The meaning of expectation value

[eastbay]

In the "winner take all dogma" thread I started, it seems clear that the meaning of expectation value is in dispute. Maybe I'm crazy or stupid, but let me put forth what I think it means, and then you guys can attempt to tear me a new one if I deserve it.

The problem came up in the context of $EV of a certain play in a tournament. What I was saying in that thread, which I believe is strictly correct in one sense, is that almost every hand of a tournament has 0 $EV, because no money changes hands. If you define $EV in terms of the money changing hands, I think that's correct.

But I think it is better to say that the $EV of any play in a tournament that does not win or lose the tournament is indeterminate. It doesn't have one. Not enough information.

To make this more clear, let me ask a question which I think is entirely analogous:

Q: In a ring game, what is the $EV of a preflop play X that does not put you or any opponent all-in? (you can fill in X with anything you like. Saying open-raising QTs 3xBB at a full table from UTG.)

I think the answer is "it doesn't have one." It only has one when you specify how additional streets are to be played. If I fold every flop no matter what comes, that gives it an EV (or more correctly, it gives the hand strategy an EV, not just the preflop play). If I push-in on every flop no matter what comes, that gives it a different EV. Both of those "strategies" have an EV. The preflop play simply does not have an EV unless you complete the strategy for playing the hand.

This is the same reason that I consider any given hand in a tournament, by itself, not to have a $EV. It only takes on an $EV in the context of a strategy for playing the rest of the hands in the tournament. This is why trying to compare chip EV for a hand with $EV for a hand, is a nonsensical endeavor. Because you can't put $EV on a tournament hand, just as you can't put $EV on a preflop play (that leaves opponents with additional decisions).

What you can do is put an $EV on a way to play the whole hand, and you can put $EV on a way to play the whole tournament. Anything less, and $EV is indeterminate.

eastbay

[rodeoclown]

I think what you're forgetting is the key
concept of conditional probability. When you
raise 3xBB with QTs, there is a conditional
probability that arises...
1) everyone folds with probability (1-x)
2) you see a flop
a) you "hit" with conditional probability (y | x)
b) you "miss" with conditional probability ((1-y) | x)
3) you bet when you hit (z | y)
you don't bet when you hit ((1-z) | y)
etc. etc.

To get the true $EV of the play, you need to multiply
all of the conditional probabilities by their respective
payoffs.

So, it might not be easy to compute $EV, but it is there.
When you make a play in a tournament, it has an $EV, not
because money changes hands, but because it changes the
probability that you will win the tournament. For example,
say you know that, when you get to T4000 in a s&g, you win
60%, get second 35% and third 5%. Knowing this, a play
that would put you at T4000 half of the time has a REAL
$EV of .5(.6(first place prize) + .35(2nd) + .05(3rd)), in
addition to any $EV you would have if you managed to
survive the situation and NOT be at T4000. I think it's
pretty common for people to refer to their ROI as nearly
equivalent to $EV when they sit down at a tournament, but
by your argument, there is not enough information to know
whether or not you are going to win, so there can't be $EV.

A further note... even if you're not sure you will make the
money at a certain level, as long as you can estimate your
probabilities, you can compute $EV. For example, say you
have KK on the first hand, and someone goes all-in in
front of you.
You know that, with T1000, your wins are (20%, 20%, 10%)
and with T2000, you are (30%, 25%, 10%). Now you estimate
your win percentage with KK, maybe 60% against a hand that
would go all-in in front of you. Your EV of folding
is (.2(1st) + .2(2nd) + .1(3rd) and the EV of calling is
.6(.3(1st) + .25(2nd) + .1(3rd)) - .4(EV folding).
Note that you need to subtract the opportunity cost of
folding rather than the real loss of the buyin.

In conclusion, it may be difficult, but $EV is ALWAYS
determinate, by definition.

[eastbay]

[ QUOTE ]
I think what you're forgetting is the key
concept of conditional probability. When you
raise 3xBB with QTs, there is a conditional
probability that arises...
1) everyone folds with probability (1-x)
2) you see a flop
a) you "hit" with conditional probability (y | x)
b) you "miss" with conditional probability ((1-y) | x)
3) you bet when you hit (z | y)
you don't bet when you hit ((1-z) | y)
etc. etc.

