Musings on variance.

duggs

duggs

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Just had a random thought in my class today, variance is an all encompassing concept that applies to pretty much everything from seat draw to opponents to hands dealt to flops to which part of your opponents range you run into and our opponents specific leaks relative to the population etc etc etc.

now onto EV v winnings, its fairly easy to show that EV is a strictly better measure of performance than amount won. what i got thinking about was the distribution of winnings around EV.

obviously in all poker samples EV = Expected (winnings) for any given number of trials but I'm more interested in the distribution around that.
if we used a coin flip as an example, where we have 50% equity its fairly intuitive that we would have a normal distribution with equal volatility both above and below the EV.

its not immediately obvious to me as to how winnings would distributed if we were to run a hand where we have 80% equity and whether the sample, while still having the same mean, would me skedastic or not.

thoughts?
 
BluffMeAllIn

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tried to have thoughts did some googling to figure out what skedastic was led me to conditional variance, read a little more and all this thinking is now hurting my head.

Only done one statistics course in my life about 8 years ago maybe so nothing is overly fresh in my mind, however i guess it would have to work out the conditional variance give X being the 80% equity if I am correct in referencing this link in relation to what your referring to:
http://en.wikipedia.org/wiki/Conditional_variance

Some heavy thinking topic of discussion you have here duggs but certainly intrigueing in getting a better understanding of variance. I mean just as a thought obviously like you said with 50/50 distribution would be even but when you are looking on the equity side of things the higher your equity I think the more skewed the distribution is going to be on the lesser side of 80% equity......just as a look it could distribute perhaps from the upside of 90% to the lower side of 60% in looking at it from a skew.

I am probably making no sense but thought your though on thinking and I guess essentially looking to wrap your head around the specifics of variance interesting, and just some general thoughts given I'd probably need a week to read up on and refresh my brain into algebra again let alone the statistic/probability side of it lol.
 
duggs

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Doing a course on econometrics atm, which is what got me thinking, I'd imagine the sample would be skewed downwards, like half of the sample will still be above, but I imagine above would be more tightly clustered as its harder to deviate as far as lower half where it could have a bigger spread, I don't really have any reasoning to back this up but i feel like that's what it will look like
 
U

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I'm trying to picture what you are saying. You think on a scatter plot that half the results would tightly clustered above the line and half would be spread below the line?

I think that you would actually have two major clusters. You would have a larger sample of winning that should be fairly tight. But, you should also have a smaller cluster of losing that should also be fairly tight.

The reason is that most of the time when you are an 80% favorite you're likely to be getting more money in the pot - I guess I am thinking 80% favorite postflop. So you should have fairly strong divergences from the mean. Your wins and losses should be almost diametrically opposed. Though you will have some small wins (which would come closer to approaching the mean) and some split pots (which will fall below the mean). Over all I still the the two groups should be fairly tight.

I might be misunderstanding what you are asking though.
 
duggs

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Remove the poker stuff from it, imagine we get to make 80% winning bets for a large amount of bets, say 1000 times. Now lets say we get a meaningful sample of people to do the same. Everyone's Eexpected return will be the same, but what would the distribution look like
 
U

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Remove the poker stuff from it, imagine we get to make 80% winning bets for a large amount of bets, say 1000 times. Now lets say we get a meaningful sample of people to do the same. Everyone's Eexpected return will be the same, but what would the distribution look like

Well you have to have some of the poker stuff in it, because that is going to skew the results. The street you win/lose on is going to change the value of the win/loss.

So if you were to look at preflop win/loss at 80% equity then you will likely have a bigger range of values. Even on a very large sample size this will have a huge impact on your mean, and will have increased variance.

If you look at river bets when you are 80% then you are much much more likely to have very divergent distributions. The pot is much larger so your win/loss is much larger. You variance will be much more tightly clustered and (I think) would be much less bell curvey than a preflop bet.

Again, this could just be me not understanding what you are looking for. Most of the statistical stuff I have done has been for investment instruments, so my view might be goofy.
 
duggs

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Yea but I'm imposing that the bets are all of the same size, so then there is only one street and it doesn't need to be ultra realistic ( we have a situation where we get it in as a huge favourite for a large sample, if this were realistic everyone would crush)
 
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Then the distribution would be pretty fixed. You would have a the majority of your results above the line (approaching 80%) and the minority below the line (roughly 20%). The values are fixed.

