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I just watched Rounders again on FX and have a question/comment about the last hand.
Why did KGB Teddy make such a large bet at the end? He thought Matt Damon had a busted straight draw. He states that the ace couldn't possibly help him. Why couldn't Matt ( I forget the character's name)have had A9 suited ( like the earlier big hand hand he lost)?
I realize KGB Teddy is on tilt but this still seems like a huge error on his part. What could have Matt had that would make him call the bet but not have a better hand? Did KGB Teddy want to make Matt fold to make him look bad?
I think the biggest mistake KGB made was not factoring in the difference in stakes. KGB was playing for money and Matt was playing for his life.
I really didn't like the scene where he told KGB his tell. He should have just folded and carved him up from there. The movie tries to mitigate the scene by having Mike say that he didn't have the kinda time needed to break KGB, and he wanted to put him on tilt, but I don't buy it. It just made for a good scene. Exactly how much time do you need in pot limit heads up? One hand can make and break someone, as it did here, and I don't feel that his "tilt" status changed the situation, as he was always playing smarmy-aggressive,( new style type, see MM's new book).
Have any of you guys ever considered that it was just two actors in a movie wrirren by screen writers (not poker players, for damn sure they were not no-limit poker players) for a non poker playing audience?
I actually thought the film did a good job of showing the ups and downs of poker life. Actually a great job. That first scene with Damon and KGB when he loses his bankroll?? Perfect, real. The way we can fall in love with a hand, wow! The ups and downs of the second and final game? When you regret like hell you kept playing and didn't leave with your win? Limit or no limit, anyone that says they haven't been there is lying.
Still, there were a lot of things that could have been done better: The one pointed out in post below about the tells was one. Putting the man on tilt is a lot less of guaranteed winner then playing to his weakness. Second, and one that has always bugged me, can anyone tell me--are there all night banks in NYC where you can go cash a $10K check? I sure as hell doubt it. Also, while card reading is an important skill, I don't know anyone that can look at a single play at a table and tell you 'exactly' what everyone has, the way Damon does of the Judges game.
Still, great movie and Mikey McDermott's girlfriend has one of the best butts in filmaking!
;-)
I thought the movie was reasonably realistic as well. I just hated the end. I wish Damon would have quit after he won the first game, having learned his lesson and not wanting to gamble so big anymore. After all, anyone can lose a single session and he was reasonably out of debt. A much better ending (but less romantic) would have been had he moved to Vegas and paid off his debt monthly through grinding the middle limit games.
Nonetheless, his sidekick's character was right on. The dead beat degenerate who only lives for the moment and takes others down with them. Unfortunately, I know lots of those types (they owe me a fortune).
Evidently you've never seen David Chiu deal. . .
I was called a tight unimaginative player by a respected player...Is that a comment or an insult? Sounds like a Kenneth Ng'ism...based on his posting history, it was both, he always talks like that. :-D
Tight unimaginative players can make lots of money, but they can't make it from tight imaginative players.
I assume they will make money from weak loose players. Then again, any reasonable player will as well.
Will the TUP make "lots" of money or just a samller pecentage of what an strong player would make.
Hello,
A while back I was discussing a hand with a dealer where he lost a large pot because the dealer burned and turned before action was complete.
That has happened to many of us and I shrugged and said "What can ya do?"
He forwarded an idea that I thought made some sense. It sort of borrows the 'All In Protection" concept from online poker.
He said that it is not right that dealers mistakes cause so much money to be pushed in a different direction, that the dealer had too much effect on the outcome.
His proposal was that if a card is burned and turned before action is complete, then anyone who has not yet completed their action be declared all in for the rest of the hand and all further betting between players who did complete the action is done in a side pot.
No board cards would ever be shuffled back into the deck. The dealer's mistakes would never affect the 5 board cards that come out.
I know this solution is not perfect, but he stated that this solution is more equitable than board cards being reshuffled and very likely changing the value of hands.
I was wondering if anyone had any comments on this proposal?
I think it has it's drawbacks - like the players declared all-in having no opportunity to protect their hands and pots may still be pushed the wrong way because someone draws out when they would have folded to a bet or raise.
However, I think this 'All In Protection' proposal may ensure that more money gets pushed to the 'rightful' owner than the current rule. The current rule is 'fair' in that you are just as likely to be hurt as helped by it, but it is unfair in that the dealer's mistake causes so many pots to go a direction they would not otherwise have taken.
What do you think?
PS: Yes, I made trips on the turn last night, but action was not complete and the card came back in and I lost a pot I would have won. The All In Protection proposal would have got me a good chunk of 'my' pot. Instead I got another "Gee, I'm sorry sir."
The Early Burn and Turn Rule as it sits now has only cost me money, it has never netted me a pot I would not have otherwise won.
it may not be perfect but it does seem alot better than the way the situation is [normally] handled now.
While we are on the topic, this is one "dealer error" that is, IMO, close to unforgivable under all but the most unusual circumstances.
It is almost always indicative of a dealer who is not paying attention; there is no excuse for a dealer having a cavalier attitude regarding a pot that often contains more than he or she earns in a day (sometimes)more than he/she earns in a week.
When I dealt in a game where it was common for five or more players to stay in past the flop I would often say, "pot right" - not as a statement but as a question.
I made my share of errors during that brief stint but I NEVER "burned and turned" with the action not being completed.
Nor did I ever expose the river prematurely.
I would have been ashamed to look the "rightful" winner in the eye if I ever did.
I see dealer's do it only to look at the victim with a s--t happens expression; it makes me cringe.
I have only been involved in two incidents as a player where a card was exposed early - 1 win, 1 loss, both relatively small pots.
The dealer who was responsible for my loss appologized profusely and admitted to having "drifted off".
Later that evening he approached me and INSISTED I allow him to make good the amount of the pot. ($61)
He is currently the manager of that room.
- Probably just a coincidence, don't you think ?
Ok, maybe I am losing my mind, but i was in a tournament today, no limit, about 3/4 stage, with middling chip position. I rasied in from the button, using 1/3 of my stack in doing so. The button reraised me, and i went all in. As i turned out, i lost when my flush hit, but 3 8's hit the board, giving his pocket 5's a full house. As I said something to him after the hand, he said he was ahead all the way, to which, i replied, "yes, in straight hand value terms, you were, but 55 is NOT worth a RR unless trying to abort a steal attemp". He disagreed, i said that had he knew what i had, AKs, he would not have reraised with 55. He again said yes, he would, and that 55 is equal to AKs heads up. I disaggreed, and had two other people jump in on his side, saying thatm heads up, ANY pair is at least equal to AKs. Now i know about running hot and cold hands, and differing hand values, but small pair Vs. AKs is equal-to-favorite heads up? That would go against all logic. You're never supposed to raise small pair, unless for deceptive value preflop, or just to buy the button or see the turn cheap if all check to you. Why is Aks almost always considered a raising hand if, isolated, it's noi better than 22? Am I losing my mind? In HEP, Sklansky says that 99 is a small favorite against ak, and I have always heard 10-10 is about equal to AKs. These both make sense, as they are VERY close on the Hand group scale, but A group 1 hand being behind a group 6, even 7? I find this very hard to believe. Someone please set the record straight. These players all seem very knowledgeable who are siding against me, and I find it hard to believe they could all SO wrong. So then, is it me?
Ok, if I was behind by about 2-3% with final holding knowledge, this still makes his reraising me a VERY bad move, any table persona aside,(which mine was definitely NOT a blind stealing crap raiser). His reraise put 80% of his chips in play, at which time I went all in,(he had about 10% more chips than I did B4 the hand) Am I at least permitted being right here?
If you have been showing down quality hands, he is in the classic "small favorite or HUGE underdog" trap; he is a slight favorite if you have 2 big cards while he is an enormous undergog if you have a higher pair.
- When there is a confrontation between pairs, the larger pair wins an average of (approximately) 80% of the time. AA vs KK is an example of the higher pair being an even bigger favorite (~82%) - 33 vs 22 would be a situation where the smaller pair has a better chance (I don't know the exact %) but is still a very large underdog.
Since you said that you had been showing down "real" hands I think his choice was somewhere between bad and horrendous. I reserve the right to retract if there were any major "financial" concerns but if this was in the early to middle stage of the tournament I can only think of a small number of situations where his play is something I would endorse.
However, there is a question you might want to ask yourself. Was HE showing down alot of trash ? In other words, did you consider the possibility that most players would only make the play he made with a big hand ?
If he has "KK" you are in a jam; I don't play alot of no-limit so I don't know these figures by heart but I think you are about a 3-1 underdog.
If he has "AA" the situation is near hopeless; the odds are more than 10-1 against you - perhaps as much as 15-1. (I really should learn these numbers but as a limit player, "hot and cold" percentages are of very little use to me.)
The bottom line is this.
There is no way I re-raise with his hand (55).
If I hold your hand and DO get re-raised I am probably going to stop and think, but I'm not going to be thinking about which he is more likely to have 55 or QQ. I'm going to be asking myself how likely it is that he holds a pair.
If this thread gets any additional responses they may be from those suggesting you should have factored the chances of him having a monster into your decision.
He didn't, but that does not mean he couldn't have.
- Better luck,
- J D
P.S. There is one more possibility to consider - one that should flatter you. If he knew that sooner or later he was going to have to "confront" you anyway, he might have been saying to himself, "let me see if I can get rid of this guy with a hand that is [probably] the favorite, even if it's only slightly favored".
The only problem with this theory is that he picked a bad hand to do it with. I'll leave it to those with more experience in these matters to come up with one, but I do NOT believe that "55" was a good choice.
even if it did work.
The tables had just converged, this was probably 3/4 of the way through the tournament levels, with about 3/5 of the participants gone. He had only been on the table for 5 minutes or so, so didn't have alot of time to gauge my play. Somewhere, i could swear that i heard that AA vs AKs was only a 6/5 favorite, but i figure that's gotta be wrong. Also , we were probably both stacked better than about 80% of the field, so the "don't fight another big stack" theory goes out the window, as does "attack the small stacks". This brings to mind something i read in "Zen and the art of Poker"; "In the battle of two tigers, One is killed, while the other is severely wounded". In other words, if you don't have to battle strength, don't. Plenty of rabbits out there.
You may want to say the same thing but it's better to clarify this, in order not to give out incorrect odds for consideration by players: Heads up, in Hold'em, AA wins 82% of the time (not an "82% favorite") against KK, so the pocket rockets' edge is 82%-18%=64% approximately. That gives pocket Aces, in this instance, odds of about 4.3-1 as a favorite.
The KK hand, by the way, fares very slightly worse when one (or two) of its suits match one (or two) of the Aces'.
this may not be the biggest factor, but is one---very tight players may get less action because the other players have seen your play--just tiny show of "not so good" hands can change the image. Jim
There could be a few reasons for your kind of results:
1. You might be tilting a little bit when you are losing. Tight players are awful when they loosen up just a bit because they become calling stations. Maniacs, on the other hand, sometimes become even more dangerous when stuck. If you do tilt a little, then at least do it in an aggressive manner (note: I am not advocating tilting).
