Maxaon, its better to follow the proof of the second post because its simpler and doesn't assume the person plays many many games during their lifetime. It works even if played only once. Regarding sigma if this means standard deviation then it doesn't decrease, it keeps increasing as n^(1/2) but the difference of the sum of the max ev always minus the alternative strategy grows like n .So in effect the difference becomes as big as you want it with respect to the standard deviation of the strategies difference . So my point is in the end the difference of the two is always positive (luck is irrelevant) and this makes the max EV sums always clearly larger than all else in the large n limit (actually it makes it as close to 100% as you want it only at infinity it is 100%) . The relation between that difference of the 2 sums and the sd is critical in claiming that the one strategy is better than all alternatives because it tells you that in the end luck is irrelevant, the best strategy will be ahead by a lot more than the typical standard deviation. That argument renders it safely a better strategy.

But focus more on the second post that requires almost no extra assumptions about what the guy does in his life .


Jewbinson ; I made no unnecessary assumptions on post n2 as you claimed. None other than the one you quoted actually ie that i can create a sequence of trials played by some other person using the logic of the original guy that didn't select f1 thinking the other choice was better. I just used central limit theorem results to argue that one sum is clearly ahead of the other in the large n limit making it contradictory that the f2 is the best way to go. Also in no way you are talking out of your a... Of course not. The intuition to go to multiple trials one way or another is the right direction. Just read my response to you before to see what the differences are and why you need a few more details to get to the answer.

The reason i explain in detail things in my posts is to make everyone happy and myself maximally exposed to criticism without hiding behind obscurity because the target is the truth not my ego or whatever else.Those that get it fast are ok to begin with and probably get bored but its small loss. Those that are new to the ideas learn something (or get hints to look up on the web in literature) and do not feel often left out or unable to get it which helps boost their self confidence to read more , think more and get the idea they can go out and do things like that and better with their minds without ever having some moron put them down either by keeping them at distance or snubbing them or making them think they are not enough somehow. This is why my explanations are long and i use many examples. I want to be thorough to cover holes in logic that are lost in fast non detailed arguments (read literature in modern theoretical physics and see how pathetic the situation has become with ideas like string theory and other mental monstrosities that are never true paradigm changing theories brave enough to innovate in human philosophy which is where true breakthrough starts always - ie explain Quantum Mechanics and spacetime rather than force QM on strings - and at the same time audacious enough to claim in papers all kinds of things with confidence without rigorous proofs and never covering all angles or even acknowledging the existence of problems in logic as if its dogmatically imposed like religion often) .

OK, BUT I'm saying you could have used the phrase "without loss of generality" a few times, although I'm not sure I fully understand some of your sentences. What does mean?

And I'm not saying long theorems are a bad thing, but I personally prefer smaller/denser ones because they make me research each little bit and that's when I learn, but hey, that's just me.

My query in the previous post was...

Why is it that ? [Prove this mathematically].

Your trials are 'similar' [how do we define similar here? Do we say that "They are the same experiment but at different times"?] and 'independent', which we also need to be careful how we define to answer this question. Any ideas?

Okay, I think you could...

label the trials as T1, T2, T3, ..., Tn, and prove by induction that the quoted statement is true. We just need the trials to be "time independent" and "experimentally independent", which they are. And I think that's done... amirite?

I want the 2 choices f1 and f2 that have ev f1 and ev f2 respectively to have finite variance for my arguments to work after that (i said i dont care what other differences the distributions may have ) . Clearly this is satisfied in all poker and all games played basically everywhere. I want the sds of the 2 or more choices to be finite because i am going to later use an argument as n becomes real large and i dont want the sds to be a problem. They are not in fact ever in almost all human bets . The trick is the differences of 2 strategies grow as n but the sd as n^(1/2) . So thats where i use the central limit theorem. I can construct large enough sums that are always better for the higher EV choice than any other choice.