To get the true $EV of the play, you need to multiply
all of the conditional probabilities by their respective
payoffs.

So, it might not be easy to compute $EV, but it is there.
When you make a play in a tournament, it has an $EV, not
because money changes hands, but because it changes the
probability that you will win the tournament. For example,
say you know that, when you get to T4000 in a s&g, you win
60%, get second 35% and third 5%. Knowing this, a play
that would put you at T4000 half of the time has a REAL
$EV of .5(.6(first place prize) + .35(2nd) + .05(3rd)), in
addition to any $EV you would have if you managed to
survive the situation and NOT be at T4000. I think it's
pretty common for people to refer to their ROI as nearly
equivalent to $EV when they sit down at a tournament, but
by your argument, there is not enough information to know
whether or not you are going to win, so there can't be $EV.

A further note... even if you're not sure you will make the
money at a certain level, as long as you can estimate your
probabilities, you can compute $EV. For example, say you
have KK on the first hand, and someone goes all-in in
front of you.
You know that, with T1000, your wins are (20%, 20%, 10%)
and with T2000, you are (30%, 25%, 10%). Now you estimate
your win percentage with KK, maybe 60% against a hand that
would go all-in in front of you. Your EV of folding
is (.2(1st) + .2(2nd) + .1(3rd) and the EV of calling is
.6(.3(1st) + .25(2nd) + .1(3rd)) - .4(EV folding).
Note that you need to subtract the opportunity cost of
folding rather than the real loss of the buyin.

In conclusion, it may be difficult, but $EV is ALWAYS
determinate, by definition.


[/ QUOTE ]

I think we are mostly agreeing with one another, but I disagree with your conclusion. Here's why.

I think expectation value averages over the stochastic variables in the problem, but not the non-stochastic variables. The reason it doesn't is because there aren't any probabilities associated with the non-stochastic variables.

The non-stochastic variables are the actions you are going to take depending on context, i.e., your strategy.

So maybe your strategy is going to be probabilistic in some sense, and that's fine, you can include that in the caculation, given the strategy.

But if you haven't specified the strategy, how do you specify a probability of taking action X? You can't, and that's why you can't complete the calculation. And that's why the EV is indeterminate.

Conversely, if you've completed the calculation, you've filled in numbers for the P_n associated with your actions, and therefore you've specified a strategy for your play. And that EV you've got is relative to that strategy, i.e, those P_n. There is no EV associated with the play by itself; only that play combined with the rest of the strategy for the hand. That's why to speak of the "$EV of that move" is nonsense. What you really mean is the $EV of a strategy that contains that move. But there's an infinite number of strategies which contain that move, so to speak of it in isolation is meaningless.

Likewise for tournaments.

eastbay

[CrisBrown]

Hiya eastbay,

Think about how deal calculators work. There are several mathematical models, and they differ slightly, but they all assign a given relative value for each stack ... assuming equal skill and equivalent opportunities for the remainder of the tournament. In theory, at any point in a SNG, you could compute the fraction of the prize pool to which you'd be entitled if the tournament were to end immediately (i.e.: your share of a deal).

Given enough time (or a very slick deal calculator), you could also compute:

* The prize pool fraction to which you'd be entitled if you fold;
* The prize pool fraction to which you'd be entitled if you call and win; and,
* The prize pool fraction to which you'd be entitled if you call and lose.

You could then compute the probabilities for winning and losing if you call, and establish the $EVs for calling and folding.

The existence and accepted use of deal calculators seems to disprove the notion that only the final hand has any $EV. You don't win or lose the money immediately, but each pot won or lost affects your likelihood of winning real money.

So what is the relationship of $EV to CEV? I don't have a deal calculator, and I don't know how they work. Others with more experience and better mathematical skills than me have suggested a CEV:$EV ratio of ~11:10 early on, and ~7:5 on the bubble. I'm willing to take their word for it until someone proves otherwise.

Cris

[Utah]

I think the answer is "it doesn't have one." It only has one when you specify how additional streets are to be played. If I fold every flop no matter what comes, that gives it an EV (or more correctly, it gives the hand strategy an EV, not just the preflop play

There is a correct strategy for each respective player for the entire hand. You can use game theory to define those exact strategies (lookup minimax theorem and the principle of indifference).