2. You might be subconsiously leaving early when ahead and staying longer when losing. Thus, if you get out, you will leave a little ahead or if you get off to a good start. You probably don't like to leave stuck a few hundred so you stay until you get ahead a little or you are stuck enough to make you "throw in the towel" and want to quit.
Just some possible reasons for your results.
Thanks Russ, You have hit the nail on the head!!(my head).
Playing 1500-3000 I was down to $3000 in chips in the limit omaha tournament after putting in my $1500 blind. A player raised and I just called instead of moving all in, leaving myself with $1500. What parameters would make this play correct?
I can't think of a hand where I wouldn't just call before the flop in omaha in this situation (I can think of some that I would fold).
You have a hand that is very likely a clear fit or fold type hand, e.g., a low/mid straight wrap.
There is a third player in the hand who is already all-in.
There are others who will be forced all in before you within one table orbit.
Having a good but not great hand would certainly be one. Another parameter would be if you are very close to the money - perhaps enough players may bust out before the blinds come around to you again.
If there is no way the raiser is going to fold if you do go all in, I think it would be justify this play anyway unless you have an exceptional holding.
David;
Is not being focused on a 1st or 2nd finish, in favor of anywhere in the money another parameter for you?
Heads up, AKs beats JTs and JTs has a 5% edge over 55. Then, 55 has the edge over AKs!
A > B > C > A
Paper, rock, scissors.
Sorry if I have extended your pain Steve! LOL I actually like Shakespeare, but not sure I follow your reference to it? Mind explaining?
Anyway, as to your post on Making it Easy, I'm not sure why you would want it deleted. It obviously struck a nerve and was something that needed to be said. To some degree the mathematical discussion here (it's worse on the Internet Forum) has reached the point of being so much psuedo-intellectual BS.
A basic understanding of pot odds, odds, etc. is very important, but I just can't imagine Amarillo Slim or Johnny Moss sitting at a high limit game in the back room of a Texas saloon, jotting notes and doing calculations in their little black book. (Please, for all the people that will attack this post: I only said 'I can't imagine it,' I grant that I may be wrong.)
Best wishes.
Ray
Ray,
The source is Macbeth: "Life's but a walking shadow...a tale told by an idiot, full of sound and fury, signifying nothing."
John
Sklansky, et al you smarties:
Why doesn't it make more sense to actually play tighter as the bully against the short stack in a tournament, and let the blinds, which are relatively larger, whittle away at him, rather than taking stupid shots at him, and thereby giving him opportunities he can actually choose from, between blinds?
When somebody is extremely short-stacked, once they put any money in the pot they are (or should be) committed. So, if somebody is half-way all-in with the big blind, or the like, I don't raise with my big stack unless I have AT LEAST an above average hand. His cards are random, so why give him the edge when I know he'll call? However, if he has the 1 bet in the blind and 3 or more bets behind, and it's late in the tourney, NOW he might fold with anything but a very good hand, because he's afraid of being eliminated.
So, you judge each situation, and you pick your spots to be aggressive. It might seem like a stupid shot when he calls me and I end up showing down J7o to his AK, but if his stack was the right size, then my initial raise may easily have been correct with ANY two cards.
Later, Greg Raymer (FossilMan)
You do not need an above average hand to make raising with the big stack correct play in your first example. Since the short stack's chips are worth more than yours in a percentage payback tournament, you are getting the best of it by forcing him to gamble with you on a coin toss. Further, and this is important, even if you have slightly the worse hand it is likely to be correct to make the same play--for the same reason. As the stack vs. stack ratio changes the cutoff point for how much the worst hand the big stack can correctly play vs. the small stack also changes. Interestingly, it is further influenced by the number of players remaining in the tournament, since the less players remaining the greater equity you are entitled to from the tournament EV the short stack is losing in these confrontations. So when you and the small stack are the final two players remaining in the tournament you can force hinm to gamble even more, and be correct to do so, since you are now the sole beneficiary of the tournament EV he is losing.
The concept of chips being worth more in the short stack doesn't apply to heads-up situations, nor to winner-take-all events (and once you get heads-up, it is winner take all, since you're now only playing for the difference between first and second place).
You are correct about being able to play a SLIGHTLY below average hand if you're going to be heads-up against an all-in player. However, don't take that too far.
Also, in a real event, unless you are the small blind, you need to worry about that player also (which means you need a somewhat better than average hand, maybe). Or, if you know they're going to fold anything but AA, you will be playing against a short stack with valuable chips plus the dead money from the small blind, meaning again you can play a below average hand and do so correctly.
Later, Greg Raymer (FossilMan)
This does not seem to be correct to me. My sense is that once you get heads-up in a percentage payback tourney it is not changed to winner-take-all, and the difference in chip values still applies. You are correct that you are playing for the difference between payouts at that point but I do not think that this changes it to a winner-take-all tournament structure. I think chips should still have different values based upon stack sizes even heads-up between the last two contestants in a percentage payback tourney.
You do bring up some interesting other points, and you are correct that any dead money in the pot can correctly influence your starting hand selection.
M,
You wrote: "I think chips should still have different values based upon stack sizes even heads-up between the last two contestants in a percentage payback tourney."
You might want to think some more about this, for Greg is indeed correct. When percentage payback tournaments become heads-up, they essentially become winner-take-all for the remaining two players.
If you understand why the percentage payback effects chip values when more than two players remain, it should be easy to understand why that effect disappears when only two players remain.
I still don't see it. Granted they are both guaranteed second-place money at a minimum, but when it is 3-handed they are all guaranteed 3rd place money at a minimum. Heads-up they are playing for the difference, but 3-handed they are playing for the differences, and so on.
Finally, to extrapolate further, when it is heads-up and one contestant has merely one lone chip remaining, are you saying that one chip is not more valuable than any single chip in his opponent's stack? This would seem to be what you are implying if you say that chips do not change value when heads-up for the final two spots in a percentage payback tournament. Yet unless I am seriously mistaken, my understanding is that that one lone chip is worth more than any single chip in the opponent's huge stack.
Mark,
You wrote: "You might want to think some more about this, for Greg is indeed correct. When percentage payback tournaments become heads-up, they essentially become winner-take-all for the remaining two players.
If you understand why the percentage payback effects chip values when more than two players remain, it should be easy to understand why that effect disappears when only two players remain." -------------------------------------------------------
I believe there is a fundamental fallacy at work here. This fallacy appears to me to be confusing the fact that they are playing for the difference between payouts with a winner-take-all tournament structure.
This should not be hard to illustrate. Let's say that the two final contestsants have: Player A=T3000, and Player B=T1500. The payouts for first and second place happen to be $30,000 and $15,000. If this were a winner-take-all tournament, the correct settlement/deal prices at this point would indeed be $30,000 for Player A and $15,000 for Player B. Since this is however a percentage payback tourney, the correct settlement/deal prices are as follows: Player A has a 2/3 chance of winning first place = $20,000, plus a 1/3 chance of winning second place = $5,000, for a total fair settlement price of $25,000 for Player A. Player B has a 1/3 chance of winning first place = $10,000, plus a 2/3 chance of winning 2nd place = $10,000, for a total fair settlement price of $20,000. From this we can see that each of the short-stack Player B's chips are worth more than each of Player A's chips on an individual chip basis.
Since we can figure this alternatively by taking the base guaranteed amount, $15,000, and to this adding the product of each player's chances of winning the difference between the payouts, i.e., $15,000 + 2/3 ($15,000) for Player A, and $15,000 + 1/3($15,000) for Player B, it is tempting to think that since they are "playing for the difference" they are now playing a winnier-take-all structure. But this is controverted by the above math which clearly shows that Player B's chips are each worth more than face value. The fact is that they are still within a percentage payback structure even though they are "playing for the difference." The guaranteed floor payout is what makes the difference and what changes the math of the situation, and what causes the chips to still change value.
I wrote: "The guaranteed floor payout is what makes the difference and what changes the math of the situation, and what causes the chips to still change value."
To this let's add the reduced first place payout as well.
The $15,000 for second place is already in their pockets. Ignore it. They are playing for the $15,000 difference, and their chips are worth the corresponding portion of $15,000. The chips of Player A in your hypo are worth $10,000, and those of B $5,000. If Player B doubles through, he and A switch places. At all chip counts the value of each chip is constant, T4500 in play divided into $15,000, no matter who has more, or how much more.
With 3 players, yes, you're still playing for the differences, but it is differenceS, plural. To extend things, let's give player A T1800, player B T1800, and player C T900, and instead of $30,000 and $15,000 remaining, we also have a third prize of $10,000. Now, the differences are $5,000 and $20,000. Players A and B each have 40% of the chips, so a 40% chance of winning. C has a 20% chance of winning. Let's estimate that C will finish 3rd 50% of the time, and 2nd 30%. That means A and B finish 2nd and 3rd 35% and 25%, respectively. Doing the math, their equity in the differences is Player A = 40%x$20,000 + 35%x$5,000 = $9750 Player B is also $9750 Player C = 20%x20,000 + 30%x$5,000 = $5500
Note that although A and B each have double the chips of C, their equity is not double. Thus, the average value of the chips in C's stack is higher than that in the stacks of A and B, even though we're only looking at the differences, and not the total prize.
Later, Greg Raymer (FossilMan)
Greg,
When they are heads-up, the short stack's chips are worth more on a real-dollars-per-chip basis than the large stack's chips are worth on a real-dollars-per-chip basis. When we do the math taking into account the total of both places this is clear. It should also be clear that this correct math produces a different result than it would produce if the tourney were truly a winner-take-all event. If it were truly a winner-take-all event the math would show that all chips have the same value in real-dollars-per-chip. These two values (the two real-dollar-per-chip values produced in this situation for a percentage payback tourney vs. a winner-take-all tourney) are not the same; therefore you cannot call playing off the final two places in a percentage payback tourney the same as a winner take all event. If it were, the math would show that the real-dollar-per-chip values were identical in both cases. But it doesn't. It shows that the real-dollar-per-chip values change (the short stack's real-dollar-per-chip value is higher than the large stack's real-dollar-per-chip) when playing off the final two spots in a percentage payback tourney. In a winner-take-all event these real-dollar-per-chip values would be the same for both parties regardless of stack size.
The fallacy is in throwing the second place prize out of your calculations for this purpose. While the fair deal/settlement price can be figured accurately starting from this premise, the real-dollar-value-per-chip becomes distorted if you begin the calculation from this starting point. Do you see why? (this has to do with fractions).
but you're wrong.
There is a reason why you should be concerned with the value of a chip. It teaches that if you're the short-stack and go up against a bigger stack, you need more than a very slight edge to justify the risk in the latter stages of the tourney, because the dollar value of the chips you might lose is greater than the dollar value of the chips you might win. Likewise, this concept teaches the big stack that they should go ahead and be more aggressive, becuase what they lose when they lose is not as valuable as what they win when they win. In terms of chips, the big stack can take slightly the worst of it, yet in terms of dollar values still be taking slightly the best of it.