I think that you would see mostly tight clusters on either side, since the value of the bet is the same. The difference is binary, on off switch where it will be on 80% of the time and off 20% of the time.

I think I am clearly missing something here. It's not like it's a weighted die where there are potentially 6 outcomes that are weighted towards one, its just on off.

You don't have conditional variance here, because the game is reduced to a single obvious bet. I guess the only condition is if they call or fold, in which case you could have some variance in your results.
 
BluffMeAllIn

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Yeah i had essentially assumed it was being looked at in essence being allin for even stacks preflop like AA vs KK for example.....but the more I'm reading around on it all the longer I think it would take me to have any value of input into this discussion at present

I like the thinking on it though as a better way of understanding the variance aspect.....did a google on the course your doing duggs, seems interesting and certainly has application as per your train of thought in poker related situations.

Used to abs love math, got towards statistics a little in university but then switched around a little with my degree focus and hard to get the head wrapped around things as quick as I once could on all that but do plan on doing some reading on the math side of things a little more in depth so it might spark up my brain again lol.

It's currently hurting from reading all kinds of stuff on deviations and statistics and variance and egonometrics here this morning lol.
 
duggs

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Yea well I study economics so it's pretty closely linked. But scourge is stronger at stats than me so waiting for him to jump in
 
BluffMeAllIn

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Then the distribution would be pretty fixed. You would have a the majority of your results above the line (approaching 80%) and the minority below the line (roughly 20%). The values are fixed.

I think that you would see mostly tight clusters on either side, since the value of the bet is the same. The difference is binary, on off switch where it will be on 80% of the time and off 20% of the time.

I think I am clearly missing something here. It's not like it's a weighted die where there are potentially 6 outcomes that are weighted towards one, its just on off.

You don't have conditional variance here, because the game is reduced to a single obvious bet. I guess the only condition is if they call or fold, in which case you could have some variance in your results.

I was thinking it would be looked at in regards to expected value as opposed to simply wins/losses. Basically look at it as an example of 1000 people who play 1000 hands and have 80% equity in each of those hands. If each hand is a 1$ bet, how many of the 1000 people fall above winning 80% of the the hands compare to how many fall below winning 80% of the hands. This would give the distribution I think duggs is trying to refer to if I'm thinking correctly.

We know the expected value is 80% but due to variance there is going to be a deviation whereby its not going to be exactly 80% all of the time and so how it would fall is where I thought he was trying to get at.
 
duggs

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I was thinking it would be looked at in regards to expected value as opposed to simply wins/losses. Basically look at it as an example of 1000 people who play 1000 hands and have 80% equity in each of those hands. If each hand is a 1$ bet, how many of the 1000 people fall above winning 80% of the the hands compare to how many fall below winning 80% of the hands. This would give the distribution I think duggs is trying to refer to if I'm thinking correctly.

We know the expected value is 80% but due to variance there is going to be a deviation whereby its not going to be exactly 80% all of the time and so how it would fall is where I thought he was trying to get at.

This is exactly it, if I had a computer handy I could muck around and try approximate the experiment, but I feel like I'm missing something obvious, like I'm not sure if central limit theorem will just impose normality in a case like this or not
 
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Yea well I study economics so it's pretty closely linked. But scourge is stronger at stats than me so waiting for him to jump in

That's kind a me too. I got my MBA in finance, so you need to know some statistics (I had to take around 3 courses), but I mostly worked on financial instruments. I don't use it much these days, outside of poker, especially since I don't work in finance anymore and do legal mumbo jumbo in a court room.
 
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I was thinking it would be looked at in regards to expected value as opposed to simply wins/losses. Basically look at it as an example of 1000 people who play 1000 hands and have 80% equity in each of those hands. If each hand is a 1$ bet, how many of the 1000 people fall above winning 80% of the the hands compare to how many fall below winning 80% of the hands. This would give the distribution I think duggs is trying to refer to if I'm thinking correctly.