This will give you a theoretical EV for the hand assuming each player plays according to best strategy and it will take into account both the stocastic and non-stocastic.

For example (using a NL ring game 5/10 blinds with all players having $1000 - to keep it simple), receiving A,Ks in 7th position with everyone folding to you has an exact theoretical EV and it should be played with an exact strategy assuming everyone plays correctly (if everyone doesn't play correctly then the EV can only rise or stay the same for the players playing correctly).

See the post last may called - "way tougher two round game theory problem" - for an example

[eastbay]

[ QUOTE ]
Hiya eastbay,

Think about how deal calculators work. There are several mathematical models, and they differ slightly, but they all assign a given relative value for each stack ... assuming equal skill


[/ QUOTE ]

I think that's a non-starter of an assumption if one is trying to determine how to play. Correct play is a strong function of your opponents' play. I don't think that can be in dispute. There aren't entire chapters in HPFAP on "loose games" for no reason.

Moreover, the difference in skill between you and your opponents is probably the dominant factor in deciding how much to weigh risk in your decisions in a tournament setting. For the reason everyone intuitively understands: you can pass on small edges now if your opponents will hand you bigger ones soon.

So anything based on that assumption I consider very questionable if not useless if it is being used to try to determine how to play. There may be other questions that could be asked for which it is a reasonable assumption.

eastbay

[eastbay]

[ QUOTE ]
I'm willing to take their word for it until someone proves otherwise.

Cris

[/ QUOTE ]

Just to be obnoxious:

"The fact that an opinion has been widely held is no evidence whatever that it is not utterly absurd; indeed in view of the silliness of the majority of mankind, a widespread belief is more likely to be foolish than sensible."
-- Bertrand Russell

eastbay

[eastbay]

[ QUOTE ]
I think the answer is "it doesn't have one." It only has one when you specify how additional streets are to be played. If I fold every flop no matter what comes, that gives it an EV (or more correctly, it gives the hand strategy an EV, not just the preflop play

There is a correct strategy for each respective player for the entire hand. You can use game theory to define those exact strategies (lookup minimax theorem and the principle of indifference).

This will give you a theoretical EV for the hand assuming each player plays according to best strategy and it will take into account both the stocastic and non-stocastic.

For example (using a NL ring game 5/10 blinds with all players having $1000 - to keep it simple), receiving A,Ks in 7th position with everyone folding to you has an exact theoretical EV and it should be played with an exact strategy assuming everyone plays correctly (if everyone doesn't play correctly then the EV can only rise or stay the same for the players playing correctly).

See the post last may called - "way tougher two round game theory problem" - for an example





[/ QUOTE ]

You've presented one way to derive a strategy so that EV can be calculated. So what?

eastbay

[Utah]

You've presented one way to derive a strategy so that EV can be calculated. So what?

No, I did not derive a strategy so that EV can be calculated. I am telling you there is an inherent EV - which is in direct contradition to your statements. Therefore, your hypothisis that EV is indeterminate is 100% wrong.

However, if you are going to be a smart ass I will go no further and I will not respond to your posts anymore.

[Guy McSucker]

I'll give this a try. In my world, the concept of EV mid-tournament works out this way.

At any given point in the tournament, there is (in theory) a certain probability that you will eventually come first, second, third etc. This is a function of your stack size, the other players' stacks, your skill levels, and so on. Darned hard to know what it is, but I think it's fair to say that there is such a number.

Multiplying these probabilities by the payouts and adding them all up, you get your overall EV for the tournament; many call this your equity.

A given hand changes the stack sizes, and therefore everyone's equity. The difference between your equity before and after the hand is a dollar amount which you have "won" or "lost" that hand.

During the play of a hand, you make decisions which will affect the chip EV for that hand. To get them really right, you need to work out not how these chips that you will win or lose will affect your stack, but your equity. And that's how you work out the dollar EV of a given play.

For fun, consider a case where everyone is equally skilled, so the only factor in the equity calculation is stack size. If it's winner take all, the only thing you need to calculate is chance of winning. As Craig said, it's generally accepted that this chance is proportional to chip stack, under these conditions. Hence dollar EV is linearly related to chip EV in this case.

