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Old 03-19-2012, 10:58 AM   #26
yaqh
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Re: Theory terminology

Thanks Rusty. Could you edit OP to fix the definition of equity (per posts 16-18) and to indicate that the use of the word 'optimal' is controversial?
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Old 03-19-2012, 12:36 PM   #27
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Re: Theory terminology

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Originally Posted by yaqh View Post
Thanks Rusty. Could you edit OP to fix the definition of equity (per posts 16-18) and to indicate that the use of the word 'optimal' is controversial?
I changed the defn of equity, I feel like my wording is a little awkward though.

I added a short para about the origin of GTO = Nash in poker and an explanation that this isn't universal in poker. I moved the paragraph about how optimal != maximally exploitative one para closer to the end also, so that those 2 would be together, since they bookend the discussion a bit.
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Old 03-19-2012, 01:02 PM   #28
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Re: Theory terminology

Looks good. Thanks.
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Old 03-30-2012, 03:49 PM   #29
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Re: Theory terminology

When we first started talking about game theory and poker, there was some literature using "optimal" and some literature using other stuff. I think we picked "optimal" because we and others we knew were pretty heavily influenced by Koller's paper, which used that terminology. I think that saying that "optimal" is nonstandard is kind of wrong, especially in math/computer science applications of game theory. And who cares what economists say anyway?

If I had known that our book would have influence on the way that people ACTUALLY talked about poker as opposed to only in some pipe dream than anyone at all would read it, I would have used "equilibrium" and probably saved everyone a lot of heartache. Alas.
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Old 03-30-2012, 04:00 PM   #30
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Re: Theory terminology

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If I had known that our book would have influence on the way that people ACTUALLY talked about poker as opposed to only in some pipe dream than anyone at all would read it, I would have used "equilibrium" and probably saved everyone a lot of heartache. Alas.
I pretty much agree that what terminology mathematicians use is not really all THAT important, because mathematicians are used to that sort of ambiguity. I remember that in my field there were various schools of thought on terminology (and worse, notation) and you just got used to it.

The real "problem" is that "optimal" has an everyday usage that leads people to believe when they first see it that they understand what's being discussed. But anyway, I agree that it's a good problem to have, in the sense that people actually talk about game theory in relation to poker, rather than not talking about it.
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Old 03-30-2012, 04:05 PM   #31
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Re: Theory terminology

Oh, additionally, von Neumann proved the minimax theorem, which is all that people need to talk about the things in poker they usually associate with game theory. In fact, when people call the jam/fold strategy "Nash" it's kind of funny, since Nash proved the existence of such equilibria for almost every game except this type! (which has been proved earlier).

EDIT: I know that "Nash equilibrium" applies to twoplayer zero-sum games also, I'm just saying. We don't call the identity on the disk a "Riemann map" even though it is.
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Old 03-30-2012, 06:34 PM   #32
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Re: Theory terminology

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Originally Posted by Jerrod Ankenman View Post
If I had known that our book would have influence on the way that people ACTUALLY talked about poker as opposed to only in some pipe dream than anyone at all would read it, I would have used "equilibrium" and probably saved everyone a lot of heartache. Alas.
It's too bad that your use of the sveltely economical "exploitive" was not similarly followed by those inclined to to the bloated -- but, alas, equally dictionary-approved -- "exploitative." That extra syllable, over the course of the next hundred millennia, will consume sufficient electrons to leave the internet a sere, useless utility.
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Old 05-17-2012, 08:09 PM   #33
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Re: Theory terminology

Great work. Very useful.
I would say a "dominated strategy" is playing oop the most of your hands.
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Old 07-20-2012, 08:33 AM   #34
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Re: Theory terminology

Nice instruction. Anyone will be happy to see the simple demonstration.
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Old 07-20-2012, 09:07 AM   #35
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Re: Theory terminology

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Notice that nothing about these definitions implies that the players will break-even on average. However, it turns out that in poker, if all players are playing their equilibrium strategies, they will break even in the long-term average sense when we average over all positions in the game.
I only play/think about heads-up, but it strikes me that this is probably not true in 3+ player games... oops.
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Old 08-24-2012, 05:00 AM   #36
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Re: Theory terminology

I'm new to this subforum and struggling while trying to read it because of unfamiliar terminology. Note that these are terms being used in this >theory< forum and thus I'm asking for clarification in this thread.