Once you get heads-up, this concept disappears. Even if you do the math and include the second place money, here is something you'll notice. The incremental value of each chip is identical.
Let's say we have 100 chips in play. First prize is $200, and second prize is $100. Using your method, if I have 20 chips, my chips are worth $100 + 20%($200) = $140. If I win 10 chips from you, my value goes up to $160. In the first instance, you're saying my value/chip is equal to $7, while in the second state the value/chip has now become $5.33, so that chips are worth more in a short stack. It looks that way. However, what I really did was gain 10 chips and add $20 to the value of my stack.
now, let's say you had won that pot. You had 80 chips worth $100 + 80%($200) = $260. Average value = $3.25. You win 10 more chips from me, now worth a total of $280, or $3.11 per chip. Again, you might say that the value of each chip went down in the bigger stack, but notice that 10 extra chips increased your total value by $20.
Therefore, no matter how big or small your stack, every chip you add to it adds $2 in value, and every chip you lose costs $2 in value. Therefore, if you have any edge in terms of chips, you have the same edge in terms of money. This is very much unlike the situation described above (where losing 20 chips costs you $x, while gaining 20 chips gains you some amount less than $x).
Later, Greg Raymer (FossilMan)
With the above parameters, the short stack holding 20 chips has equity of $120. ($100 + 20% of the $100 difference which = $120,
or alternatively, (20% of $200) + (80% of $100) = $120.
If this were truly a winner-take-all event, the short stack holding 20 chips would have equity of 20% of $300, which is $60.
In the first case above, a percentage payback tourney, the short stack's chips are each worth 120/20, or $6.00 apiece. The large stack's chips are each worth 180/80, or $2.25 apiece.
If this were a winner-take-all tourney, each chip in the tourney would be worth $3.00, regardless of who held it.
Since these values are not equivalent, it shows that heads-up for the final two places in a percentagfe payback tourney is not the sdame as a winner take all event with regard to chip values , and that chip valures still do change even heads up for the final two spots in a percentage payback tourney.
These chip value differences point up the fact that a bullying tactic by the large stack in a percentage payback tourney may be correct even with some holdings which are worse than "slightly worse" than the average holding. In fact there is a sound mathematical basis for such tactics (to a point).
As I mentioned in another post, when it gets heads-up, the large stack is the sole beneficiary of the tournament EV the small stack is losing on coin flips, and therefore should be even more inclined to bully the small stack when vying for the final two spots than when at a full table where the tournament EV lost by the small stack is distributed to all remaining participants. This is in direct conflict with the theory put forth by Greg (and supported by Mark Glover).
After re-reading this post I must admit I am starting to now become a bit confused. I thought I had a clear picture of everything involved but now I am questioning some new aspects.
If I am indeed wrong then I apologize to you Greg, and to Mark Glover, if I seemed brusque. I can see how your point that value-per-chip when it changes hands may remain constant, although it is clear that value-per-chip in the stack varies. It is also clear that if this were a true winner-take-all event, that value-per-chip in the stack would not vary, so how can we reconcile this apparent dilemma? Again this seems to somehow be related to whether you look at the whole chip picture or just the difference in chips.
Although my thoughts and statements regarding this have been rather definite, now I must re-evaluate. And before I do that I am going to catch up on some poker-playing this evening before it gets too late.
'Til later,
Mark ("M")
Well after playing some hold'em this evening, I am still somewhat unsure of some things.
It seems pretty clear that the math shows that chips do change value when heads-up for the final two places in a percentage payback tournament. However perhaps Greg is right, in the sense that at this point it no longer matters, because any chips which actually do change hands have a constant value. I don't think this necessarily fully complements Greg's assertion (and Mark Glover's concurrence) that this means it is winner-take-all at this point, but it is important. This aspect also seems to cast doubt on my assertion that the large stack can correctly bully the small stack much more freely even with subpar hands because his chips are worth less individually.
"It teaches that if you're the short-stack and go up against a bigger stack, you need more than a very slight edge to justify the risk in the latter stages of the tourney, because the dollar value of the chips you might lose is greater than the dollar value of the chips you might win."
Fossilman and M,
I know I'm getting in very late in this (interesting)discussion but I have a point to make. Well, you know me. I somewhat agree with Fossilman's statement that I quoted above. But there is another factor that is of prime concern in later stages of a tournament when you are short stacked. A factor that Fossilman does not seem to consider in his statement. Fossilman does not consider the possibility of "doubling up" and the effect that it will have on your making the money or winning the tournament. His staetment only considers the "edge" one needs to risk their chips when short stacked. Current dollar value of chips alone is not the only factor to consider when evaluating your risk. That's my point. Thakns.
vince
Wrong Forum, Man.
P.S. No more coaching newbies on the rail of my 40-80 7CS game :)
Maven
"His cards are random, so why give him the edge when I know he'll call?"
Minor correction: If you know he'll call. There have been plenty of times when I've seen bad players fold hands to which they should have been committed. Give your opponents every opportunity to make mistakes.
My own initial conclusion is that all of your ideas, plus many more not even polled here, are valid in isolation. But taken together, they pretty much cancel out, no matter how you cut it.
It's all illusion. But ain't life fun?
Anyone know where this phrase comes from? I am simply curious.
The expression is used in stud when a player receives a "picture card" (King, Queen, or Jack).
It is used a little more often in HI-LOW stud. In this game, "catching paint" (especially on 4th street) is usually an undesirable event since most of the quality starting hands contain either three small cards or a small pair with an Ace kicker.
I am not an old-timer so it is possible that there is or was another source from which this expression came, but this is the only place I still hear it used.
I hope that was of some help.
- J D -
Yes, thank you.
I had only heard this used in hi low, and in reference to a high card - not necessarily a picture card. But then, maybe I just wasn't that observant and it was a picture card.
I am not the greatest math person, nor am I the greatest at expressing myself here on these forums, so i hope that what i intend is clear.
I too am a people person and find my greatest strengths in poker are reading hands and situations, and going with intuition on some points. Yet at the same time, the medium level mathematical points are invuluable in my opinion. The combination of the two skills can do nothing but improve your profits. While situations and game selection may make the need to master one or the other sides of the spectrum unnecessary there are nothing but benefits to be gotten from at least an intermediate knowledge. Good examples would be tournament situations where you are laying odds to an opponent (NL or PL) or taking odds, or weighing chip value at varying stages, another would be PL omaha ring games where you are correct in certain instances to fold the nuts on the flop due to combined outs against you. In Limit games you can take the following thought process to make a mathematically correct play VS. a tricky player you dont know: Player A i have seen play these hands in this way my hand VS. those hands is an X% favorite, therefore i can expect a profit from a raise here.... or i should fold given the information i have.
Again i apologize if this is unclear, the point is dont underestimate what the statistical information can do for your decision making, it can do nothing but compliment a good intuitive gut instinct, and even when its just not clicking at the tables, math is something to fall back on when all else fails..
Best of luck....
-Ray
I agreee 100% fwiw
The Bellagio has many high stakes games, 80-160 through 1500-3000 on a regular basis (especially on weekends or hollidays). If you want to play big enough, probably in the range of 1000-2000+ its possible that they will arrange a game to suit your preferences.
Shawn
which reminds me--think I read it here on 2+ 2--Belagio recently had big 20 & 40 game-- 20,000 & 40,000 that is !!!
I bought this book recently and am halfway through the 400 or so pages. I find it enjoyable, and packed with interesting ideas.
There is something I am not quite clear on which is probably obvious to many of you ... Cooke repeatedly stresses the formula of volume multiplied by edge equals expectation. Obviously you want your money going in when you have edge - what I am trying to work out is whether or not you should be committing less money if you believe your edge is small, and more when your edge is large ?? In a way this doesn't seem right, as surely even if your edge is small you want as much volume on it as possible to maximise expectation??
It's not just the size of your edge, but also your certainty. If you think your edge is 10%, but occasionally you'll be wrong and your opponent will actually have you drawing dead, then you want to invest less money. If your edge was 5% but it was pretty much a lock at that figure, then you're safe to invest more.
However, in actual play, Roy's concept really isn't very helpful at all. I wouldn't spend too much effort working with it. That effort is better spent on tracking your opponents and figuring out their hands.
Later, Greg Raymer (FossilMan)
I've always taken Roy's edge motto as a kind of mantra for staying cool and always trying to make the right plays. If you concentrate on playing with an edge then after enough volume of play you will be ahead.
So I think of it more in a psychological way rather than as a tactical concept.
Regards,
Paul Talbot
If you think your edge is 10%, but you might be wrong and you might be in the negative zone, then isn't there also the chance that your edge could be greater than 10%. I'd rather invest more in an edge centered at 10% with high variance than an edge of exactly 5%. In the long run you'll make twice as much.
If, on the other hand, it was most likely that my edge is 10%, but this is a max, and I could be drawing dead (-100%), then there are many instances of this kind where I don't want to play at all. A perfect example is stud poker when an opponent pairs his door card early. You may think it's most likely that opponent has nothing else, making your overpair a small favorite, but it's also a reasonable probability opponent has trips, making you a giant dog. In this case, a fold is in order even though you may have the better hand greater than 50% of the time.
You're right. When you can identify a small edge with great certainty, limiting your risk by betting smaller tends to result in a lower expectation than risking a greater amount. This kind of marginal advantage is often confused with situations where your edge is less clear, and has a potential range that overlaps negative and positive but you're not sure where the boundries lie. Basic poker strategy involves avoiding these situations either by using information and analysis to clarify them or by limiting the amount risked.
Short answer: Pushing a small edge will increase volatility, which is something that you might not want.
Long answer: The question is about what you want to do, coincidentally. Theoretically, pushing small edges just as much as large edges is not the best way to make money, because it incurs too much risk.
Poker players don't get into this much (it's more common to blackjack players, and probably some other gambling types who are more free to choose their bets), but it is interesting. It's also directly related to what you're asking. Anyway, what you're really interested in doing when you're betting is making your bankroll grow, not maximizing your expectation. A simple example demonstrating this would be a wager where you bet your entire bankroll or net worth when you are a small favorite. You wouldn't do this because the risk is too great.
Mathematically, you also aren't maximizing your chance of doubling your bankroll. Too often you bust out. There's an area of mathematics that deals with "utility functions", I believe, that tries to maximize the rate of growth of your bankroll over repeated bets. The "Kelly Criterion" provides a formula for how to do this. If this is interesting for you, you may want to look at some explanation of the Kelly Criterion; you should be able to find one on some blackjack site.
Whew ... so this is surely more info than you wanted or needed. It's not all applicable or useful, either. Your statement about pushing your edge to maximize expectation is correct, by the way - which is why I don't believe there's a simple explanation to your question.