However, if you are better than your opponents, then your chance of winning with a 1000 stack is better than anyone else's, so an individual chip is worth more to you than it is to your opponents. I don't know, but it seems likely, that even in this case the relation of chip stack to equity is linear, so chip EV is still the way to go.

In a proportional payout tournament, the fact that there are some payouts to lower places means that chips are not linearly related to equity. E.g. if one player has 10k chips and eight others have 100 each, the leader will not have 100x the equity of the others, since it's impossible to win more than, say, 40x what everyone is currently guaranteed. So in this situation, equity and chips are not linearly related, and you have to be careful when calculating hand EV.

Does that make any sense at all?

Guy.

[eastbay]

[ QUOTE ]
You've presented one way to derive a strategy so that EV can be calculated. So what?

No, I did not derive a strategy so that EV can be calculated. I am telling you there is an inherent EV


[/ QUOTE ]

If you want to attach the word "inherent" to an EV associated with your game-theory derived strategy, you're welcome to do that. But the point is that you have to have a strategy for playing the rest of the hand to calculate the EV. You've chosen one based on game theory. I could choose one based on something else. But any EV is associated with some such choice. Correct?

[ QUOTE ]

- which is in direct contradition to your statements. Therefore, your hypothisis that EV is indeterminate is 100% wrong.


[/ QUOTE ]

I disagree. I think you're trying to play some kind of word game by calling the EV associated with your strategy "inherent." I don't see what is "inherent" about it. It's simply one possible strategy of many, with its own assumptions and flaws (that your oppponents "play correctly", etc.)

eastbay

[AleoMagus]

I have been reading all of the posts in this thread, and there is a question that keeps popping up in my mind.

What are you trying to say?

Lets suppose you are right. Lets suppose that all but the last hand in a tourney has zero EV. What would this imply about how I should play? Can I only rationally make decisions based upon chip EV?

Meaningful decisions could only come from chip EV?
...This is the answer that keeps popping into my head and that answer more than anything is what makes me think your reasoning is flawed.

I can accept that $EV is sketchy info in the middle of a tourney, but to suggest that it is completely indeterminate or meaningless altogether is something I find highly unlikely.

There are many times that we all find ourselves folding good hands the bubble or when two large stacks push in against one another. Why do we do this? Is it because we have negative chip EV? No, it is becasue by avoiding confrontations where we can move ahead in the tourney, we are making positive $EV plays by folding.

Perhaps I am missing something. I think by suggesting that there is no $EV, you will run into some bizzare conclusions.

Or are you making a more subtle point about how hard it can often be to calculate $EV?

Regards,
Brad S

[AleoMagus]

[ QUOTE ]
I could choose one based on something else. But any EV is associated with some such choice. Correct?

[/ QUOTE ]

Sure. Choose away

[ QUOTE ]
I disagree. I think you're trying to play some kind of word game by calling the EV associated with your strategy "inherent." I don't see what is "inherent" about it. It's simply one possible strategy of many, with its own assumptions and flaws (that your oppponents "play correctly", etc.)

[/ QUOTE ]

I don't think this suggests that EV is inherent in the game theoretic strategy assumption. I think that this suggests EV is inherent becasue you can always make a strategy assumption (whatever it is) and that each assumption will have EV associated with it.

The point is, EV is determinate. With no other info, I'll take the Game theory EV calculations based on perfect play. If my opponent is a total idiot, I'll come up with other strategy assumptions, but I CAN draw conclusions about $EV that will affect my decision making process.

But maybe I can never make a perfect assumption? How do I know what assumptions about strategy to make? Wouldn't this mean that $EV is indeterminate?

If you want to go down that road, be my guest, but it is a very slippery slope.

Regards,
Brad S

[eastbay]

[ QUOTE ]
[ QUOTE ]
I could choose one based on something else. But any EV is associated with some such choice. Correct?

[/ QUOTE ]

Sure. Choose away

[ QUOTE ]
I disagree. I think you're trying to play some kind of word game by calling the EV associated with your strategy "inherent." I don't see what is "inherent" about it. It's simply one possible strategy of many, with its own assumptions and flaws (that your oppponents "play correctly", etc.)