Anyway, going back to terminology, what do the following expressions mean:

toy game
villian (I understand it to be "the other player", but why such an evil-connotating term? Does it imply "another player that's playing badly", or "a player I loathe"?)
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Old 08-24-2012, 05:44 AM   #37
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Re: Theory terminology

Hero/villain is a story telling convention - think of the old silent movies and vaudevillian plays: The essence of story telling is conflict, and the basic structure is a "hero" which does NOT mean a virtuous soul, but means the "one we should root for", and the villain, which is the rival of the hero.

In story telling, the hero could be evil, and the villain could be good. For instance, in the film "Heat" you cold say the HERO is the Robert De Niro character - a bank robber. The VILLAIN is the Al Pacino character, the police detective.

Or in a better example, Dog Day Afternoon, the HERO was Al Pacino as the bank robber, and the VILLAIN was Charles Durning as the police detective.
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Old 08-24-2012, 10:04 AM   #38
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Re: Theory terminology

A toy game is basically just a small and simple game that we can solve exactly and otherwise play around with.

For example, the 1/2 street AKQ game: there are two players, the deck has just three cards: an ace, a king, and a queen. High card wins. Both players ante one unit. Hero can bet pot or check and go to showdown. If he bets, Villain can call or fold.

Super simple game, but its solution has some interesting properties (balanced, polarized betting range) that might give us some idea about how to play real poker...
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Old 11-30-2012, 05:28 AM   #39
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Re: Theory terminology

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Originally Posted by yaqh View Post
Quote:
Notice that nothing about these definitions implies that the players will break-even on average. However, it turns out that in poker, if all players are playing their equilibrium strategies, they will break even in the long-term average sense when we average over all positions in the game.
I only play/think about heads-up, but it strikes me that this is probably not true in 3+ player games... oops.
Actually, it is true in zero-sum games. In an n-player game, let e_k be the expectation for an equilibrium strategy in position k, 1<=k<=n. Then

sum_k e_k=0 (due to the zero-sum property),

and thus

(1/n) sum_k e_k =0

which is the expectation of a player who plays each position with the same frequency.
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Old 12-03-2012, 12:07 AM   #40
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Re: Theory terminology

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Actually, it is true in zero-sum games. In an n-player game, let e_k be the expectation for an equilibrium strategy in position k, 1<=k<=n. Then

sum_k e_k=0 (due to the zero-sum property),

and thus

(1/n) sum_k e_k =0

which is the expectation of a player who plays each position with the same frequency.
I think the issue is that e_k isn't well defined here. Unlike in the heads-up case, in a 3+ player 0-sum game, there can be multiple equilibria with different expectations for each position.
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Old 12-03-2012, 02:41 AM   #41
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Re: Theory terminology

Even if this were true (I don't see it, but I don't have a refutation neither; examples always appreciated), we assume that "all players play their equilibrium strategy," so we are at a (fixed) equilibrium. And for each such set of strategies the outcomes are defined.
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Old 12-03-2012, 03:39 AM   #42
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Re: Theory terminology

Actually my reasoning only uses that there is a fixed strategy for each position.
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Old 12-03-2012, 03:14 PM   #43
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Re: Theory terminology

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Actually my reasoning only uses that there is a fixed strategy for each position.
It does, and there's not.
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Old 12-04-2012, 04:17 AM   #44
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Re: Theory terminology

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However, it turns out that in poker, if all players are playing their equilibrium strategies, they will break even in the long-term average sense when we average over all positions in the game.
Ok, then the expression "their equilibrium strategies" does not make much sense in the multiway game. If some players choose strategies from one equilibrium, and others from another one with different expectations, then we're not in an equilibrium anymore and someone can exploit the situation.

Of course you could imagine that when I'm is on the button, then we play equilibrium A, and when I'm UTG we play equilibrium B. Then my argument doesn't work anymore.
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Old 12-04-2012, 10:32 AM   #45
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Re: Theory terminology

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Of course you could imagine that when I'm is on the button, then we play equilibrium A, and when I'm UTG we play equilibrium B. Then my argument doesn't work anymore.
Yea, that's pretty much the situation I had in mind, and I think that can happen in multi-handed play but not in heads-up. Or rather, if there are multiple equilibria in heads-up, they all have the same value.

