Problem with Game Theory and Bluffing (long)
Renaud
Bluffing (Ch 19) in his "The Theory of Poker"book. I'm enthralled
with the idea of turning a -EV proposition into a winning one by
applying the correct bluffing strategy. However, I don't believe it
practically EVER turns an unprofitable situation into a profitable
one, as the author implies (p 189). It only works if the pot odds are
very low and your chance of winning is over 38% to start with.
To use two examples from his book, say you are on a flush draw with 4
bets in the pot (20% chance of winning). Your opponent is getting 5-1
pot odds if you bet on the river, so according to Sklansky, you should
bluff 20%/5 = 4% of the time (you could use 2 cards you haven't seen
to randomize the bluffs - 9/5). Regardless of what she does in
response to your bet, this is a long-term losing situation for you,
even with the optimal bluffing strategy.
e.g. heads-up and she calls each time*: (I'm assuming you both had 2
bets in the pot)
You win 3 bets 20% of the time (when you hit and she calls)
and lose 2 bets 76% of the time(when you don't hit the flush or your
bluff cards)
and lose 3 bets 4% of the time (when you hit your bluff cards and she
calls).
Total: .2(3) - .76(2) - .04(3) = -.96 -> losing almost a full bet per
hand!
* It doesn't matter how she calls or folds when you bluff like this
(as explained in the book), the result is the same.
His other example is different. You have 18/42 (or 43%) chance of
winning and you both have 1 bet in the pot. Now she is getting 3-1
pot odds if you bet on the river, so you should bluff 43% / 3 = 14% of
the time (or use 6 cards you haven't seen to randomize the bluffs --
18/3)
You win 2 bets 43% of the time (when you hit)
and lose 1 bet 57% - 14% = 43% of the time (when you don't)
and lose 2 bets 14% of the time (when you bluff).
Total: (1)43%-2(14%) = +.143 -> winning bets per hand
So somewhere between these two examples is the magic winning
percentage you need to turn an unprofitable situation into a
profitable one.
OK, let me get even more geeky for a second. Let a = # of bets you
have in the pot (which we'll say is the same as your opponent's).
Then it turns out that the percent chance to make your hand you need
to have a positive expectation using this method is
a(2a+1) / (8*sum of the first a numbers),
which converges pretty quickly to 1/2 at large values of a. I was
extremely psyched to get such a compact result! Anyway, when you both
have just one bet in the pot, you need a 38% chance or better to make
your hand. By the time you get to 3 bets each, it has already jumped
to 44%! There aren't many draw hands in poker that give you those
kind of chances. The best you can say is that bluffing like this
"takes optimum advantage of the situation (p. 187)."
What do you think?
Vince lepore
I'm not going to go to Sklansky's book to check this out because your
examples don't make sense. First you cite randomizing a river bluff.
Then you describe situations where you make your hand. If you bluff
the river it is because you have all of your cards and didn't make
your hand. Something is fishy in your example. If you are speaking
of betting a flush draw on the turn in Holdem or 4th street in stud
that's a different matter. Most of the advice I've seen from Sklansky
in these situations state that there must be some chance of your
opponent folding on this street. Please clear this up and I will
answer you.
Vince
PacPalBuzz
From: rest...@earthlink.net (Renaud)
Date: Sat, Jan 18, 2003 6:28 AM
Message-id: <5fe4cffa.03011...@posting.google.com> >>
Renaud - Game theory offers a philosophy of bluffing that may be different from
the way you have previously looked at bluffing.
I think you may be missing a key concept. As I see it, the object of bluffing
in game theory is to collect an extra bet from an opponent on those occasions
when you bet with the goods.
Therefore, you bluff just often enough that your opponent is forced to call you
*all* the time (or forfeit pots). You actually *plan* to lose when you bluff.
(Bluffs are "advertising"). However, you more than make up for those losses by
collecting on all the hands when you *do* have the goods.
Of course, if someone is chasing all the time anyway, then you should cut way
down on your bluffing frequency. Or if you are bluffing at game theory
frequency and usually getting away with your bluffs, then you should consider
bluffing more often.
Just my opinion.
Buzz
Tom Weideman
> Renaud - Game theory offers a philosophy of bluffing that may be different
> from the way you have previously looked at bluffing.