Thanks to all who have responded.. What I have taken from this is that I am right to think that you should still get your money in on small edges - AS LONG AS you are prepared to accept large swings.. This pretty much mirrors my game because while I am comfortably ahead I have experienced quite large swings both ways.
Funnily enough, reading more of the Cooke book yesterday it became a lot clearer that he advocates a style which will lead to swings, and stresses the importance of being psychologically able to handle that.
Luke, suppose you have a choice of two games. The first game is the Bellagio $30-$60 game where your earn is $22 per hour and your standard deviation is $600 per hour. The second is the $10-$20 game at the Mirage where your earn is $20 per hour and your standard deviation is $200 per hour.
Theorectically you should prefer the $30-$60 game but from a practical standpoint, the $10-$20 game might be better. In the first game, you would need about $3,000 over a nine hour playing session. In the second game, you probably won't need more than a $1,000. It is it worth exposing another $2,000 to win an extra $18?
While you may have an edge in both games, your edge in the $30-$60 game is only marginally better than in a $10-$20 game. But your risk is much higher in the bigger game. It is true that volume times edge equals expectation. But marginal increases in edge may not be worth the big swings that frequently go along with an increased edge.
So what does your hourly earn rate at the $30-$60 game need to be for it to be worthwhile tolerating the additional swings? $30? How do you decide?
It depends upon each person's financial situation and their tolerance for risk. Some people are risk takers and others are risk avoiders. Personally, unless I can beat a $30-$60 game for at least $30 per hour, I prefer to play in a lower limit game where my earn might only be $20-$25 per hour. But this depends upon the individual.
These ideas are drawn a bit from an earlier post. How can you determine what betting limits you should be playing at?
If you're bankroll is a million dollars and you are playing $1-$2 then you are wasting your time. If you're bankroll is $100 and you are playing $20-$40 you stand to lose your entire bankroll too fast. How can you determine what level is correct? Can you use your hourly rate & hourly standard deviation to determine what level you should be playing at? If you assume that your bankroll cannot be replenished what is a safe level to be playing at? What if you are will to accept a 5% chance your bankroll will be wiped out in the next year?
I suppose if your bankroll is small you would rather play at a lower limit, even if your hourly expectation is a bit lower than the next higher level because you cannot tolerate the swings at a higher level.
For example. Let's say you invest $100 to play at an online poker site. That $100 is all you will ever have.
Assume the limits there are $0.50-$1.00, $1-$2, $2-$4, $3-$6 and $5-$10.
If you start at $0.50-$1.00 when should you move up? When your bankroll reaches a certain level? When your hourly rate reaches a certain level? At some point you have to feel like you are wasting your time at that level. But you can't start at $5-$10, you could lose everything in 1/2 hour.
What would be a good reason to move back down?
Are there strategies (that normal people can understand) that will maximize your profit, but minimize your chance of busting out?
Sammy you need to purchase a book entitled "Gambling Theory and Other Topics" by Mason Malmuth. It is the definitive work on this subject. You can order it through this site and Conjelco by clicking on to the "order form" or "books" under "Directory" on the left hand side of the screen.
I think a good strategy is to start taking selective shots at the next limit when you have 150-200 big bets at that limit in your bankroll and then move up more permanantly when you get to about 300 big bets.
Using your example, with $100 (assuming you can't/wont replace it) start at $0.50-$1.00. When you get to $300 start playing some $1-$2 but only enter soft games and be prepared to move back if you hit a 25+ big bet losing session. At some point you'll cross the line and be playing $1-$2 as your regular game when you are somewhere around $600 and you feel comfortable with it. You can now take some selective shots at $2-$4 while you build up at $1-$2.
Repeat until you can't beat the next level or you run into the situation of Jim's 10-20 vs 30-60 example. Where you end up will have a lot more to do with what you get out of poker and why you play than it will have to do with any calculation you can perform.
I agree with Jim though, check out Mason's "Gambling Theory and other Topics."
Regards,
Paul Talbot
You make an interesting point about Jim's $20-$40, $30-$60 example.
Let's take an example at lower limits.
Say I am sure I can beat $3-$6 for $10/hour. And (currently) I can beat $5-$10 for $9/hour.
Unless I plan on playing $3-$6 for the rest of my life, wouldn't it be smarter for me to play $5-$10? That way I can learn to beat $5-$10 for more and eventually move to higher limits? In the long run, giving up $1 per hour now, could maybe result in more money in the future.
Any comments?
It ain't rocket science.
Play in games you are favored to win in. It's the game more than the limits. Learn to scout out the best games for YOU.
It is a lot better than setteling on a limit.
let's send the player in question - the one with the lifetime bankroll of $100 to the greatest 5-10 game that ever was.
There is probably better than a 60% chance that he will maul the game; there is at least 1 chance in 3 (perhaps higher) that he will take just one or two bad beats and be forced to retire forever.
He could even be blinded broke; in fact either of these could easily happen in a game as small as 2-4.
Do you even take the time to read the question before you answer it ?
Just wondering -
J D
- I see you are still posting without a return address. While I will admit to taking a certain amount of [misplaced] pleasure in pointing out those of your responses that are clearly made without giving any thought to the subject at hand, I - and I suspect at least a few of the other regulars - would at least consider contacting you privately if we knew where to do so.
Several games here have 4 or 5 people capping the betting on every card with little or nothing. These games seem insane, but how does one practice reading a table full of maniacs?
You don't. These games are not for players who thrive on identifying unique situations and taking advantage of them (say with dramatic steal re-raises on the river).
Lets play a different game. This game is a straight 8/16 with blinds of only 1-2. There is no checking and no raising; you either "call" or fold. 7 players routinely take the flop and 3-4 routinely show down. Some players don't even look until the show-down.
How would you play this game? Piece of cake, eh? Nice and easy, nice and predictable. Play tight and invest after the flop when you have a good chance to win the showdown. Yes, you will show down losers fairly often but so what? If you win 1/3 of the show downs against 3-4 players plus dead money, you are way ahead.
Well, this IS your "wild" and "insane" game, so long as you know its GOING to be capped.
- Louie
You don't read maniacs (unless you're a genius at it). You simply call, fold, or raise based upon your hand, and the presence of others in the pot, compared to their essentially random hand. With multiple maniacs capping it on every street, you don't really play poker. You wait for a hand that it is highly likely to be the best hand going in, and you decide on the flop whether the hand is still good enough to call them down to the river or should be folded now. A game like this is really just all math. And very profitable if played correctly. Also very boring.
Later, Greg Raymer (FossilMan)
I am a full time mathematician and just a recreational poker player. Some posts criticize the use of math in analyzing poker situations. But what else are you going to do, if you actually want to KNOW something? In any case, these questions are interesting (mathematically) in their own right.For example: players A, B, C remain in a tournament and A has a big stack. If A gets into a heads up confrontation with B, who has the best of it? Answer: player C! That's interesting, even to a non-poker player. You might say that this is intuitively obvious, but that would just mean your intuition happens to be right this time --- how could you really know this without actually analyzing the situation?
Here's something even more interesting. There exists an game-theoretically `optimal' poker strategy such that your expected profit is exactly zero (ignoring rake) REGARDLESS OF WHAT YOUR OPPONENTS DO --- there is nothing your opponents could do to throw you off that zeroEV even if they intentionally tried to lose to you. That's really interesting! But how could you know that without high-level math?
Have fun!
Dirk(MildManneredMathMan)
Regarding your second point...
I assume you are referring to von Neuman's Minimax Theorum. That theorum only applies to *2 player* zero-sum games, however. How did you come to that conclusion about a game with more than 2 players?
-Anon
I guess I only had a vague familiarity with the theory, and didn't know what the theorem actually said. Perhaps you could tell me what is known about the case with more than two players. (Assume zero-sum.) In any case, I'm sure that whatever turns out to be true, will also be very interesting, and that that truth could not be discovered without mathematics, which was really my point.
Dirk(MildManneredMathMan)
Sorry, I misread your post; I have no idea what you're talking about. If you play minimax strategy in a 2 player zero sum game, your ev is greater than or equal to 0 (equal to if opponent plays minimax strategy also). I don't know what this exactly equal to zero thing you're talking about is. I am not a mathematician and have only a rudimentary math background.
Anon is absolutely correct. Game-theory strategies only produce the maximum expected value (which, by the way, would be negative because of the rake) that is possible when your opponents are also using the optimum strategy. Also, what your opponents do and the choices they make greater affect your expected payoff as well as yours. If your opponents try to lose on purpose they will definetely lose big time. The concept you were confused with is that in using game theory strategies, there is no way to gain an edge on you (if all players have the same choices) even if your opponents knew exactly what your strategy was down to the letter. In poker, game theory would produce a mixed strategy which would be a strategy based on making decisions with unequal probabilities. There are many good cheap books on the subject published by Dover.
Dirk,
You wrote: "Here's something even more interesting. There exists an game-theoretically `optimal' poker strategy such that your expected profit is exactly zero (ignoring rake) REGARDLESS OF WHAT YOUR OPPONENTS DO --- there is nothing your opponents could do to throw you off that zeroEV even if they intentionally tried to lose to you. That's really interesting!"
You might want to think some more about that claim. Perhaps you are referring to particular sub-strategies within the game of poker. Even in a two-player game, your expected profit can depend very much on what your opponent does (it will be zero, if she also plays optimally). This should be intuitively obvious.
A more accurate statement would be: In a two-player, zero-sum poker situation, optimal game-wide strategy ensures that your EV will not be less than zero. Even this is technically incorrect, unless you assume all potential advantages are equalized (e.g., randomly draw for the button).
Good post. Yes, it would be easy to give your money to a "perfect" player: throw away the good hands, play the bad hands, and don't show anything down.
I think your statement that the perfect strategy "ensures that your EV will not be less than zero" is correct.
- Louie
What Dirk is referring to is game theory strategy where you are indifferent to your opponents actions. I know this applies to certain situations but not sure that it applies to a multiplayer poker game in all situations. When it does apply, your EV using an equilibrium strategy will be 0 for that play, regardless of your opponents play unless he folds after you check and does not show down his cards.
PS: More accurately, if you use the equilibrium strategy, decisions won't matter but your EV may not be 0, depending on the situation you are already in. Assuming the play up to this point in a hand had EV = 0 for both players, then the EV of the equilibrium play would = 0.
As `anon' noted above, what I said does not apply to three or more players. But I am curious as to what the situation actually is.
Dirk(MildManneredMathMan)
Yes, a "n" person game has a solution as well. The two player case was proven in the late 20s I belive (1928 I think) and the n person case was proven in 1944 I think.
-Anon
Your second point, MathMan, is utter hogwash. "But how could you know that" when you are bogged down in high-level math?
If what you are stating is true, apart from the case of simply NOT PLAYING being your strategy, then God could not have created a universe that would support life forms.
I can't explain this thoroughly here, but understand that a "strategy" is an arbitrary simplification or a broad set of policies (as compared to the academic situation of perfect knowledge and anticipation of all possible events, assets, and contingencies). As such, your "optimal" strategy would be INFINITELY COMPLEX, and as such would consume the entire information capacity of the universe, and any quantum trapdoors into nothing, to store and execute.