My understanding of 3+-player poker games is pretty much all from MoP, Ch 29. It does a good job of explaining why they're theoretically interesting but doesn't give much in the way of practical advice or application.
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Old 01-21-2013, 06:17 PM   #46
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Re: Theory terminology

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I think the issue is that e_k isn't well defined here. Unlike in the heads-up case, in a 3+ player 0-sum game, there can be multiple equilibria with different expectations for each position.
Please correct me where I go wrong (starting from whether this terminology thread is the correct place for this...). If there were multiple equilibria with different e_k in a 3-handed game:
-Button gets to choose the strategy first, so he chooses the one s_1 that maximises e_1 against any possible equilibrium strategies the remaining players can pick against s_1.
-SB goes next. If he still has several equilibrium strategies to choose from, he pics the one s_2 that maximises e_2 against BTN and any equilibrium strategies BB can pick from.
-BB, with the information given by previous actions, picks the one s_3 that maximises e_3.

Why does that not lead to predefined e_k for each position? I realize collusion can mess this up: BB picks a strategy that hurts him a bit but hurts SB even more, bleeding the value to BTN. But by definition, BB didn't pick an equilibrium strategy in this case (he hurt himself, so he could have unilaterally picked a better strategy for himself).
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Old 01-23-2013, 12:39 PM   #47
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Re: Theory terminology

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Please correct me where I go wrong (starting from whether this terminology thread is the correct place for this...). If there were multiple equilibria with different e_k in a 3-handed game:
-Button gets to choose the strategy first, so he chooses the one s_1 that maximises e_1 against any possible equilibrium strategies the remaining players can pick against s_1.
-SB goes next. If he still has several equilibrium strategies to choose from, he pics the one s_2 that maximises e_2 against BTN and any equilibrium strategies BB can pick from.
-BB, with the information given by previous actions, picks the one s_3 that maximises e_3.

Why does that not lead to predefined e_k for each position? I realize collusion can mess this up: BB picks a strategy that hurts him a bit but hurts SB even more, bleeding the value to BTN. But by definition, BB didn't pick an equilibrium strategy in this case (he hurt himself, so he could have unilaterally picked a better strategy for himself).
Hm, so if you get to act first in a game, you get to choose an action with a particular hand, but I don't think it's really right/meaningful to say that you don't get to choose a strategy until you act. And it's certainly not that you get to choose an equilibrium, which would imply that you choose other players' strategies as well.

Again, a strategy is something that tells you how to play in any spot you could ever be put in in a game. So you can imagine picking it before even starting to play. And then if you're in a hand, and someone acting before you takes an action, a strategy already has an answer for that.
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Old 01-23-2013, 12:52 PM   #48
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Re: Theory terminology

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Hm, so if you get to act first in a game, you get to choose an action with a particular hand, but I don't think it's really right/meaningful to say that you don't get to choose a strategy until you act. And it's certainly not that you get to choose an equilibrium, which would imply that you choose other players' strategies as well.
Why not? BTN gets to choose the strategy first, he doesn't need to look at his cards before picking one. Say it's shallow stack holdem, so his equilibrium strategy is likely (all in with xx or better, fold the rest). That defines the equilibrium strategies for all players, resulting in fixed e_k, imo. Sit&go players are good with this... and I guess you can use their calculators to find the equilibrium strategies for each player, just set chipEV=$EV.
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Old 01-23-2013, 01:48 PM   #49
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Re: Theory terminology

Consider the following game: Three players, A,B,C. Everybody antes $2 and picks a number 0 or 1. Payouts are as follows: If everybody picks 0, then A gets 1, B gets 2, C gets 3. If everybody picks 1, then A gets 3, B gets 1, and C gets 2. If not all votes are the same, then the majority vote splits the pot. So if A picks 0 while B and C pick 1, then B and C get $3 each.

Here we have two equilibria when all players pick the same number. These equilibria have different values for the players.
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Old 01-23-2013, 05:13 PM   #50
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Re: Theory terminology

They "vote" in the dark? If so, I don't see anyone having the strategic option of "use my equilibrium strategy of picking the same as the others".
If the vote in the order A,B,C, then A gets, well, screwed.
Is there a solution to this game, or can they pick at random (which is a solution in itself)? If there is a solution, I guess it's a mixed strategy?

This is different from, say, 3-handed 1 card poker (or holdem), in the sense that neither a 0 or 1 is inherently the nuts. I'm sure there's a name for this kind of game.

How sweet would a 3-handed stud game be, where after the 7th and all betting is done, the remaining players vote whether this was Stud-Hi or Razz
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