>
> I think you may be missing a key concept. As I see it, the object of bluffing
> in game theory is to collect an extra bet from an opponent on those occasions
> when you bet with the goods.
Sort of. The way Sklansky explains it is actually quite good. To
paraphrase: If you bluff only one time in a million, then your opponent
doesn't have the pot odds (unless the pot is around a million bets in size,
which for real examples it never is) to try to catch your bluff, so you get
a free pot theft once every million opportunities. That's better than never
trying to steal. Of course, attempting to steal twice every million hands
gets you two free pots out of every million, because the opponent still
doesn't have odds to try to catch you. How far can you push this? Well,
you can keep adding to your steal frequency until it gets to the point where
your opponent has the pot odds to call you down, but then you better not add
any more - the gravy train of free steals ends. But by this time, you've
added a lot to your overall ev.
> Therefore, you bluff just often enough that your opponent is forced to call
> you *all* the time (or forfeit pots). You actually *plan* to lose when you
> bluff. (Bluffs are "advertising"). However, you more than make up for those
> losses by collecting on all the hands when you *do* have the goods.
No. You bluff just often enough so that it doesn't matter whether your
opponent calls you or not. That is, you provide him/her with the
probability that you are bluffing such that his/her pot odds give them an ev
of zero to try to pick off a bluff. You don't force them to call you every
time - it just doesn't matter whether they do or not.
> Of course, if someone is chasing all the time anyway, then you should cut way
> down on your bluffing frequency. Or if you are bluffing at game theory
> frequency and usually getting away with your bluffs, then you should consider
> bluffing more often.
Yes, that would be what's called an "exploitive strategy" that (if your
judgment is correct) returns more than the "optimal strategy" provided by
game theory. Of course, playing exploitively also risks being exploited,
which would of course return less than an optimal strategy.
Tom Weideman
PacPalBuzz
From: Tom Weideman zwi...@attbi.com
Date: Sat, Jan 18, 2003 2:25 PM
Message-id: <BA4F145F.2429D%zwi...@attbi.com>
>>
<<"No. You bluff just often enough so that it doesn't matter whether your
opponent calls you or not. That is, you provide him/her with the probability
that you are bluffing such that his/her pot odds give them an ev of zero to try
to pick off a bluff. You don't force them to call you every time - it just
doesn't matter whether they do or not."
Tom - Thanks. I can clearly see that is a better description of the game theory
approach.
"Of course, playing exploitively also risks being exploited, which would of
course return less than an optimal strategy."
Well put.
Thanks for your very helpful (to me at least) response.
Buzz
tadperry
news:20030118190401...@mb-fy.aol.com...
It was a very good response.
The thing to remember at this point is that the best players do play
exploitatively without being themselves exploited.
It's not a mathematical impossibility to be able to do that. You do have to
be careful however, because, as pointed out, you leave yourself definitely
*open* to that. If you watch for that exploitation to develop in response to
a tendency, you can re-exploit. If you follow the line of reasoning more
clearly than your opponent, you're usually in a position of advantage. If
not, the opponent has forced you into the equilibrium strategy where you can
await further deceptive opportunities against this particular opponent in
the future or other inroads to the player's stack including drunkenness,
tilt, boredom, or you-name-it.
tvp
Renaud
> examples don't make sense. First you cite randomizing a river bluff.
> Then you describe situations where you make your hand. If you bluff
> the river it is because you have all of your cards and didn't make
> your hand.
Hi Vince. These aren't my examples, btw, but I apologize for not
presenting them as clearly as Sklansky did. The situation is that
there are no more cards to come, and you have to decide whether or not
to bluff if you didn't make your hand. Tom Weideman nailed it. But to
reiterate, the basic idea is to bet on the end more often than you
would be expected to make your hand, but not so much more that your
opponent is getting good odds to call each time. My (sort of trivial)
point was that even this ideal bluffing strategy won't hardly ever
turn a -EV situation into a + EV one, it just minimizes your losses.
It's not really relevant to Hold Em, since it will usually be obvious
to everyone if you made your hand or not.
Speaking of fishy though, I'm having trouble with the calling
frequency thing. An implied premise of the bluffing strategy is that
your opponent knows the odds of you having made your hand, and should
start to call more when you bet more often than these odds would
indicate. How much more? Well, like Tom said, if you bluff more
often than the pot odds she gets to call, you're bluffing too much,
and vice versa.