In real life, there is ALWAYS an infinitesimally more complex strategy which will evolve INEVITABLY to displace yours in the Darwinistic consumption of the assets of the universe (that being entropy).
Love always,
SM
Steve Murray: what part of the word `finite' don't you understand?
I estimate that the number of ways a ten handed Texas holdem game can be dealt is about 10^21. There are 10 positions you could be in. There are much less than 10^24 ways for the betting to go (assuming limit and at most three raises per round, and no unlimited heads-up raising). So there are at most 10^46 possible games. This number may be big, but it is definitely finite.
(If you really want to get into it, all of the probabilities in your strategy would be rational, and therefore finitely describable.)
Of course, no such strategy could not be practically implemented, but that was so completely obvious to everyone else in this thread that no-one else bothered to mention it.
By the way, even the potentially infinite scenarios of unlimited heads-up raising and variable betting are no big drama, and all strategies worth considering would be finitely describable.
Dirk(MildManneredMathMan)
What if you have some knowledge about how an opposing player is likely to play, you smarmy, ivory-tower lack-wit?
Or are you assuming you have perfect knowledge of how an opposing player is likely to play?
AND WHERE ON EARTH DID YOU GET THAT KNOWLEDGE??
Your example is only remotely meaningful if you have zero knowledge or perfect knowledge, neither of which are remotely possible in this universe.
Of course there are a finite number of ways the hands can be dealt, but the inputs into strategies from that point on are infinite. You think I am as dumb as those asses paying $30,000.00 a year to listen to you?
Of course there are a finite number of ways even a heads up game can unfold (assuming you can "know" how many dollars and assets there are in this world which could be pulled into the raises, which you cannot know). And using your hypothetical strategy, there are many threads that never would unfold, and so on, but... Oh, why bother...
Question: You also said they couldn't change your expectation even if they tried to lose to you. How would you know if they were about to try to lose to you, and how would that have any bearing at all? I don't get why you even mentioned that?
Dude! Everyone knows the difference between theory and practice. So what's the big deal. This just happens to be a somewhat theoretical thread. Other posters seem to be perfectly comfortable with this. Just go with the flow, man! Peace!
Dirk(MildManneredMathMan)
But your theory misses the central problem not only of poker, but of life on Earth: information acquisition.
From quantum uncertainty to the following essay,
http://www.virtualschool.edu/mon/Economics/HayekUseOfKnowledge.html
the problem is not what to do when you have the information which, to quote the essay, "is implicit" in the data. Of course!
The problem is getting the data upon which to base your decisions. If it were possible to acquire this data perfectly, the microstructure of empty sapce would smear and collapse, and we would never live to witness it. So....
SM
What if your opponent turns over one card and winks at you?
What if it's a really hot day?
Oh, I get it, you're assuming you're planning on playing some "optimal" strategy, and that your opponent WILL know exactly what you are doing, and yet WON'T be able to beat you? And you will know...
You mean... Wait, now I REALLY get it!
YOU HAVE MISSED THE GAME COMPLETELY.
In fact, you've missed the game of poker so completely I am laughing out loud. I will leave it at that.
Thanks for a good laugh, and see you at the table.
How could you "know" your computer wouldn't crash?
Because I "know" your opponent could compute the probabilities of this happening, and compute a counter-strategy to your probable contingent strategy and, after perhaps only 50 years of play, he'd have ALL your money. ALL OF IT.
Love ya, silly!
What table? As I said, I'm a mathematician, rather than a poker player, and I have these discussions for the fun of it. Why would I sit down at the poker table with you? On the other hand, maybe I should consider it. You wouldn't happen to have a bit of a TILT PROBLEM would you?...
Dirk(MildManneredMathMan)
I respect your opinion that math is not an appropriate model for poker. There are people who don't know the game theory, but have an intuition for the game and the hand reading skills to be a winning player. Winning is more than knowing the math.
However, even if the model isn't appropriate, that doesn't mean that the model is in itself flawed; it's just misapplied. In other words, the math isn't wrong. Mathematical analysis can be done on a poker hand, regardless of whether you think the analysis is enlightening or not. It's the same thing with competing strategies. You can evaluate whether or not a theoretical strategy will win or lose against another. While no one will use such a strategy at the table, there will be people who resemble the strategies. You can use your hand and people reading skills to determine what strategy a person tends to use, and the conclusions drawn from theory might just provide the answer of how to play.
Dirk,
here's a question for you, can you prove that there exists at least one Nash Equilibrium in a nine handed texas hold'em game? assume no rake.
I have no idea about the answer to this question.
Funny, why doesn't the name of this "Nash" guy ring a bell? He must not be feeding too many people's kids.
`can you prove that there exists at least one Nash Equilibrium in a nine handed texas hold'em game? '
As is already, clear, I am not familiar with the literature. Could you give me a reference? By the way, is that an open question?
Dirk(MildManneredMathMan)
As I understand it...
A Nash equilibrium exists if each player’s strategy is optimal when compared to the strategies of the other players. Taking into consideration the strategies of the other players, a player’s optimal strategy is the strategy that will maximize the player’s expected return. In other words, a Nash equilibrium exists when each player makes his or her best response, taking into consideration the responses of the other players. Therefore, a Nash equilibrium in theory can exist in a nine handed game and can proven assuming adequate computational power.
As an aside, I would suspect that a Nash equilibrium is much more likely to occur in tight high-limit games than in loose low-limit games.
William
Okay. That's what I suspected a Nash equilibrium might be. So, does anyone know the answer to Boris's question:
`Is there at least one Nash Equilibrium in a nine handed texas hold'em game? (assuming no rake.)'
This might drive the anti-math guys nuts, but I think this is a really fundamental question to ask about texas holdem, or any other particular game.
Dirk(MildManneredMathMan)
PS: anyone got a simple example of a game with no Nash equilibrium (or at least a reference to the literature).
"`Is there at least one Nash Equilibrium in a nine handed texas hold'em game? (assuming no rake.)'"
Answer: yes
"anyone got a simple example of a game with no Nash equilibrium (or at least a reference to the literature)."
We can create an example. First let’s create a simple Nash equilibrium and then we will modify the example so that no Nash equilibrium exists.
Example 1 (Nash equilibrium):
Suppose I offer the following proposition to two people, Joe and Bill. I explain that they must choose either White or Black and they will be rewarded in the following manner. If both pick White, they will each receive one dollar. If only one of them picks White (it does not matter which one), Joe alone will be given a dollar. If they both pick Black then Bill alone will receive one dollar. They are then given five minutes to secretly decide their choice, write it down on a piece of paper and hand it to me.
A Nash equilibrium exists when neither Joe nor Bill would want to change their choice even after being told of the other parties choice. Therefore, in this example, a Nash equilibrium exists when they both pick White.
Example 2 (no Nash equilibrium exists)
Suppose I now modify the game just slightly. All is as above except if both pick White, then only Bill will receive one dollar.
Now there is no Nash equilibrium. No matter what the original choices are, once the choices are exposed, one of the participants will want to change their choice. Having changed one choice, the other player will then wish to change theirs, which will in-turn cause the first player to want to change again which will... Therefore, there are no choices that are optimum for each player; consequently, no Nash equilibrium exists.
William
I have not seen the definition of a Nash equlibrium, but it would seem that each player should choose a playing strategy that is a probability distribution of choices, rather than a single choice (which corresponds to making a particular choice with probability 100%).
Then, in your example, there would be a Nash equilibrium when each player chooses black or white with probability 50%.
Dirk(MildManneredMathMan)
Hi Dirk,
Actually, there was no probability involved in that example because it is a version of a classic example often used to teach the Nash equilibrium. It has been a while, but as I remember it, a Nash equilibrium with more than two parties involved is also deterministic when, or if, it exists.
William
DEFINITION: Nash Equilibrium… If there is a set of strategies with the property that no player can benefit by changing their strategy while the other players keep their strategies unchanged, then that set of strategies and the corresponding payoffs constitute the Nash Equilibrium.
This definition seems to require a deterministic solution to me. In any case, it does confirm that the examples I gave were correct in that example #1 was a Nash Equilibrium and #2 was not.
William
P.S. That definition came from Drexel. I intended to indicate the source when i wrote the message but didn't, sorry.
William
I just asked one of my colleagues, and he thought the strategies were probabilistic (in general). He also said some games don't have a Nash Equilibrium (although a two-player zero-sum game definitely does). Now I feel I know the definition. I would be surprised if Texas holdem didn't have on but who knows.
Dirk(MildManneredMathMan)
Dirk,
Your colleague may be correct, but the example I provided is a widely accepted way of demonstrating the Nash equilibrium and is deterministic.
From previous:
“DEFINITION: Nash Equilibrium… If there is a set of strategies with the property that no player can benefit by changing their strategy while the other players keep their strategies unchanged, then that set of strategies and the corresponding payoffs constitute the Nash Equilibrium. “
Possibly, I can give an example and then we can use that example to further discuss this topic.
For simplicity sake, assume there are three players who see the river. Player 1 has a pair of Queens, player 2 has a busted flush and player 3 has three Ts. Now we expose the hands and start out search for a Nash equilibrium by making the following test assumptions: player 1 will check, player 2 will check, player 3 check. Now, if you change any of the decisions “no player can benefit by changing their strategy, while the other players keep their strategies unchanged.” Therefore, we have found a Nash equilibrium and it is deterministic in nature, not probabilistic.
Now let’s take a look at a second Nash equilibrium. We make the following test assumptions: player 1 will check and fold, player 2 will check and fold, and player 3 will bet. Again, “no player can benefit by changing their strategy, while the other players keep their strategies unchanged.” This is a second Nash equilibrium in this situation and it too is deterministic in nature.
There may be a third Nash equilibrium in this example, but I can not find it at this time.
I am not saying that the Nash equilibrium has no possible probabilistic applications, but I am saying that its application is purely deterministic in these examples. Possibly, your colleague could provide some supporting reference for his belief that it is probabilistic.
William
My problem at the moment is that I haven't seen a printed definition of Nash equilibrium printed in a book (I will try to look when I get time), so I am in the awkward position of having to speculate as to the definition.
My impression (from asking a colleague) is that strategies are *allowed* to be probabilistic, so a Nash equilibrium exists, given a probabilistic strategy for each player, if no player can increase expectation by changing to another probabilistic strategy (wile the other strategies remain the same).
For example, in the game of paper-rock-scissors it seems to me that the strategy where each player chooses one of the three options independently at random with probability 1/3, is a Nash equilibrium. (And it is probably the only one.)
But when I get a chance to go to the library and pick up a game theory book, I will be more comfortable with this.