We know the expected value is 80% but due to variance there is going to be a deviation whereby its not going to be exactly 80% all of the time and so how it would fall is where I thought he was trying to get at.

Over a large enough sample the two should converge, so your win/loss will ultimately reflect (to a great degree) the EV. Unless I run this through one of my stats addons in excel, I guess I can't give anything other than conjecture. But EV should approach and then converge with results at a certain point. This may not be true in actual poker though, because there are so many variables. In the hypothetical we have only 2 potential outcomes, as opposed to a continuum of outcomes along the win/loss range.

I think I am missing the point of the exercise, and probably am not adding anything useful at this point. Sorry for the derail.
 
duggs

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Yea you are saying Expected winnings will = EV which is definitely true, but I want to see the distribution of the sample around EV.

The more I think about it the more I'm convinced that in large samples it will approximate a normal distribution but in smaller samples it will have a wider spread below EV and above Ev it will have tighter tail
 
BluffMeAllIn

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Over a large enough sample the two should converge

yeah think this is probably what duggs is saying regarding the central limit theorem aspect as well in regards to the distribution normalizing where it will fall on a bell curve around the 80%.

Again, I had to google and look up the central limit theorem.....this is fun i quite enjoy learning especially when its interesting, and probably not really contributing here either but its got some mice trying to spin the wheel in my brain :D

In probability theory, the central limit theorem (CLT) states that, given certain conditions, the arithmetic mean of a sufficiently large number of iterates of independent random variables, each with a well-defined expected value and well-defined variance, will be approximately normally distributed.[1] That is, suppose that a sample is obtained containing a large number of observations, each observation being randomly generated in a way that does not depend on the values of the other observations, and that the arithmetic average of the observed values is computed. If this procedure is performed many times, the central limit theorem says that the computed values of the average will be distributed according to the normal distribution (commonly known as a "bell curve").
 
BluffMeAllIn

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Yea you are saying Expected winnings will = EV which is definitely true, but I want to see the distribution of the sample around EV.

The more I think about it the more I'm convinced that in large samples it will approximate a normal distribution but in smaller samples it will have a wider spread below EV and above Ev it will have tighter tail

This would occur in smaller sample I take it because the effects of variance would be a skew on the results, which I guess is basically what you are wondering in your OP lol. However based on same could it not turn out due to variance and essentially an insignificant sample to be skewed above the 80% as well?

I feel like i just went around in a circle basically back to the question you posed for discussion initially, hmmm I feel like a theorist lol.
 
Matt Vaughan

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A few thoughts jump to mind, though they're not super organized:

First of all, if we're talking solely about the distribution of # of times you win for an 80-20 spot that we repeat a bunch of times, then this is basically the definition of a binomial distribution, no? (http://en.wikipedia.org/wiki/Binomial_distribution)

You can think of whether we win or lose the hand as "success" or "failure." So let's say that we are only looking at spots where we are an 80:20 favorite. And let's say we want to look at the distribution of running that spot 100 times.

P(X=x) = (n C x)(p^x)(1-p)^(n-x)

Where n = 100, P(X=x) is the chance that you win x of the n times, p = 0.8, and 1-p is of course 0.2. You could start to plot this pretty easily using statistical software, and probably begin to incorporate a lot of different poker "spots."
 
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Yup of course it is, looking at Skewness in Bernoulli distributions they skew left when p greater than .5 and are unskewed at .5. Dammit can't believe i didn't figure this out on my own
 
Matt Vaughan

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I mean it's weird right, I'm honestly annoyed I didn't think of it instantaneously. But when you just break it down as a single spot like that and want to see how you're wins will be distributed in 80/20's, that's really all it is. Could be a nice exercise to make some plots of 80/20's, 70/30's, etc just to see how we're actually doing.

You could also make it more poker-related by making it more related to risk vs. reward. Since in basically every realistic situation we're getting better than 1:1.
 
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And as n goes to infinity Skewness goes o zero, right on the money with my guesses.
 
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Skewing left means its easier to run far below EV than it is to run way above EV. Which makes sense
 
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Take an extreme example of 99 % sample of 100. Max above ev you can run is 1 while max below is 99
 
Matt Vaughan

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duggs "posting aloud" again. :p
 
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