Now i say if it is true that the choice f2 with inferior ev is better than f1 in this one trial the guy is playing its obviously self evident that an identical trial with exactly the same conditions played should have us choose again f2 if he is right because its the clone of the one before and nothing is different. Absolutely nothing. Like saying to a guy what do you prefer to go all in QQ vs TT-AA,AK or AKo vs 44-AA,AJ+. The EV of the first all in is better . We can repeat that trial indefinitely identical each time by reshuffling . Nothing changes if you repeat that all in . So we can try this 1 trillion times and we will always select AKo which is the worse EV because we think it was the right choice initially for that one trial. So every time we have a new trial nothing has changed. We still believe its right to select AKo rather than QQ all in. But now if in each trial AKo is the best i can do clearly an accumulation of identical trials which ought to be better than an accumulation of identical trials that select QQ instead. If the AKo choice is better it will always be better. And now i have you where i wanted! In the large n limit but without you the one time trial tester doing the large n limit, instead someone else that is using your idea that this is the right choice and applies it as many times as needed to demolish it.

At the large n limit we have central limit theorem that ensures the best EV sum is really ahead 100% of the time in the n-> infinity limit. That makes the AKo choice inferior to the QQ one. So our idea to have selected AKo for that one trial is now proven to lead to a contradictory outcome.

Obviously the trials i propose for the other guy ,that trusts the first guy's choice is better , are all identical and time independent and they require nothing special and unrealistic. Its like flipping a coin that is biased every single time the same way . Nothing changes. It will continue to offer bias heads probability say >50% and every new trial is identical to the one before (with proper mixing of the coin i mean which can be always ensured)

So in essence the first guy cannot try more than once if he so decided but the other guy is not prevented from doing it as many times as he wants and using the logic of the first guy create large enough sums that the central limit theorem becomes applicable where we couldnt use it in one trial.


Look to most people a higher EV distribution is self evident to be a better choice. All i say is if you dispute that allow me to replicate your choice a billion times and obviously the more i do it the better, i should be ahead by a lot in the end if you are right, well the opposite proves true and in the large n limit 100% of the time this choice proves wrong so there has got to be something wrong with the initial assumption that not taking the best EV was preferable.

Ok. I was just asking if you can mathematically prove why it is that if you have lots of trials which are the same, but at different times or places or both, and they are all independent of each other, then the outcomes of each trial are equally likely? I know this seems intuitively obvioius, and I think you can prove it, but it really depends on your definition of space-independence and time-independence- anyway, this is a bit off-topic.

"The EV for a single trial of what" I would ask? What kind of information do you have before the trial?

If you have information before the trial, then you make the highest EV decision with that information. If you don't have any information, you have to do the highest EV decision. EV is flexible and dependable, it's not an average static measurement.

All the identical trials obey the same distribution. Their results will be stochastic and all over the place in general but the distribution is the one that they have in common. Exactly like AA vs KK , you have win tie and loss always with the same probability each but you dont know which will come just how often expected. I am using the addition of random variables that obey the same probability distribution (that version of the CL theorem). It is actually the simplest version. Nomatter what distribution the one trial has in the end the addition of the outcomes of many such trials behaves like a normal distribution with average the EV*n and sd n^(1/2)*Sd if the single trial has EV and sd given. It is because the resulting sum has the properties the theorem says that i can exploit that argument and turn the 1 trial to many to get to the contradiction.

It can be anything really from say the choice to call and all in preflop with QQ when their range is TT-AA,AK to folding instead. The trial here is the repeat of the all in decision if you call . If you fold the distribution is trivial its always 0 outcome . But in many cases at poker we have choices between all in and flat calling say to go to the river and then play more . Sometimes one may be able to find the EV of both choices and decide which is better . These 2 choices correspond now to 2 distributions with different averages and standard deviations. The trial is always a well understood process. Clearly only when we know the distribution we can make arguments because we can calculate things like EV. If there is uncertainty then we have no clue what the EV is.

But you have a huge number of cases we daily face where certain lines of action have different EVs that are fairly well understood (maybe with some error but basically well ordered in strength). Say you debate between raising all in 20bb to steal or raising 4bb instead if prepared to fold to an all in reraise. Here you have 2 choices the instant all in or the raise. Each will have a different EV based on what the opponent tends to do and how often they have the hands they shove with or flat call.