It seems, to my nubbly self, freakishly hypothetical that anyone could
come close to estimating an opponent's betting frequency on the end
for each type of hand they might have in a real game. But Mr Sklansky
says you don't need to bother -- just invert the pot odds you are
getting to call and fold that many hands. For example, if there are 5
bets in the pot, fold 1 time and call 5 times. Who plays like that?
Do you really call most of the time even if your hand only beats a
bluff? I think my folding percentage is probably the inverse of this
-- I mean I fold more than I call, and I only have the vaguest of
notions when someone is betting too many hands.
Maybe that's why I still need a day job ;)
bk...@aol.com
difference. In the lower limit games, there are nearly allways too many
players who just call and call. When this is true, bluffing would seem to be
a waste of time and money.
Not that they are playing badly, because in limit games the pot odds generally
are justified. In such games, IMHO, it's best to call, raise, etc only when
your pot odds are greater than the actual odds against improving your hand.
Good luck.
In article <20030118160545...@mb-fy.aol.com>, PacPalBuzz
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Jerrod Ankenman
Renaud wrote:
> It seems, to my nubbly self, freakishly hypothetical that anyone could
> come close to estimating an opponent's betting frequency on the end
> for each type of hand they might have in a real game. But Mr Sklansky
> says you don't need to bother -- just invert the pot odds you are
> getting to call and fold that many hands. For example, if there are 5
> bets in the pot, fold 1 time and call 5 times. Who plays like that?
> Do you really call most of the time even if your hand only beats a
> bluff? I think my folding percentage is probably the inverse of this
> -- I mean I fold more than I call, and I only have the vaguest of
> notions when someone is betting too many hands.
Optimal strategy on an isolated river is often different than the river
portion of optimal multi-street strategy.
Here's a toy game exercise for astute readers:
Let's say there are 4 bets in the pot on the river. It's holdem. You
have aces, and you know through the magic of toy game technology that
your opponent either had a flush draw or a hand that was drawing dead on
the turn. There was no betting on the turn. On the river, the flush is
completed. Your opponent bets. With what frequency must you call to
prevent your opponent from exploiting you in this situation?
Now assume the same situation, except that you bet the turn and your
opponent called. There are still 4 bets in the pot on the river. On the
river, the flush is completed. Your opponent bets. With what frequency
must you call to prevent your opponent from exploiting you in *this*
situation?
You probably don't need to calculate these things, just figure out what
the difference is between these two situations and why it affects the
calling frequency on the river.
Jerrod Ankenman
Michael Maurer
> Now assume the same situation, except that you bet the turn and your
> opponent called. There are still 4 bets in the pot on the river. On the
> river, the flush is completed. Your opponent bets. With what frequency
> must you call to prevent your opponent from exploiting you in *this*
> situation?
In making your point, you may have inadvertently chosen a more
complicated example than you intended. Are you suggesting that your
opponent would have folded on the turn had he known he was drawing
dead? Don't jump to conclusions!
Try solving the following toy problem, first posed to me by Paul
Pudaite:
Pudaite's Two Street Game
Start with 4 bets in the pot. Player 1 has a made hand; player
2 has, with equal probability, either a drawing hand or a dead
hand. The drawing hand has a 25% chance of drawing out on
player 1's made hand.
There is a round of betting.
On the second (last) street of play, a community card is dealt.
25% of the possible community cards make player 2's drawing
hand. Both players know which cards these are.
There is another round of betting.
PacPalBuzz
From: bk...@aol.com
Date: Wed, Jan 22, 2003 7:19 AM
Message-id: <b0mcq0$5qb$1...@news.netmar.com> >>
BJK - Thanks for responding.
"I may be wrong, But it seems to me, that the size ($) may make a big
difference. In the lower limit games, there are nearly allways too many
players who just call and call. When this is true, bluffing would seem to be a
waste of time and money."
Nearly always too many players who just call and call? Sounds like heaven.
Your suggestion that size ($) makes a big difference makes sense to me. There
are some players who do seem to just call and call and I imagine they are more
predominant at lower limits than higher limits.