[/ QUOTE ]

I don't think this suggests that EV is inherent in the game theoretic strategy assumption.


[/ QUOTE ]

Take it up with the other poster. That's what he said.

[ QUOTE ]

I think that this suggests EV is inherent becasue you can always make a strategy assumption (whatever it is) and that each assumption will have EV associated with it.


[/ QUOTE ]

So you agree with me.

[ QUOTE ]

The point is, EV is determinate. With no other info, I'll take the Game theory EV calculations based on perfect play. If my opponent is a total idiot, I'll come up with other strategy assumptions, but I CAN draw conclusions about $EV that will affect my decision making process.

But maybe I can never make a perfect assumption? How do I know what assumptions about strategy to make? Wouldn't this mean that $EV is indeterminate?

If you want to go down that road, be my guest, but it is a very slippery slope.

Regards,
Brad S

[/ QUOTE ]

I don't see any slope. I never asked for a "perfect assumption" or any such thing. I wouldn't even contend one exists. What I did try to get people to recognize is that an incomplete strategy (for a hand, or a tournament) does not give an EV. Only a complete strategy does (complete in the sense of determining all future actions required to complete the hand, or the tournament). Choose it however you like, just don't tell me that the incomplete strategy is enough to determine EV. It is not. This is what many are insisting. You're not, however.

eastbay

[eastbay]

[ QUOTE ]
I have been reading all of the posts in this thread, and there is a question that keeps popping up in my mind.

What are you trying to say?

Lets suppose you are right. Lets suppose that all but the last hand in a tourney has zero EV. What would this imply about how I should play? Can I only rationally make decisions based upon chip EV?

Meaningful decisions could only come from chip EV?
...This is the answer that keeps popping into my head and that answer more than anything is what makes me think your reasoning is flawed.

I can accept that $EV is sketchy info in the middle of a tourney, but to suggest that it is completely indeterminate or meaningless altogether is something I find highly unlikely.

There are many times that we all find ourselves folding good hands the bubble or when two large stacks push in against one another. Why do we do this? Is it because we have negative chip EV? No, it is becasue by avoiding confrontations where we can move ahead in the tourney, we are making positive $EV plays by folding.

Perhaps I am missing something. I think by suggesting that there is no $EV, you will run into some bizzare conclusions.

Or are you making a more subtle point about how hard it can often be to calculate $EV?

Regards,
Brad S

[/ QUOTE ]

What I was originally asking about is the meaning of so-called "expectation value" of an incomplete part of a wager. That incomplete action might be a single bet in a hand of ring hold 'em, or it might be a single hand in a tournament.

What has become crystal clear from the ensuing discussion is that, by itself, those incomplete actions aren't enough to determine an EV. So that a certain raise with certain cards, for example, has no single $EV associated with it, without making further decisions about how the rest of the hand is to be played. Only by specifying the rest of the actions needed to complete the outcome of the wager, can you determine an EV associated with that action. And even then, it will be relative to those other actions. The $EV could be wildly different depending on the other actions that must be chosen to complete the calculation.

That's it. It's pretty simple.

(From some other discussion, someone asked me to state a hand in a tournament for which some action's chip EV is not the same as $EV, or something like that. I stated that one such hand was not enough to give a $EV. I stand by that for the reasons given.)

eastbay

[Cosimo]

EV means Expected Value, not Expectation Value.

During the middle of a tournament, each player has equity in the prize money that depends on their stack, their skill, and other factors. Because this prize money is cash, the equity also has a dollar value. Because the decisions made during the tourney affect the player's stack (and therefore their equity), each option they are faced with can also be assigned a dollar value. Hence, EV is expressed in dollar units.

EV is an imprecise measure used to influence choices that are somewhat dependent on chance. S&M give countless examples of EV calculations throughout many of their books. Almost all of these examples start with "assume that you know that your opponent will always and only make this move if he holds AA, KK, or AKs." This, of course, is an unreasonable assumption, but it's made to simplify the discussion. At the end of the math, S&M conclude with "the EV of this play is X." The concept of EV hence has a very practical value.