Dirk(MildManneredMathMan)
“For example, in the game of paper-rock-scissors (p-r-s) it seems to me that the strategy where each player chooses one of the three options independently at random with probability 1/3, is a Nash equilibrium. (And it is probably the only one.) “
Your phrase “And it is probably the only one” is, in my opinion correct. In any specific p-r-s game, the one-third choice is not always optimal when you are analyzing a specific game outcome with complete information. I do not believe that p-r-s has a Nash equilibrium outcome for any specific game with complete information and I have never seen Nash defined or demonstrated in such a way that it would.
However, we are not talking about the same application of Nash. In your example, you are describing mixed strategy Nash equilibrium with .33333 probability of applying each possible strategy (p-r-s), based upon incomplete information. As my thinking is evolving, I realize that I am describing a dominant strategy Nash equilibrium based upon complete information. After giving your statement some thought, I agree that p-r-s does have a Nash equilibrium if analyzed based upon a mixed strategy/incomplete information model, but at the dominant strategy/complete information level, it does not.
Obviously, I understand that poker is a game of incomplete information, but I understood your question to be asking if a Nash equilibrium could exist. I therefore used the dominant strategy/complete information model in order to determine whether one “could” exist or not.
Once again, yes you can apply Nash as you described. It is my fault for not mentioning (or even giving it significant thought) initially to the fact that I was working with Nash in a complete information environment and assuming a dominant strategy model. I just automatically gravitated to that approach without giving it much thought because of what I thought you were asking.
Please let me know if you find any startling revelations when you go to the library.
Thanks, William
.
Kim, they could fold a million times in a row, but could you "expect" them to fold the million-and-first time?
Meaning, they could give you money, but could you ever "expect" it?
But then, this is the type of discussion you'd expect to be carried on by people who can lose at poker, for amusement, and still eat. I think we should give them a free buffet!
Assume they can only fold if they are bet into. If evryone checks, they must show down.
Dirk(MildManneredMathMan)
Kim's proposal is relevant because in real poker games, Dirk's amendment cannot be applied. For example, before the flop and on 3rd street, respectively, in Hold'em and in 7 card stud, there is a forced bet (respectively, the blind and the bring-in), after which the opponent is not allowed to check, but must either fold, call, or raise.
More generally, in real two player poker games it is impossible to guarantee that your expectation will be exactly zero because optimal strategy exploits (achieves strictly positive expectation) strictly dominated actions by your opponent. By playing optimally, you can only guarantee that your expectation will be non-negative.
Here's another example of how an opponent can play in a manner that forces you to achieve strictly positive expectation if you play optimally: if you are first to act on the river and the optimal play is for you to check and call if your opponent bets, then if your opponent holds the nuts and checks, you will gain one bet compared to what you would achieve against an optimal opponent.
On another note, when playing heads up Hold'em optimally, you are only guaranteed non-negative expectation if you play an even number of hands. If you play a single hand, the game is not symmetric, and your worst-case expectation is almost certainly not precisely zero.
I agree with this post. There were errors in my initial post. What I said did not apply to games with more than two players. But I also now realise that a so-inclined heads-up opponent could force you to have positive (and so, non-zero) expectation, essentially by failing to make cinch plays. This would have followed from the original two player game theory theorem.
Dirk(MildManneredMathMan)
Dirk,
You wrote: "But I also now realise that a so-inclined heads-up opponent could force you to have positive (and so, non-zero) expectation, essentially by failing to make cinch plays."
Technically, your statement is correct. But you still might not understand the point that some of us are making. The "optimal" player still could have a positive expectation even if his opponent always made the cinch plays.
I think it's safe to say that the "optimal" player would have a positive expectation even if his opponent is today's best heads-up poker player. That's because it is unlikely that today's best heads-up poker player plays an "optimal" strategy.
(n/t)
I am extending my thoughts below on game theory with three or more players. (I made an inaccurate comment regarding multiplayer games in my earlier post `MATH IS FUN'. I have not read the literature on this.)
Consider the following three player zero-sum `game,' with three players named U (you), H and T. Each `round' the players simultaneously choose `Heads' or `Tails'. If one player chooses differently from the other two, then that player receive $1 from each of the other two players. If all choose the same there is no payout.
You, player U, need to choose a strategy to maximise your profits. (Not playing is not an option --- you must play.) Now suppose your opponents are very nice to you and tell you exactly how they will play.
H says `I will always choose Heads every round'.
T says `I will always choose Tails every round'.
Then no matter what your strategy, you will lose a dollar every round. (You only get to choose who wins that round.) Note that there is no collusion between your opponents, and they are not `playing well' in any sense. They are simply each playing a preset strategy.
QUESTION: COULD YOU BE IN A POKER GAME LIKE THIS: Each other player at the table has some `playing style' which corresponds to a (possibly probabilistic) strategy. Could it be that there is some combination of playing styles, so that you have negative EV, no matter what your strategy, even if you know all the playing styles of your opponents and you have unlimited barinpower.
Dirk(MildManneredMathMan)
(I had a friend who used to play this game. He and his buddy had an uncanny ability to declare differently, but I digress).
The opponent's strategy is exactly the same as if they WERE colluding. It really doesn't matter if they know it or not. A duck is a duck...
Although I cannot put my finger on it exactly, this game is fundamentally different than poker. The outcomes of this game is based on arbitrary combinations of the opponents actions and not based on any relative "weight" of the opponent's position.
The analogy would be if an opponent deliberately chose a strategy that was BAD for himself but even more to the benefit of one other opponent. Even if they were not sending signals they would still be "colluding". Such a strategy would be to raise player A on the river when he bet into B in a situation where A is much more likely to have the better hand, which is usually the case. Hero's raise is bad for him since he loses one bet but even better for A who wins most of 2 bets. This is colluding even if hero "declares" this to be his strategy ahead of time.
No. Combinations of non-colluding styles cannot cause a "perfect" strategy to fail, although combinations CAN affect one's EV.
- Louie
Thinking about Mason's discussion of non-self weighting strategies it seems to me that this situation could not present itself. This is because you DON'T always have to play.
Paul Talbot
Assuming each player is equally motivated to win, that there is no collusion whatsoever and understanding that I have no choice but to play, my best strategy would be to create a fellow victim. Upon hearing my competitors strategies, I would inform both H and T that I hoped H had deep pockets because I was also going to choose heads each time. It is amazing how quickly fixed strategies can change under the correct circumstances.
Your example is fun to think about, but I do not believe there is a similar possibility in poker because the best hand can win no matter what H and T plan to do.
William
"Upon hearing my competitors strategies, I would inform both H and T that I hoped H had deep pockets because I was also going to choose heads each time."
Won't work if T and H are playing out of the same bankroll (even if they are not colluding in the game). This situation is all too common in real poker.
JQ Writes:
> Won't work if T and H are playing out of the same bankroll > (even if they are not colluding in the game)...
It should not matter if they are using the same backer’s bankroll provided they are only concerned about their own individual profit and there is no compensation from the overall profit of the bankroll. If they shared the bankroll’s profit, then they would be colluding by proxy because they would be motivated not to change their strategy no matter how many times they individually lost in a row. Dirk specifically excluded the possibility of collusion in his original post. The point is that I will force a change in strategy by putting either H or T on the spot because neither H nor T wants to be individually unprofitable.
> ...This situation is all too common in real poker.
Provided there is no collusion: How does backing two poker players from the same bankroll negatively affect a third party’s expectation? It seems to me that the third person’s long-term expected return derives from their individual poker playing skill relative to the two other opponents and not from the source of the opponents money.
William
"If they shared the bankroll’s profit, then they would be colluding by proxy because they would be motivated not to change their strategy no matter how many times they individually lost in a row."
That's not necessarily true. Let's say they are married and one of them just sucks, but the other can't do anything about it (make them stop playing, etc.).
They won't go broke because one spouse's winnings will more than offset the other spouse's losses. This might even be viewed as a justification for the winning spouse putting up with the losing spouse's losses ("He/she loses at poker, but I win even more than he/she loses, so it isn't a big deal.").
JQ, you used referenced the following quote from my previous message:
"If they shared the bankroll’s profit, then they would be colluding by proxy because they would be motivated not to change their strategy no matter how many times they individually lost in a row."
To which you replied:
> That's not necessarily true. Let's say they are > married and one of them just sucks, but the other > can't do anything about it (make them stop playing, > etc.).
JQ, there is not such thing as sucking or not sucking when it comes to picking heads or tails of a coin toss.
> They won't go broke because one spouse's winnings > will more than offset the other spouse's losses. > This might even be viewed as a justification for the > winning spouse putting up with the losing spouse's > losses ("He/she loses at poker, but I win even more > than he/she loses, so it isn't a big deal.").
The problem here is that you have equated my sentence with poker, but it was specifically in reference to the heads/tails game. In the case of the heads/tails game, what I said would most certainly be true. To play against a couple who share in the outcome of their joint bankroll, and therefore have no need to change their individual strategies, would be the same as playing one opponent and giving them two choices for each round. If you gave one opponent two choices then that opponent would be motivated to choose one each of the two possible outcomes and they would therefore be guaranteed a win for each round.
In the case of poker, money from a common source does not necessarily constitute collusion. That is why, in my previous post that, I stated: “It seems to me that the third person’s long-term expected return derives from their individual poker playing skill relative to the two other opponents and not from the source of the opponents money.”
In other words, you are quoting my statement out of context when you associate it with poker rather than with the heads/tails game. Additionally, I did specifically address poker later in the post and I also asked a question which you have not yet answered.
In your previous post you stated:
>> Won't work if T and H are playing out of the same bankroll >> (even if they are not colluding in the game). This situation is >> all too common in real poker.
Which caused me to ask:
> Provided there is no collusion: How does backing two poker players > from the same bankroll negatively affect a third party’s expectation?
Since you did not answer my question, I still do not understand what you mean by “This situation is all to common in poker.” What is wrong with players playing out of the same bankroll if there is no collusion?
Thanks, William
I think you missed the whole point of the H/T game, which was to illustrate a point about real poker.
All of my comments about shared bankrolls, married couples, etc. were referring to real poker, not coin tossing.
JQ Wrote: “I think you missed the whole point of the H/T game, which was to illustrate a point about real poker. “
Actually, the H/T game was used to ask a question about real poker, not to illustrate a point. Here is Dirk’s question before which he used the H/T game to set the stage.
Dirk Wrote: “QUESTION: COULD YOU BE IN A POKER GAME LIKE THIS: Each other player at the table has some `playing style' which corresponds to a (possibly probabilistic) strategy. Could it be that there is some combination of playing styles, so that you have negative EV, no matter what your strategy, even if you know all the playing styles of your opponents and you have unlimited barinpower. “
I am aware of no “constantly negative EV” corollary between poker and the H/T game except that their context specific situational characteristics are mutually exclusive. Further, I do not believe I missed the “whole point”, or any portion thereof. In fact, I responded specifically to Dirk's question in my initial post to him.
JQ, you have still not answered my question about why you believe community poker bankrolls are unfair in and of themselves without collusion being involved. Your husband and wife example did not illustrate anything that I would construe as unfair to a third party. I have sat in many private games and watched one spouse try to cancel out the other spouse’s loses. Although the spouses were playing from a shared bankroll, those games were still perfectly fair.