Even so, it would be a mistake for anyone to assume that low limit games are
inhabited by players who nearly always just call and call and that bluffing is
a waste of time and money. That’s just my opinion, based on my observations
over the years.
I can’t speak for big money poker. However, I regularly encounter bluffing in
(1) low limit home games, (2) low limit casino games, and (3) low limit casino
tournaments. Thus my personal experience at lower limits is that players do
bluff - and often very successfully.
I can assure you that you will give up a lot of potential profit playing low
limit poker in casinos if you assume your opponents are never bluffing - and
thus always fold to save that last big bet when someone has bet into you -
since in that case many of your opponents will think you an easy target for a
bluff. I can also assure you that you will lose a lot of money if you assume
your opponents are always bluffing and always call that last bet - since in
that case you won’t see any bluffing directed at you, except by fools.
I don’t run across many opponents I consider fools; some, but not many.
However, I regularly encounter some very intelligent and proficient poker
players at low limits, both in home games and in casino games. You might wonder
how that can be true if the main object in playing poker is to make money.
Wouldn’t it seem that all the intelligent and proficient poker players would
migrate to higher limits? But they don’t. Whatever your own object is when you
play poker and/or whatever you think should be the main object in playing
poker, perhaps optimizing their poker incomes is not the main object of all
those who enjoy low limit poker. But while my own object is not to make money,
I do like to win. (It’s confusing to my wife since the way you keep track of
winning in poker is with money - and she knows for certain I can make more and
risk nothing doing something else).
At any rate, in the lower limit games in which I play, some players seem to
bluff too much while others seem not to bluff often enough. Conversely, some
players are easy to bluff while others are almost impossible to bluff. That’s
just my opinion, based on my observations over the years.
From my viewpoint, the trick when someone bets into me is distinguishing the
players who will try to bluff me too often from the players who never (or
hardly ever) bluff. The trick when I am considering a bluff is knowing who I
can bluff and who I can’t bluff.
However, often I don’t know if a bluff might work or not. I don’t want to start
bluffing too much, but neither do I want to bluff too infrequently.
Enter game theory.
<<"Not that they are playing badly, because in limit games the pot odds
generally are justified. In such games, IMHO, it's best to call, raise, etc
only when your pot odds are greater than the actual odds against improving your
hand.">>
Thanks for the thought. I may be wrong, but I think of calling as related to
having pot odds greater than the odds of winning. I think of betting and
raising mainly as related to either (1) having fresh money odds greater than
the odds of winning, or (2) bluffing. I don't think you should bluff much -
not much at all - but I do think you do better if you sometimes bluff.
Just my opinion.
Buzz
PacPalBuzz
Date: Sat, Jan 18, 2003 10:50 PM
Message-id: <NUrW9.45113$Yq3.9167@sccrnsc02> >>
Todd - Thanks for your response.
<<"The thing to remember at this point is that the best players do play
exploitatively without being themselves exploited.">>
That seems a worthwhile goal. However, sometimes I find myself up against an
opponent who seems better able to read me than I am to read him/her.
For example, in the upcoming ESCARGOT, I expect to encounter some very fine
r.g.p. opponents who are much better poker players than I am. I want to be able
to randomize my play so that they will not be able to know what to expect. I'm
thinking that the equilibrium game theory strategy at least somewhat dulls
their edge.
Buzz
PacPalBuzz
Date: Sat, Jan 18, 2003 10:50 PM
Message-id: <NUrW9.45113$Yq3.9167@sccrnsc02> >>
Tad - I addressed you as Todd in my previous response. but your name is Tad. I
knew that, but it's late and I'm tired. My humble apologies for my error.
Buzz
A. Prock
>For example, in the upcoming ESCARGOT, I expect to encounter some very fine
>r.g.p. opponents who are much better poker players than I am. I want to be able
>to randomize my play so that they will not be able to know what to expect.
I've found this is much easier if you first randomize your entire
being.
- Andrew
Lou Krieger
"A. Prock" <jeffy...@yahoo.com> wrote in message
news:3e306655$0$7860$8026...@spool.cs.wisc.edu...
Isn't cosmetic surgery rather expensive these days?