In addition to EV, other variables are also used to decide whether to make a play or not: eg if a better situation will come along later, if there are other SnGs that you could sit at, and if getting the bracelet from this event would get you more book sales. Considerations, such as (in sit-n-goes) whether $/tourney or $/hr are of primary importance to the player, also change the set of equations that are relevant for a particular decision.

I recall (possibly imperfectly) that S&M (&Z, etc) don't explicitly distinguish the EV calculated from the theoretical model with the "real, factual" EV of the same play in a real, flesh-and-blood game. In the forum, most posters are aware that these mathematical models provide guidelines for assessing play; they are not calculations of "true EV". Indeed, the notion of "true EV" is meaningless except as a theoretical exercise, never discussed outside of these self-indulgent philosophical threads. Note that I reject the analytic/synthetic dichotomy.

[eastbay]

[ QUOTE ]
EV means Expected Value, not Expectation Value.


[/ QUOTE ]

Synonymous. See, for example: http://mathworld.wolfram.com/ExpectationValue.html


[ QUOTE ]
In the forum, most posters are aware that these mathematical models provide guidelines for assessing play; they are not calculations of "true EV". Indeed, the notion of "true EV" is meaningless except as a theoretical exercise, never discussed outside of these self-indulgent philosophical threads. Note that I reject the analytic/synthetic dichotomy.


[/ QUOTE ]

I'm not sure what exercise of "true EV" you're talking about, except possibly by the guy who said game-theory derived strategies provide "inherent" EV's. I of course disagreed with him on this point.

eastbay

[PrayingMantis]

[ QUOTE ]
What has become crystal clear from the ensuing discussion is that, by itself, those incomplete actions aren't enough to determine an EV

[/ QUOTE ]

That was obviously "crystal clear" to *you*, before any discussion took place. There are so many faulty logical arguments in your reasoning, here and in other posts, that there's a feeling you're either high on some serious drugs, or that you are heading in some very very strange way - that might actually lead to something very interesting, only I can not see how, and in what poker universe.

Even from this one sentence above - "those incomplete actions aren't enough to determine an EV " - one *can not* conclude that $EV does not exist for a specific decisions on a particular hand. All you can conclude is that there's not enough information to do so *accurately*, a fact we all agree upon.

Claiming that there's not a way to determine $EV for a specific move, because all future action is not known, hence $EV=0, is completely nonsensical. It will lead you, pretty smoothly, to the conclusion that there's not such a thing as $EV. And I think that's where you're heading.

Good luck in your quest,

PrayingMantis

[eastbay]

[ QUOTE ]
[ QUOTE ]
What has become crystal clear from the ensuing discussion is that, by itself, those incomplete actions aren't enough to determine an EV

[/ QUOTE ]

That was obviously "crystal clear" to *you*, before any discussion took place.


[/ QUOTE ]

Well thanks for letting me know what my own thoughts are. But actually it wasn't. I wasn't clear one way or the other. Now it is.

[ QUOTE ]

There are so many faulty logical arguments in your reasoning, here and in other posts, that there's a feeling you're either high on some serious drugs


[/ QUOTE ]

Care to point out just a single one? I'm not being facetious or a wise-ass, and I'll take it on good faith that you aren't either, for the time being.

eastbay


PS You're confusing two different things when you say that I conclude $EV=0 since future actions are unspecified. If you're interested in differentiating those two things, I can talk you through it. But my guess is that you're not, so I won't.

[AleoMagus]

[ QUOTE ]
Note that I reject the analytic/synthetic dichotomy.

[/ QUOTE ]

Go Quine! WooHoo!

Regards,
Brad S

[PrayingMantis]

From your original post in this thread:

[ QUOTE ]
But I think it is better to say that the $EV of any play in a tournament that does not win or lose the tournament is indeterminate. It doesn't have one. Not enough information.


[/ QUOTE ]

The argument here is the basis for your whole reasoning. It goes like this: in order for $EV to exist, it needs *enough* information. If there's not *enough* information, $EV does not exist. What is *enogh* information? Enough information, is the "amount" of informatoin, needed for $EV to exist. $EV exists? no. Conclusion: there's not enough information. Thus: $EV does not exist. End of argumentation.


PrayingMantis

[eastbay]

[ QUOTE ]
From your original post in this thread:

[ QUOTE ]
But I think it is better to say that the $EV of any play in a tournament that does not win or lose the tournament is indeterminate. It doesn't have one. Not enough information.