Often alot of good professional players (including my non-professional self) use 'gut' calls and natural instinct to make their decisions, that is, along side their well calculated long term +EV decisions.

Good to check this out also http://forumserver.twoplustwo.com/15...allacy-792330/

I know all of this!...I am asking WHY the bold is true...and I think the answer is beacause of time independence, and the fact that all the trials are independent. These two facts allow all the corollaries you state in all your posts to be true. The thing is, I can't find a good definition of "trials being time-independent". This is really what my point boils down to, I think. Can you give a good definition of "trials being time-independent"??

WTF? If someone shoves on you on your first and last hand of poker ever, and you wake up with AA, you're in a +ev spot if you call, and it won't repeat.

Isn't it clear you should call? even though its just one time? Why would you need proof for this, it's obvious that you are more likely to win than lose.

Plus, that "if the K rolls and you lose, you actually didn't maximize value for that outcome" is so ******ed, of course you did, maximizing value doesn't mean winning 100%

Rusty,

To answer your question: [Ask yourself a simple question: if choosing the accepted long term EV choice is not what you want to do, then by what means should you choose your action? Is there a better way to choose, for one instance only?]

It is preferable to trade mistakes with a bad player...make a -EV play in order to obtain a greater -EV play from your opponent in the future.

grunching....

maxi-max-gain strategy = make the choice that maximizes the chances of realizing the best possible result, ignoring all risks

mini-max-loss strategy = make the choice that minimizes your chances of realizing the worst possible result, essentially mitigating all risks

sometimes we are lucky in that both strategies suggest the same course of action, so we do that.

but usually, these strategies will suggest opposite courses of action, in which case we need to weight all outcomes by their probability of occuring, and add up the wins and losses for each course of action to find a balanced solution. this is what EV is essentially doing.

so when you make a +EV play, even in isolation, you are essentially balancing all the positives and negatives associated with the play and realizing that it is still the correct choice.

Mate this is really simple...

You have a situation where you believe if you make action a or action b you either win x amount of money, or you lose it.

If you believe action a has a 70% chance of winning the amount x and action b has a 30% chance of winning the amount x. However, you knew you were only going to have this situation occur ONCE.

Then does it ever make sense to choose action b, while you might lose because your odds aren't that great, if you did lose, would it make it the wrong choice?

Of course not, even though this situation only occurs ONCE, we make the decision based on the odds in our favor. If we don't make the decision with the odds in our favour, chances are we lose, even in the short run.

Why would you ever choose an option that gives you worse odds?

Imagine 1,000,000 people did the above test and half chose action a and half chose action b. Which half of the 1,000,000 people do you think made more money? Which half would you prefer to be in?

Just think about every situation in terms of your odds. If you are making the most likely action to win you more money (or lose you less), it's intuitively the one and only correct decision.

Even though you missed the point even in your example it's not that simple. Let's say you have an option to take 1% chance to win $1M or 100% chance to win $8k. If you take 1,000,000 ppl having made each choice the first group will have an average of $10k and the second group will average $8k. Yet most ppl in the first group won't have any money at all. So it's often better to take the second option even if it's lower EV.

Secondly imagine someone else marked the cards and know which cards are going to come. You get AA and go all in preflop because you have an 80% chance of winning vs w/e. Yet the guy who marked the cards knows you're going to get sucked out on. To you it looks like you made the correct decision but to the guy who knows which cards are going to come the decision you made is obviously incorrect. From this you can see that you're simply falling back on using EV of a decision because you lack information. Instead of making the "correct" decision for this one particular case you're falling back on a strategy that will have positive outcome in the long run. What I asked for is proof that the "pick Max EV" strategy converges over many different games each played only once. It's not as simple as you make it sound as there is some math involved.

Yes, we all know that it is intuitively obvious that choosing the most +EV option is in our best interest when given the choice...
But OP wants someone to prove this mathematically (read OP), and this is more difficult, although certainly doable, and I think masque de Z has proved it in his long post on page 3 (I think it was page 3).