William
P.S. This is yet another example of why my wife believes that no one likes to talk to me on this forum.
"JQ, you have still not answered my question about why you believe community poker bankrolls are unfair in and of themselves without collusion being involved."
I think this discussion illustrates why they are potentially unfair. It is not a given that such "unbeatable style combinations" exist in real poker, but nobody has shown that they don't either. If they do, then community bankrolls could represent a form of indirect collusion where a style of play which would individually lose becomes a net winner in combination with another player.
JQ,
It seems to me that the community bankroll would not have any effect, provided there was no joint sharing of the overall bankroll profit. However, I do also see that you are making a very good and valuable point about community bankrolls and the likelihood of there being subtle cooperation in and between the community players.
Thanks for the explanation, William
"It seems to me that the community bankroll would not have any effect, provided there was no joint sharing of the overall bankroll profit."
My understanding of playing with a shared bankroll is that people doing it share in profits and losses. Otherwise, they aren't really sharing a bankroll, but are just loaning each other money.
However, I've never done this, so I could be wrong.
If they do share in the joint profit from a common backer then I agree with you completely, that it is a bad thing. However, if they just happen to have a common backer, and nothing more, then I think that is okay.
William
What about a game with 6 other ultra tough players, and three maniacs to your left. (Or maybe change the numbers --- the key point is you have bad position relative to the playing styles.) Could you be doomed in this game?
Dirk(MildManneredMathMan)
What about a game with 6 other ultra tough players, and three maniacs to your left. (Or maybe change the numbers --- the key point is you have bad position relative to the playing styles.) Could you be doomed in this game?
Maybe not on topic for this thread, but I think this is what you want, actually. If the maniacs are on your right, then even when you raise to isolate, tough players will recognize what you're doing and still play good hands.
When the tough players are on your right, then you have position on them. You get to see if any of them has a real hand before taking on the maniacs. It's the tough players who you need to worry about.
If I had three untra tight guys on my right, three maniacs on my left and it was the only game in town, then I would play. I would change my opening hand requirements and my expected standard deviation would go up, but my long-term expectation should still be positive.
William
In the game you described hero has the PRIME position. Since the maniacs are predicable hero would routinely check thus reserving late position on almost all betting rounds. Getting to see what the tough players do is worth much more than providing them with opportunities to toss their inferior hands by raising the maniacs.
- Louie
Does anyone have an example of an arguably unbeatable opponent style combination. Remember, maniacs need not be predictable.
Dirk(MildManneredMathMan)
If I have the nuts on the river in multi-way action and the player on my right bets out, I generally can improve my expectation either by raising (and hoping the bettor calls or re-raises) or by calling (and hoping for overcalls).
If I ignored the differences in these expectations and always raised in this situation, you would have a negative expectation if you sat on my immediate right (assuming I otherwise play optimally and the other opponents always play optimally).
There are many other situations like this as well.
Mark, your example does not illustrate a combination of playing styles and is hand specific. The original question was: "Could it be that there is some combination of playing styles, so that you have negative EV, no matter what your strategy, even if you know all the playing styles of your opponents and you have unlimited barinpower.“
I believe the answer to Dirk’s question is no, but it would indeed be interesting to learn otherwise. If it were otherwise, then strategic collusion would portend a dark future for poker.
William
William,
You wrote: "Mark, your example does not illustrate a combination of playing styles and is hand specific. The original question was: 'Could it be that there is some combination of playing styles, so that you have negative EV, no matter what your strategy, even if you know all the playing styles of your opponents and you have unlimited barinpower.'"
More fully explained, the original question was: "Each other player at the table has some 'playing style' which corresponds to a (possibly probabilistic) strategy. Could it be that there is some combination of playing styles, so that you have negative EV, no matter what your strategy, even if you know all the playing styles of your opponents and you have unlimited barinpower."
I offered an example of a strategy that would yield a negative EV to a person no matter how she played.
You also wrote: "I believe the answer to Dirk’s question is no, but it would indeed be interesting to learn otherwise. If it were otherwise, then strategic collusion would portend a dark future for poker."
I believe the answer to Dirk's question is yes. Fortunately, most colluders whom I've observed are not particularly bright. I do wonder, however, how many colluders have escaped my detection.
Your opponents only had a negative EV for one hand because you had the nuts, not because of your playing style created a negative EV for them in general.
I already understand that a superior hand can create a negative EV for the other players at the table.
William
William,
You wrote: "Your opponents only had a negative EV for one hand because you had the nuts, not because of your playing style created a negative EV for them in general."
Previously, you seemed to have misunderstood Dirk's question. Now, you seem to have misunderstood my example. I will try to make it clearer.
Assume three people are playing. (It works for more, but this helps simplify matters.) Hero is in seat #1, knows her opponents' strategies, has unlimited brain power, and is trying to maximize her expectation. Nasty is in seat #2. Lucky is in seat #3.
Assume there is an "optimal" strategy for a three-player, zero-sum hold'em game. Lucky adopts this optimal strategy.
Nasty also adopts this optimal strategy, except for one modification. If he ever has the nuts on the river and Hero bets (or raises), then Nasty always will raise (or re-raise)--regardless of whether or not Lucky is still involved in the pot.
Do you understand that whatever strategy Hero adopts, that strategy will have a negative EV?
Do you understand that Nasty's strategy also will have a negative EV?
Do you understand that Lucky's strategy will have a positive EV?
Do you understand that the "three-way river action/Hero bets/Nasty has the nuts" scenario should arise multiple times over the course of many, many hands?
Do you understand that even if this scenario occurred only once in a billion hands, Hero's EV still would be negative?
Do you understand that even if this scenario never occurred, the *expected* value of Hero's strategy still would be negative?
Do you understand that there are many additional strategy adjustments that would cause Hero's strategy to have an even worse EV?
Sorry Mark,
Mark, you are still giving specific case examples and not general case examples. Specific case examples are too limited to effectively answer Dirk's question. He was asking a general case question regarding playing styles that, by their specific nature, would create a negative EV for one of the styles.
Let's not waste each others time any further on this one Mark, you are talking apples and I am talking oranges and it is you who does not get the point.
It is interesting how you always try to insult people when you get in over your head or when you discover you are wrong. That’s it Mark, throw up some insults and maybe you can obscure the fact that you were incorrect in the first place.
Let's not waste each others time any further on this one, you are talking apples and I am talking oranges and it is you who does not get the point.
You do have my permission to post one more insulting post in order that you can have the last word. Do not expect me to read that post however.
William
William,
I don't expect you to read this, but it might benefit others who do.
You wrote: "Mark, you are still giving specific case examples and not general case examples. Specific case examples are too limited to effectively answer Dirk's question. He was asking a general case question regarding playing styles that, by their specific nature, would create a negative EV for one of the styles."
Apparently, you still misunderstand my example and Dirk's question.
My example provides a general game-wide strategy that answers Dirk's question about whether an infinitely smart person can be forced to have -EV at a table of three or more players.
You also wrote: "It is interesting how you always try to insult people when you get in over your head or when you discover you are wrong. That’s it Mark, throw up some insults and maybe you can obscure the fact that you were incorrect in the first place."
Pappy Glover: "That's sort of like the toad calling the frog ugly."
I'm sorry if you felt insulted. It's sometimes hard to know at what level to talk to someone who seems to easily misunderstand questions and examples.
Perhaps I am wrong; I'm open to that possibility. But you haven't explained why, and I believe I have explained why not.
You seem awfully sure that I am wrong. You might benefit from being more open minded.
Mark, we will never convince each other of anything, with the possible exception that we are both potentially very insulting people. So let's admit it and get on with our lives.
Best wishes, William
William,
You wrote: "Mark, we will never convince each other of anything, with the possible exception that we are both potentially very insulting people."
I certainly remain open to the possibility that you can convince me of all sorts of things. "I'm not young enough to know everything." (J.M. Barrie)
And while I might not be able to convince you of anything, perhaps Dirk can. I encourage you to read his recent contribution to this thread entitled "Comments on 'How about this example?'"
I also encourage you to keep an open mind while you read Dirk's post. "The human mind is like a parachute--it functions better when it is open." (Cole's Rules)
Mark,
The reason I closed my mind to your comments was because of their style, not because of their content.
I'll try to do better next time.
No hard feelings,
William
Comments on this and the dialogue that followed.
First let's clear one thing up, and sorry if I appear to be taking sides, but you are certainly specifying a *general* strategy or playing style, when you say how to play in some specific situation, and then say `otherwise play optimally'. Moreover you can analyse the sign (+ or - or 0) of overall EV by analysing how play in the specific situation compares to optimal play.
So in the example Mark Glover gave, with payers Hero, Nasty and Lucky (or maybe Lucky1, Lucky2, etc.), the real question is, does Hero have have a counterstrategy against Nasty's non-optimal play, so that Hero's EV is not hurt. For example, if Nasty raises if and only if he has the nuts, and Hero knows this as we assume, then Hero can simply fold whenever Nasty raises on the river.
Note, we are assuming that Nasty's strategy is given, so he does not counteradjust. Also, what I said did not exactly describe Nasty's since he may also occasionally raise as a bluff, making Hero's decision more difficult.
So at this time, I can not tell whether or not Mark gave an example of what I was asking, but it is definitely the right way to go about looking for an example, (that is, consider strategies of the form `optimal except for this small change').
Dirk(MildManneredMathMan)
Dirk,
You wrote: "So in the example Mark Glover gave, with payers Hero, Nasty and Lucky (or maybe Lucky1, Lucky2, etc.), the real question is, does Hero have have a counterstrategy against Nasty's non-optimal play, so that Hero's EV is not hurt. For example, if Nasty raises if and only if he has the nuts, and Hero knows this as we assume, then Hero can simply fold whenever Nasty raises on the river."
I don't know what an "optimal" strategy might be in three-handed hold'em (or even if such a strategy could exist). I would be shocked, however, if it excluded raising (or re-raising) on the river with anything other than the nuts.
You already mentioned that optimal strategy may include the bluff-raise. I would say optimal strategy almost certainly would include the bluff-raise. If I took the time, I probably could prove it.
Other times Nasty might raise with the non-nuts is with the second (or third, fourth, etc.) nuts. If the final board is 2h2dKs/6h/4h, Hero bets out with 2s2c, Nasty raises, and Lucky folds, would a (slightly modified) optimally playing Nasty only raise with 5h3h and not with KhKc? I doubt it.
If raised by Nasty (whose published strategy includes sometimes raising on the river with the third nuts), should an EV-maximizing Hero fold 2s2c in the above scenario? I doubt it. And if Hero should not always fold the second nuts to these kinds of raises, then Hero's overall EV will be negative.
Mark, your comments and analysis, just go to show how complicated things can be, even after trying to simplify.