[/ QUOTE ]

The argument here is the basis for your whole reasoning. It goes like this: in order for $EV to exist, it needs *enough* information. If there's not *enough* information, $EV does not exist. What is *enogh* information? Enough information, is the "amount" of informatoin, needed for $EV to exist. $EV exists? no. Conclusion: there's not enough information. Thus: $EV does not exist. End of argumentation.


PrayingMantis


[/ QUOTE ]

LOL! Never mind... how 'bout them Patriots?

eastbay

[PrayingMantis]

[ QUOTE ]
how 'bout them Patriots?


[/ QUOTE ]

I'm actually more into European soccer, but I guess them Patriots are just fine too.

PrayingMantis

[AleoMagus]

[ QUOTE ]
What I did try to get people to recognize is that an incomplete strategy (for a hand, or a tournament) does not give an EV.

[/ QUOTE ]

That may have been your intent, but it sure seemd like you were trying to say a lot more than this. Let me give you a few examples:

[ QUOTE ]
What I was saying in that thread, which I believe is strictly correct in one sense, is that almost every hand of a tournament has 0 $EV, because no money changes hands.

[/ QUOTE ]

[ QUOTE ]
the $EV of any play in a tournament that does not win or lose the tournament is indeterminate. It doesn't have one. Not enough information.

[/ QUOTE ]

[ QUOTE ]
trying to compare chip EV for a hand with $EV for a hand, is a nonsensical endeavor.

[/ QUOTE ]

[ QUOTE ]
I consider any given hand in a tournament, by itself, not to have a $EV.

[/ QUOTE ]

These are the kinds of phrases that all the responses in this thread are rallying against. Just because you backed off on your original stance (or missworded it) does not mean that these respones are now wrong.

Even now, on the one hand you seem to make a very uncontroversial point about how the EV is defined by the playing strategies we assume will be employed throughout the hand etc, etc... THEN, you make the logical leap to 'All tournament hands have zero EV' or 'All tournament hands have indeterminate EV'.

They don't have zero EV, and they don't have indeterminate EV. Just because you need to specify strategy assumptions to determine EV doesn't mean it's not there.

Regards,
Brad S

[eastbay]

[ QUOTE ]
[ QUOTE ]
What I did try to get people to recognize is that an incomplete strategy (for a hand, or a tournament) does not give an EV.

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That may have been your intent, but it sure seemd like you were trying to say a lot more than this. Let me give you a few examples:

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What I was saying in that thread, which I believe is strictly correct in one sense, is that almost every hand of a tournament has 0 $EV, because no money changes hands.

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the $EV of any play in a tournament that does not win or lose the tournament is indeterminate. It doesn't have one. Not enough information.

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trying to compare chip EV for a hand with $EV for a hand, is a nonsensical endeavor.

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I consider any given hand in a tournament, by itself, not to have a $EV.

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These are the kinds of phrases that all the responses in this thread are rallying against. Just because you backed off on your original stance (or missworded it) does not mean that these respones are now wrong.

Even now, on the one hand you seem to make a very uncontroversial point about how the EV is defined by the playing strategies we assume will be employed throughout the hand etc, etc... THEN, you make the logical leap to 'All tournament hands have zero EV' or 'All tournament hands have indeterminate EV'.

They don't have zero EV, and they don't have indeterminate EV. Just because you need to specify strategy assumptions to determine EV doesn't mean it's not there.

Regards,
Brad S







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There's actually two issues in here, and they were confused for a time, and that's probably my fault.

The first issue relates to my $0 EV statements. That issue is related to what people mean by the EV of an action. Taken literally, I stand by my statement that most actions in a tournament, or single bets in a hand, have $0 EV. Because the actions themselves do. That's simple application of definition. But the problem is that when people say that, it's not really what they mean.

What I came to realize is that people don't mean the EV of the action literally. What they mean is the EV of a series of actions for which that one is a single part. That seeems clearly to be the way people talk about these things, and I wasn't aware of it.

Taken in that sense, the EV is then dependent on the rest of the actions (what I've been calling a strategy), which are often unstated. Then the resistance came to the idea that without the rest of actions, you can't get an EV. I'm still curious why people are so put off by that notion, which seems pretty simple and obvious.