Personally, I'd take the 1st option. This is much different to playing 1 poker hand in your life because this scenario depends on your financial situation and value or need for money. For example, if you need to have an operation now that costs 700K, or you will die in 1 year, then taking the 1st option is most life EV. The example in OP is independent to all these variables

Yes, this example has nothing to do with my OP. I was just trying to show that picking max EV is not as obvious as he stated.

There are plays in poker that are high variance and low variance plays. However, there is never anything as extreme as what you are stating up there.

.......

When playing poker, you look at everything you know, you use all your knowledge to come up with three decisions, bet/fold/call. To make your decision you analyze using all the information you know to come up with probabilities and EV decisions for each action.

All you are getting at is either:

a) suspecting everyone of cheating

b) being able to tell the future

Of course if you know you are playing someone is cheating, you don't play, because if they can see your cards, you can't win. Even if you get dealt AA and you know you are going to win the pot, if your opponent can see your cards, they are folding anyway.

You can't try and say "because you lost in that specific spot it was a bad decision" because the ONLY way to know you were going to lose is being able to tell the future.

In poker we can't tell the future, we can't tell if our opponent is cheating, all we can use is the knowledge we have on our opponent and our hand strength relative to the board to make decisions.

Honestly, bringing in the fact someone cheating against you to argue with probability has made me immediately feel like leaving this thread.

That is just a ridiculous argument. No decisions EVER will be +EV against an opponent that can see your cards. Obviously.

One other thing you are lacking in your understanding of probability is trying to discern the idea of "different games".

If you flick a Jamaican coin, then a British coin then an American coin etc etc etc you will find that your probability converges.

Find any two actions that you know for certainty have a 50% probability for two outcomes and play each of these "different games" over a long period of time and you will find convergence all the same (in this case, close to a flat line).

Probability has no bounds and knows no difference between one situation where you have 20/80 odds and another situation that you have the same odds. There is no difference between "games". Your odds don't "reset" or anything magical like this once you switch to a different "game".

Where do you draw the line between a different game. Essentially, each poker hand you play is a "different game" with "different odds". If you want proof of convergence, look at any poker graph with 1mil+ hands played compared to their EV.

I'm implying neither. I was giving you can example of how the concept of "correct decision" changes based on how much information a person has. Some who marked the cards may see going all in with AA as a bad decision because he knows he is going to get sucked out on while if he didn't have this information it would be a good decision. So clearly what the correct decision is changes based on what information you have without the situation changing in any way. This just shows that we use EV as a fallback strategy when we do not have enough information. So instead of making the correct optimal decision for this particular situation we are optimizing the long run of a particular decision. That's why it's interesting to see how it converges. A 1M hand graph does not prove convergence. It needs to be proved mathematically.

In that situation with aces we made the best decision using our available knowledge. You surely can't expect people to make a decision with more information that they know?

No, we use EV as our ultimate decision using ALL of our knowledge. If we don't have enough information we can't make a better decision, that just doesn't make sense.

Wrong, we are making the correct decision for that instant in time. While we might not win, it is the correct decision for that instant. If it wasn't then we would not converge to a positive win rate, but a negative one.

It doesn't prove it but it demonstrates it very well, if you are looking to prove convergence.

There are many situations in which it occurs...

http://en.wikipedia.org/wiki/Convergence

We are not making the best decision for that instant in time. We're making the best decision over all cases that match this information. Going all in with aces is going to show profit 80% of the time. Not going all in the other 20% would be better but we don't have enough information to tell apart the cases where it loses from the ones where it wins. We are therefore forced to either go all in all the time or fold all the time and going all in all the time is much more profitable than folding.

Wrong, when you understand that even if you lose, the decision was the best for that instant of time then your confusion will be solved.

The only way a better decision can be made is with knowledge of the future outcome.

The decision was the best available. Nevertheless it is a fall-back on EV due to lack of information. The decision is correct because the EV converges. I am not confused about this. I was asking for mathematical proof.