David et al

David et al


With regards to your EV question:


Lets forget about the blinds for a moment. This seems to be causing problems and it should really not matter.


A change in EV with seat location is only due to position.. i.e. how many people are ahead or behind you. While my intuition tells me that ones EV should be linear with seat number (if small blind is 1) I can't seem to "prove it." My idea is that one has an constant EV for people they are behind and an constant EV for people they are in front of.


Lets formalize a bit...


People seem to be buying that EV is additive over seat position. Lets further breakup EV like so


EV_(i,j) = expected value when in seat i against seat j


(I leave it to the reader to show that the sum over all i and j gives you your total EV)


I've claimed above that for i j and i>k)


EV_(i,j) = EV_(i,k)


Does anyone have ideas about this????? This assumption can then be used to "prove" my linear proposition.

Honestly, I don't think it matters. I think a good player doesn't really need to quantify via calculations the decreasing value of his hand based on position. What if you actually got an answer? It is really going to change the way you play AK UTG? What about when your opponents are loose-passive vs. tight-passive? Will that go into the EV calculation?


A good player knows that he's more likely to get into trouble with a hand in first position and more likely to win (and win more) when he has it in last position. That's all there is to it. It's that simple.


He then plays accordingly based on his opponents in the game, the mood of the table, any recent image altering plays that have gone down, and so on. There is just no way to properly quantify this via a mathematical calculation because each lineup is different, and even with the same lineup over several hours the game conditions change.


The real edge comes from being able to adjust based on all these other intangibles, rather than consulting a table which tells you your +EV for AK in first position and then acting according to that.


As an intellectual excercise in a vacuum, fine, have at it. But I really think that basing your play on the results of these kinds of calculations can often be detrimental.


natedogg

Your EV in late position increases not only because you will make more with hands you would play in both positions but also because you will play more hands. It is therefore not linear. Another reason it is not linear is that certain good hands are hurt more by being in early position than others.

I believe it is increasing and concave up, that means


EV(i) < EV(i+1) and


EV(i)+EV(i+2) < 2EV(i+1).


The first seems obvious, but can anyone PROVE that it is increasing?


Dirk

An example of increasing concave up would be EV(i)=1, EV(i+1)=1.1, and EV(i+2)=1.3. Thus EV(i)+EV(i+2) [1+1.3=2.3) is GREATER than 2EV(i+1) [2*1.1=2.2).

I didn't understand your references to EV(i,k).


Anyway, you seem to be suggesting that if EV(i)=1 and EV(i+1)=1.1, then the EV(i+2) is (very very close to) 1.2.


I don't think so. The difference between having 1 player behind you and 0 players behind you is MUCH greater than having 7 players compared to 8.


Consider the reasonable opening strategy of playing only hands that have a 50:50 or better chance to be the "best" hand. You would play few hands UTG. You would play a few more in UTG+1 and thus there would be little difference in your overall EV between these positions. But you would play LOTS more hands on the button than in the CO since there are 1/3 less players yet to act. Thus the EV difference between Button and CO is MUCH greater than between UTG and UTG+1.


- Louie

This sounds right. Saying that the difference between adjacent seats increases as you go clockwise is another way of saying that EV(i) is concave-up.


Dirk

David - part of position means that you play more hands - and I agree that individual hand values have different EV in different positions. These statement (with which I agree) does not PROVE the EV is or is not linear. Although I'm feeling like it is not linear now (-:


Can we really ever get precise about this??? Every hand has an EV in each seat. Fine. Only play hands in a seat with a positive EV - fine. In each seat take the probability weighted sum of all positive EVs and get the EV for that seat.


Are we going to agree on EV's for each hand - NOT.


Over and back to work - Hyp

Hyperbolic,


You seem to have adopted the stance that intuitively, EV should be a linear function of position. It just remains to find some elegant proof.


But it appears to me that just the opposite is true. I think it would be absolutely remarkable if EV indeed was a linear function of position. Why should it be linear??


Why should, say, seat five's EV be exactly the mean of the EVs of seat 1 and seat 9??


In fact, it would be difficult enough to prove the much more obvious theorem that EV is a monotonically increasing function of position.


In reality, you can't even isolate EV as a function of the single variable position because of the complex interaction of hundreds of variables that actually exist.


I'll give you a counterexample to your theorem. It is extreme, but it is a counterexample. What if you're in a game that, when you play a hand UTG, everyone behind you calls, but then eveyone folds on the flop no matter what they have. You will have a huge EV in this game in that position. But let's say that in every other position, they play normally.


You can see that in this game, you will have a positive spike in your EV right at the UTG position, which is certainly non-linear.

Obviously, something this extreme will never happen. But it also doesn't take much to make a function deviate from PERFERCTLY LINEAR.


-SDman

Nothing will be exactly linear, but perhaps EV(i,j) will be approximately linear in i and j. This would mean EV(i) would be approximately quadratic in i.


Dirk