I agree, that it is not even clear that optimal strategies exist (or that a Nash equilibium exists, whatever that is), but maybe this can be shown. But at the level of trying to give a moderately convincing example (to someone who is not too easily convinced), describing playing styles that are `optimal, except for...' is a good way to go. It's just that you example, even though it narrows the situation down to something more specific, the analysis is still way too complicated for me to convince myself whether or not it works.
I think I will start a new thread above, where the situation is simplified. (It'll take a bit of time to write.)
Dirk(MildManneredMathMan)
Dirk,
I do not feel slighted at all that you feel Mark had a good point. The truth is that he had so worn my out by that point that I had stopped reading his posts.
William
I know I said there was not corollary, but how about this Dirk,
Consider the EV of a passive/loose player sittin
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Posted by: Greg Raymer (FossilMan) (raymers@worldnet.att.net)
Posted on: Tuesday, 1 May 2001, at 2:42 p.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Tuesday, 1 May 2001, at 5:26 p.m.
Posted by: Mark Glover
Posted on: Tuesday, 1 May 2001, at 9:00 p.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Wednesday, 2 May 2001, at 12:13 a.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Wednesday, 2 May 2001, at 12:57 a.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Wednesday, 2 May 2001, at 1:00 a.m.
Posted by: Greg Raymer (FossilMan) (raymers@worldnet.att.net)
Posted on: Wednesday, 2 May 2001, at 2:07 p.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Wednesday, 2 May 2001, at 3:33 p.m.
Posted by: Greg Raymer (FossilMan) (raymers@worldnet.att.net)
Posted on: Wednesday, 2 May 2001, at 4:06 p.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Wednesday, 2 May 2001, at 4:27 p.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Wednesday, 2 May 2001, at 5:58 p.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Wednesday, 2 May 2001, at 6:58 p.m.
Posted by: M (mmmmmm@excelonline.com)
Posted on: Thursday, 3 May 2001, at 2:22 a.m.
Posted by: Vince Lepore (leporeva@hotmail.com)
Posted on: Thursday, 3 May 2001, at 1:13 a.m.
Posted by: Maven (neomaven@hotmail.com)
Posted on: Tuesday, 1 May 2001, at 6:14 p.m.
Posted by: JQ
Posted on: Wednesday, 2 May 2001, at 7:26 p.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Wednesday, 2 May 2001, at 1:57 p.m.
Posted by: Muffin
Posted on: Monday, 30 April 2001, at 6:51 p.m.
Posted by: J-D (johndoe36holdem@hotmail.com)
Posted on: Tuesday, 1 May 2001, at 2:13 a.m.
Posted by: Muffin
Posted on: Tuesday, 1 May 2001, at 12:34 p.m.
Posted by: zzzzzz (ray@shano.com)
Posted on: Monday, 30 April 2001, at 8:04 p.m.
Posted by: Mike (mikedahl@gate.net)
Posted on: Tuesday, 1 May 2001, at 2:25 a.m.
Posted by: Shawn Keller
Posted on: Monday, 30 April 2001, at 11:27 p.m.
Posted by: jim browder (jbrowder@yotalzone.com)
Posted on: Tuesday, 1 May 2001, at 2:24 p.m.
Posted by: Luke
Posted on: Tuesday, 1 May 2001, at 9:11 a.m.
Posted by: Greg Raymer (FossilMan) (raymers@worldnet.att.net)
Posted on: Tuesday, 1 May 2001, at 10:45 a.m.
Posted by: Talbot (talbot@colorado.edu)
Posted on: Tuesday, 1 May 2001, at 11:06 a.m.
Posted by: Russ (rgarber@stanford.edu)
Posted on: Wednesday, 2 May 2001, at 3:59 a.m.
Posted by: Chris Alger (cralger1@home.com)
Posted on: Tuesday, 1 May 2001, at 10:58 a.m.
Posted by: Muffin
Posted on: Tuesday, 1 May 2001, at 1:35 p.m.
Posted by: Luke
Posted on: Wednesday, 2 May 2001, at 5:34 a.m.
Posted by: Jim Brier (jbrier1@msn.com)
Posted on: Tuesday, 1 May 2001, at 3:34 p.m.
Posted by: Sammy Lalonde (sammylalonde@hotmail.com)
Posted on: Tuesday, 1 May 2001, at 4:36 p.m.
Posted by: Jim Brier (jbrier1@msn.com)
Posted on: Wednesday, 2 May 2001, at 1:58 a.m.
Posted by: Sammy Lalonde (sammylalonde@hotmail.com)
Posted on: Tuesday, 1 May 2001, at 2:26 p.m.
Posted by: Jim Brier (jbrier1@msn.com)
Posted on: Tuesday, 1 May 2001, at 3:40 p.m.
Posted by: Talbot (talbot@colorado.edu)
Posted on: Tuesday, 1 May 2001, at 6:49 p.m.
Posted by: Sammy Lalonde (sammylalonde@hotmail.com)
Posted on: Wednesday, 2 May 2001, at 3:16 p.m.
Posted by: Rounder
Posted on: Wednesday, 2 May 2001, at 8:14 a.m.
Posted by: J-D (johndoe36holdem@hotmail.com)
Posted on: Wednesday, 2 May 2001, at 1:02 p.m.
Posted by: Andrew
Posted on: Tuesday, 1 May 2001, at 4:25 p.m.
Posted by: Louie Landale (LLandale@EarthLink.Net)
Posted on: Tuesday, 1 May 2001, at 10:10 p.m.
Posted by: Greg Raymer (FossilMan) (raymers@worldnet.att.net)
Posted on: Wednesday, 2 May 2001, at 2:16 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Tuesday, 1 May 2001, at 6:01 p.m.
Posted by: Anon (zarchan@fas.harvard.edu)
Posted on: Tuesday, 1 May 2001, at 7:28 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Tuesday, 1 May 2001, at 7:55 p.m.
Posted by: Anon (zarchan@fas.harvard.edu)
Posted on: Tuesday, 1 May 2001, at 8:34 p.m.
Posted by: Shollenbarger (expectedvalue@cs.com)
Posted on: Thursday, 3 May 2001, at 8:43 p.m.
Posted by: Mark Glover
Posted on: Tuesday, 1 May 2001, at 8:52 p.m.
Posted by: Louie Landale (LLandale@EarthLink.Net)
Posted on: Tuesday, 1 May 2001, at 10:00 p.m.
Posted by: Russ (rgarber@stanford.edu)
Posted on: Wednesday, 2 May 2001, at 3:51 a.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Wednesday, 2 May 2001, at 10:05 a.m.
Posted by: Anon (zarchan@fas.harvard.edu)
Posted on: Wednesday, 2 May 2001, at 2:47 p.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Wednesday, 2 May 2001, at 1:42 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 10:17 a.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Thursday, 3 May 2001, at 1:21 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 1:39 p.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Thursday, 3 May 2001, at 1:47 p.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Thursday, 3 May 2001, at 1:33 p.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Thursday, 3 May 2001, at 1:39 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 2:18 p.m.
Posted by: Muffin
Posted on: Thursday, 3 May 2001, at 5:11 p.m.
Posted by: Boris (hiboris@hotmail.com)
Posted on: Wednesday, 2 May 2001, at 3:36 p.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Wednesday, 2 May 2001, at 3:39 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 9:48 a.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Thursday, 3 May 2001, at 12:59 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 1:19 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Friday, 4 May 2001, at 1:22 a.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Sunday, 6 May 2001, at 3:09 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Sunday, 6 May 2001, at 9:43 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Sunday, 6 May 2001, at 11:49 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Sunday, 6 May 2001, at 11:52 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Monday, 7 May 2001, at 11:24 a.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Monday, 7 May 2001, at 12:42 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Monday, 7 May 2001, at 1:02 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Monday, 7 May 2001, at 3:55 p.m.
Posted by: Kim Lee
Posted on: Wednesday, 2 May 2001, at 4:23 p.m.
Posted by: Steve Murray (endvolatility@yahoo.com)
Posted on: Wednesday, 2 May 2001, at 5:57 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 9:51 a.m.
Posted by: Paul R. Pudaite
Posted on: Thursday, 3 May 2001, at 3:55 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 4:35 p.m.
Posted by: Mark Glover
Posted on: Thursday, 3 May 2001, at 7:53 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Sunday, 6 May 2001, at 2:32 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Wednesday, 2 May 2001, at 10:49 a.m.
Posted by: Louie Landale (LLandale@EarthLink.Net)
Posted on: Wednesday, 2 May 2001, at 12:48 p.m.
Posted by: Talbot (talbot@colorado.edu)
Posted on: Wednesday, 2 May 2001, at 3:17 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Wednesday, 2 May 2001, at 7:05 p.m.
Posted by: JQ
Posted on: Thursday, 3 May 2001, at 2:03 a.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Thursday, 3 May 2001, at 10:37 a.m.
Posted by: JQ
Posted on: Thursday, 3 May 2001, at 3:41 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Thursday, 3 May 2001, at 5:29 p.m.
Posted by: JQ
Posted on: Thursday, 3 May 2001, at 9:14 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Thursday, 3 May 2001, at 11:15 p.m.
Posted by: JQ
Posted on: Saturday, 5 May 2001, at 2:59 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Saturday, 5 May 2001, at 4:53 p.m.
Posted by: JQ
Posted on: Saturday, 5 May 2001, at 7:19 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Saturday, 5 May 2001, at 11:37 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 8:32 a.m.
Posted by: Muffin
Posted on: Thursday, 3 May 2001, at 10:12 a.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Thursday, 3 May 2001, at 10:16 a.m.
Posted by: Louie Landale (LLandale@EarthLink.Net)
Posted on: Thursday, 3 May 2001, at 12:21 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Thursday, 3 May 2001, at 12:36 p.m.
Posted by: Mark Glover
Posted on: Thursday, 3 May 2001, at 8:29 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Thursday, 3 May 2001, at 11:40 p.m.
Posted by: Mark Glover
Posted on: Friday, 4 May 2001, at 12:37 a.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Friday, 4 May 2001, at 1:36 a.m.
Posted by: Mark Glover
Posted on: Friday, 4 May 2001, at 8:14 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Saturday, 5 May 2001, at 10:21 a.m.
Posted by: Mark Glover
Posted on: Saturday, 5 May 2001, at 12:08 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Saturday, 5 May 2001, at 4:59 p.m.
Posted by: Mark Glover
Posted on: Sunday, 6 May 2001, at 4:13 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Sunday, 6 May 2001, at 10:10 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Sunday, 6 May 2001, at 2:23 p.m.
Posted by: Mark Glover
Posted on: Sunday, 6 May 2001, at 3:54 p.m.
Posted by: Dirk(MildManneredMathMan) (vertigan@math.lsu.edu)
Posted on: Monday, 7 May 2001, at 10:40 a.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Sunday, 6 May 2001, at 10:07 p.m.
Posted by: William Seabrook (calmmaster@yahoo.com)
Posted on: Friday, 4 May 2001, at 1:40 a.m.