PacPalBuzz
Date: Thu, Jan 23, 2003 3:46 PM
Message-id: <q9%X9.9957$x9.26...@twister.socal.rr.com>
<<"For example, in the upcoming ESCARGOT, I expect to encounter some very fine
r.g.p. opponents who are much better poker players than I am. I want to be able
to randomize my play so that they will not be able to know what to expect.">>
A. Prock and Lou Krieger; two cases in point.
Buzz
jacksup
> Pudaite:
>
> Pudaite's Two Street Game
>
> Start with 4 bets in the pot. Player 1 has a made hand; player
> 2 has, with equal probability, either a drawing hand or a dead
> hand. The drawing hand has a 25% chance of drawing out on
> player 1's made hand.
>
> There is a round of betting.
>
> On the second (last) street of play, a community card is dealt.
> 25% of the possible community cards make player 2's drawing
> hand. Both players know which cards these are.
>
> There is another round of betting.
OK, I'll give this a shot.
It seems to me Player 1 wants to bet on the first betting round, and
Player 2 wants to call with all his drawing hands, plus enough of his
dead hands so that he can bluff optimally on the river.
Since two more bets will be in the pot from the previous betting
round, there will be six bets in the pot at the river. So when Player
2 bluffs on the river, he'll be offering Player 1 7-1 odds on that
bluff. Therefore, Player 2 wants to make the chances that he's
bluffing 1/8.
So, on the first betting round, Player 2 calls with all his drawing
hands, plus 1/7 of his remaining hands--8/14 of his total hands.
On the river, Player 2 just mucks when the draw misses. When the draw
hits, Player 2 bets regardless of what he holds, because he is value
betting or bluffing with optimal frequency.
So for Player 1
6/14 - wins 4 bets (P2 folds on first bet)
6/14 - wins 5 bets (P2 calls first bet, then draw misses on river)
2/14 - loses 1 bet (P2 calls first bet, draw hits and P2 plays
optimally)
P1 EV = 52/14 = 3.71 bets
similarly for P2
6/14 - 0
6/14 - loses 1
2/14 - wins 5
P2 EV = 4/14 = 0.29 bets
This seems like the optimal solution. Player 2 can't increase the
number of hands he bluffs with, so it shouldn't make sense for him to
call with any more hands on the first betting round. And if Player 1
checks the first betting round, his EV is just .75*4= 3 bets. So
clearly, P1 needs to bet on the first round.
Is this right?
Did I mess anything up?
Matt
Bill chen
important part. Now what's Player 1's exact strategy again? He bets
on the first round, and...?
Bill
mattm...@hotmail.com (jacksup) wrote in message news:<473b8d61.03012...@posting.google.com>...
Bill chen
when player 2 bets, he will have 1/7 of bluffing hands and 1/4 of made
hands. that means he will be bluffing 4/11 of the time, and player 1
has an easy call and actually has +EV from calling.
Bill
mattm...@hotmail.com (jacksup) wrote in message news:<473b8d61.03012...@posting.google.com>...
Jerrod Ankenman
> Oops, this isn't right. If the river makes the drawing hand, then
> when player 2 bets, he will have 1/7 of bluffing hands and 1/4 of made
> hands. that means he will be bluffing 4/11 of the time, and player 1
> has an easy call and actually has +EV from calling.
If the river makes the drawing hand, his range of hands doesn't
change. He played 1/7 of his bluffing hands, and 7/7 of his drawing
hands, so now he has a made hand 7/8 of the time, and will be bluffing
with optimal frequency (into a six bet pot) on the end.
Player 1's calling frequency on the end is the interesting part of the
problem.
Jerrod Ankenman
Bill chen
of P1's strategy still stands though.
Bill
Mike Garcia
This is a badly phrased question. If player 2 follows this strategy it does
not matter what player 1 does. EV(player 1) = 20/7 EV(player 2) = 8/7
Now player 1 also has an optimum strategy. But if player 1 follws his optimum
strategy is does not matter what player 2 does. EV(player 1) = 20/7
EV(player 2) = 8/7
I'm going to dervie both strategies from the same equation.
Player 1 will call if the flop hits with a probability of p. He will fold
with a probabiluty of (1-p).