I heard then "I take my assumptions from game theory" or "I take my assumptions however I like", etc. That's fine. Make them how you like. Just don't try to convince me that something else doesn't have to be decided to end up with a $EV. It clearly does. So I will still reject the idea that someone can say "action X in a tournament has Y $EV" and have it mean much of anything without stating what the other factors in that playing strategy are.

Now, if we want to get into a shades-of-gray argument about how different the $EV would be depending on those (almost always) unstated assumptions, well then I think there's not much left to do here. You can say "not much" and I can say "more than you think, maybe" but it won't accomplish much.

eastbay

[Guy McSucker]

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What has become crystal clear from the ensuing discussion is that, by itself, those incomplete actions aren't enough to determine an EV


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Didn't my post answer that to some extent, at least for the case of the EV of a single hand in a tournament?

Seems to me that everyone else is pretending it isn't there.

Guy.

[Stagemusic]

This has been a fun read. I enjoyed the circular arguments more than I ever thought that I would. /images/graemlins/tongue.gif IMO Eastbay is getting caught up in symantics to the nth degree. Of course every action has an EV and that EV is dependent upon the actions taken before or after. That only makes common sense. However, we do not recognize the dependence factor when estimating the EV of a particular action. We know because of history that certain actions have a +EV and certain actions have a -EV over the course of a given period of time. The fluctuations in this EV gives us the average. So to say that EV is static would be incorrect. It must be a variable or there would be no average. So when it is said that it is -EV to call 3 cold pre flop while holding 7 /images/graemlins/heart.gif 2 /images/graemlins/club.gif we are actually saying that more times than not we will lose the money invested, or that hand is -EV. This number is usually expressed as a % based on the number of times encountered.

The idea of EV is really very simple but I think we try to make it more complicated because it makes us look smart. To me EV is simply, on any given move, do I have the right combination of cards, pot, stack, and position that makes this move an OVERALL positive or negative one? If I make this move 100 times in a row am I more likely to win more money than I lose more often than 50 times?

Of course, I am a simple person and therefore not very bright. /images/graemlins/grin.gif

[CrisBrown]

Hi Guy,

I wasn't pretending it wasn't there; it just seemed to be a lot like what I'd written so I didn't answer. I agree that you basically answered eastbay's question: while it's difficult if not impossible to give an exact value for $EV in a given tourney situation, stack size is a major factor.

Obviously, survival is also a factor. However small your chances of winning with a short stack, they're better than your chances of winning from the rail (which are nil). So you'd rather avoid marginal all-in plays. But I think this same logic leads to another, subtler point: if folding your hand would leave you so short-stacked that you'll be forced to move all-in on any hand you play, you're probably better off taking your chances here, and especially if winning the pot would give you a solid stack.

E.g.: You have AK UTG, six-handed (12 left in a two-table SNG), where the stacks are all roughly 2200, and blinds are 100/200. You raise 3xBB to 600. MP pushes, BB calls. Now what?

You're probably a dominant favorite to one of them (a lower Ace), and a coin flip to the other (a pocket pair). And at least one of your outs is probably gone. If you win this pot, you'll have T6700, putting you in a strong position to win the SNG. If you call and lose, you'll be out. It's a marginal all-in situation for you.

But if you fold, you're down to 7xBB, with blinds hitting you on the next two hands, and in all-in-or-fold territory with your stack.

Right now, you have better than 3:1 pot odds, and you are probably at worst a 3:1 underdog (e.g.: AK vs. AQ vs. JJ: AK wins 32%). That's a very marginal +CEV. If you fold this, you're likely to face at least two more marginal all-in situations before you can get back in the chase, and very possibly with lower EV than you have here.

To me, this is a clear call, and I would only fold if I were sure I was a vastly superior player to the others left in the game (which is rarely if ever true). I may be gone on this hand, but I'd rather die on AK with a marginal +CEV here than on A5s or some such with -CEV three hands later.

Cris

[Guy McSucker]

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I wasn't pretending it wasn't there; it just seemed to be a lot like what I'd written


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Yes, it's very similar I think. I was just spelling it out a bit.

Guy.