tom Weideman and Barry Greenstein's quibbling is much like those
>> mathmeticians of the nineteenth century who claimed Newton and
>> Leibniz (like Sklansky and Lederer) were being unrigorous by using
>> infintesimals in their invention of calculus.
Mike McClain:
> LOL!! This might be the funniest thing you have ever written!!
Dsklansky:
>> I am not joking either.
I really do believe him when he says he's "not joking" here.....as after all,
Sir Isaac Newton never achieved Dsklansky's fabled SAT scores...!!
Mike McClain:
> Doh... not sure if that makes it even funnier, or if it's just plain sad.
> Look, David, you're no Newton. ...you should have defined the
> question properly.
Dsklansky:
>> Newton and Leibniz were eventually exonerated by a guy named
>> Abraham Robinson. Who will come to our rescue?
Mike McClain:
> Paging Dr. Kevorkian...
JonCooke
> almost won as much as my wife did playing 15-30 online :-)
>
> My answer was to bet half the time you improve and check the rest of the
> time.
>
> I spoke to Barry Greenstein about this hand and he did point out that there
> must be a little bluffing by all parties in this hand. He felt that if the
> pot size is N then player A should bluff every (1/2N) times he misses. And
> player B should bet every (1/2N) times he misses and player A checks.
> Likewise, I guess this would mean that player A should bluff check raise
> every (1/2N) times he misses and player B bets.
>
> Howard Lederer
tried to find the Nash equilibrium of this problem.
Howard Lederer
"JonCooke" <jco...@pmsi-consulting.com> wrote in message
news:53ebc436.02120...@posting.google.com...
> > I spoke to Barry Greenstein about this hand and he did point out that
there
> > must be a little bluffing by all parties in this hand. He felt that if
the
> > pot size is N then player A should bluff every (1/2N) times he misses.
And
> > player B should bet every (1/2N) times he misses and player A checks.
> > Likewise, I guess this would mean that player A should bluff check raise
> > every (1/2N) times he misses and player B bets.
> >
> > Howard Lederer
> Where'd those numbers come from, cos they're not what I got when I
> tried to find the Nash equilibrium of this problem.
I don't know. I think he is right on the frequency of A bluffing when he
misses if the last betting round is a dollar. I am unfamiliar with formal
game theory, but I get this number using this formula:
1/4*(1/(N+1))*2. The "1/4" is the frequency with which player A makes a
hand and bets. The "(1/(N+1))" is the bluff frequency player B needs to
call when player A bets. The "2" is the factor you need to multiply the
bluff frequency because all the bluffs happen when player A misses which is
only half the time. I am sure my answer is crude, but I think it is
correct.
I guess, since player A has announced a strategy that makes betting
unnecessary by player B, the last two bluffing frequencies aren't needed.
I have only experienced the Nash equilibrium one time in my life. I was on
a plane watching "A Beautiful Mind" and my ears finally popped.
Howard Lederer
![Mark B [Diputsur]'s profile photo](http://lh3.googleusercontent.com/a-/ALV-UjW1ukZesLHTvLryryGaLhoUTsW-4I5_N8bKNVoFKGU5T0OxggP6=s40-c)
Mark B [Diputsur]
> Inspired by the razz hand I posted here last week, David came up with a
> purely theoretical poker problem. The answer to the problem has
> implications that can be applied in many different poker situations.
>
> make a hand that beats player A's current hand exactly 50% of the time.
> But, player A will also improve his hand 50% of the time, and when he does
> improve, player B can't win. This means that both players will miss 25% of
> the time and player A will win, both players will improve 25% of the time
> and player A will win, player A will improve and player B will miss 25% of
> the time and player A will win, and player A will miss and player B will
> improve 25% of the time and player B will win.
>
> Let's further assume that the cards have been turned face up going into 7th
> street, so the players know exactly where they stand. Lets also assume that
> the pot size going into 7th street approaches infinity, and the limit on 7th
> street is $100. This means that no one will ever fold if they have any
> chance to win. Check and raise is permitted.
>
> If player A must announce his final round betting strategy before looking at
> his final card, what should it be? And what will his EV be using this
> strategy?
>
> Howard Lederer
personally, I feel like 'retarded sue' after reading it... not even going to try to follow along with the equations. Sue says: bet, positive, /thread