Case 1: Drawing hand. Half the time the flop hits and if player 1 calls
player 2 gets +6, if player 1 folds player 2 gets +5. Half the time the flop
misses and player 2 is out his bet (-1). The value to player 2 of a drawing
hand is
(.5)[ (p)(+6) + (1-p)(+5) ] + (.5)(-1) = 2 + (.5)p
Case 2: Bluff. Half the time the flop hits and if player 1 calls player 2
loses 2 bets, if player 1 folds player 2 gets +5. Half the time the flop
misses and player 2 is out his bet (-1). The value to player 2 of a bluff is
(.5)[ (p)(-2) + (1-p)(+5) ] +(.5)(-1) = 2 - (3.5)p
Case 3: Fold. Value 0.
So if Player 2 bluffs with his bust hands with the probability of q (0<q<1)
the value of the game is
value = (.5)(2 + (.5)p) + (.5)[ q(2 -(3.5)p) + (1-q)(0) ]
value = 1 + (1/4)p + q(1 - (7/4)p)
If p = 1 (Player 1 always calls) the value is 5/4 - 3/4q
If p = 0 (Player 1 never calls) the value is q + 1
This says that if player 1 always calls when the flop hits then player 2
should never bluff (q=0) and the value of the game would be 5/4. If player 1
never calls when the flop hits then player 2 should bluff as often as he can
(q=1) and the value of the game would b 2.
Taking the partial derivative with respect to q yields
d(value)/dq = 1 - (7/4)p
When p = 4/7 d(value)/dq = 0.
Plugging (4/7) in for p and chug gives us
value = 1 + (1/4)(4/7) + q( 1 - (7/4)(4/7) )
value = 1 + 1/7
Note q has dropped out and this value is less than either extreme. This must
be a minima.
If player 1 calls the flop 4 times out of 7 when it hits it does not matter
how often player 2 bluffs.
Lets do a reality check from players 2 POV
S1 (always bluff): ev = (.5)(-1) + (.5)[ (4/7)(.5)(+6) + (4/7)(.5)(-2) +
(3/7)(+5) ] = 8/7
Half the time the flop misses and player 2 is out a bet. Half the time the
flop hits. 4 time out of 7 player 1 calls and half of those times player 2
had the nuts and gets +6. 4 times out of 7 player 1 calls and half of those
time player 2 was bluffing and losses 2 bets. 3 times out of 7 player 1 folds
and player 2 gets 5 bets.
S2 (never bluff): ev = (.5)(0) + (.5) [(.5)(-1) + (.5)(4/7)(+6) +
(.5)(3/7)(+5)] = 8/7
Bust hands cost nothing. Drawing hands go to round 2. Half the time the flop
misses and player 2 is out a bet. Half the time the flop hitsand 4 times out
of 7 player 1 calls so player 2 gets +6. Half the time the flop hits and 3
times out of 7 player 1 folds so player 2 gets +5.
If player 1 calls the flop 4 times out of 7 when the flop hits any and all
mixtures of bluff (S1) and don't bluff (S2) yields exactly the same results,
on average player 2 wins 8/7 bets and player 1 wins 20/7 bets.
So the complete Player 1 stategy is
round 1 bet and Player 2 will call;
if the flop hits, check and call 4 times out 7 after Player 2 bets;
if the flop missed bet and player 2 will fold.
The complete Player 2 strategy is
round 1 Player 1 will bet; call if you are on a draw; call 1 time out of 7 if
not on a draw (bluff); If the flop hits, Player 1 will check, and you bet; if
the flop misses Player 1 will bet and you fold.
Looking at the value equation again.
value = 1 + (1/4)p + q(1 - (7/4)p)
Taking the partial derivative with respect to p we find
d(value) / dp = 1/4 - (7/4)q
when q = 1/7, d(value)/dp = 0
Do the plug and chug
value = 1 + (1/4)p + (1/7)( 1 - (7/4)p )
value = 1 + 1/7
If q = 0 (player 2 never bluffs) value (to player 2) = 1 + (1/4)p
if q = 1 (player 2 always bluffs) value (to player 2) = 2 - (3/2)p
Since player 1 wants to minimize player 2's value (q=0) -> (p=0), value = 1.
If Player 2 never bluffs, don't call him. (q=1) -> (p=1). If player 2 always
bluff, always call and value = .5. So q = 1/7 is a maxima and player 2's EV
is 8/7.
QED
Mike G