Are Winrates Normally Distributed?
 

BIG thanks to Justin A for coming to me about this. He was the one who first noticed it and most of this was his idea. I had some free time last month on my flight to Vegas so I did most of the tedious Excel stuff.

Also, Barron, I know this doesn't have anything to do with high stakes poker, but I feel that it is an interesting topic that many posters will be interested in. If you don't agree, please move it to MHHUSH.

I am not a statistician but based on my not-quite-intermediate understanding of statistics, this is what I've come up with.

Methodology: I took a database of about 150k hands at one limit. I took these hands and put them into Excel. I chopped them up into blocks of hands (in chronological order) and plotted them in a histogram.

Theory: BB/100 is the average BB you will win every 100 hands. If you've played a large enough sample of hands (technically 50 samples should be enough, so 5,000 hands), the sample should behave normally and form the shape of a bell curve.

Results: The following graphs are of BB/25, BB/50, BB/100, etc. Notice that they are skewed toward the low end. This would suggest that winrates are not normally distributed, which would mean you are more likely to run good but the bad runs will be worse. I don't remember the exactly mean but figure it's somewhere between 2 and 2.5 BB/100. for BB/10 it would be between 20 and 25, etc.

Can someone with a better understanding of statistics explain this?

http://img.photobucket.com/albums/v6...ef09/per10.jpg

http://img.photobucket.com/albums/v6...ef09/per25.jpg

http://img.photobucket.com/albums/v6...ef09/per50.jpg

http://img.photobucket.com/albums/v6...f09/per100.jpg

http://img.photobucket.com/albums/v6...f09/per200.jpg

http://img.photobucket.com/albums/v6...f09/per500.jpg

http://img.photobucket.com/albums/v6...f09/per700.jpg

Re: Are Winrates Normally Distributed?
 

i can't speak on math, but i know that people tend to play a lot when they're down and tend to leave when they're up

Re: Are Winrates Normally Distributed?
 

It looks pretty close to normal here, but I think the way you did the sampling is not quite right.

You need to draw random samples from the total group of hands. Chopping them up into blocks is easier, but not appropriate. The way you have done it, we might find a different results using a different database.

Re: Are Winrates Normally Distributed?
 

My gut instinct says this is because the raw +$/hand won is greater than the -$/hand you lose (due to some pots being multiway). Generally speaking, your small chunks of hands will break down as follows:

1) Many where you don't win big pots and you are slightly -$

2) Slightly fewer where you win enough hands to have a small +$

3) A few where you win a big hand or two & are very +$.

4) A few where you lose a few big pots & are very -$ (less however than your big +$ chunks, since due to multiway pots, the amount you WIN if big pots is larger than the amount you LOSE in the same big pot)

This should leader to a bell curve is NOT evenly distributed, but peaks on the - side, with the difference being made up by the curve coming down less steeply on postive end.

I hope I described that well, I'd do one in MS paint, but I'm a very bad artist.

if i am correct however, if you took the same data from HU play, it SHOULD look like a normal bell curve, as the effect of multiway play is completely eliminated.

jvs

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
It looks pretty close to normal here, but I think the way you did the sampling is not quite right.

You need to draw random samples from the total group of hands. Chopping them up into blocks is easier, but not appropriate. The way you have done it, we might find a different results using a different database.

[/ QUOTE ]

i could easily randomize it

Re: Are Winrates Normally Distributed?
 

one of the possible conclusions that justin a came up with is that maybe BB/100 is not the optimal measure for those who want to do tests on it. maybe we could get a more accurate standard deviation from BB/1000, though for most people, playing that many hands before getting a standard deviation would be infeasible and plain annoying

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
[ QUOTE ]
It looks pretty close to normal here, but I think the way you did the sampling is not quite right.

You need to draw random samples from the total group of hands. Chopping them up into blocks is easier, but not appropriate. The way you have done it, we might find a different results using a different database.

[/ QUOTE ]

i could easily randomize it

[/ QUOTE ]

You can easily randomize the order of the hands in your database, but that is not what I was suggesting.

You need to randomly select X number of hands and compute the win rate for that sample. Then you need to do it over and over and over again, each time selecting X number of hands from the total set of hands. Then you plot the win rates for each of those samples.

I don't know of any easy way to do that in Excel. It could be done in some statistical packages, but a data set of that size is too big for a desktop computer, it would need to be run off of a server or a mainframe computer.

There is type of statistics called Bootstrapping which does not rely on the assumption of normality. I assume there is some bootstrapping software that does this kind of thing, but I don't know enough about it.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
i can't speak on math, but i know that people tend to play a lot when they're down and tend to leave when they're up

[/ QUOTE ]

Astro, that shouldn't matter since even if Josh leaves, his next session gets grouped in the stats (if I understand correctly). This would only matter if Josh tilts.

Re: Are Winrates Normally Distributed?
 

oh i see what youre getting at. they have to be completely independent. by choosing blocks of hands, it assumes dependence since each has an equal likelihood of being picked and thus isnt random?

Re: Are Winrates Normally Distributed?
 

I think PTjvs is dead on. Also, this effect is strongest when the blocks of hands are very small. If the block is just one hand in length, then (in a 10-handed game) around 90% of your sample points will be <= 0, and about 10% will be greater than zero. As the block size gets larger, the median block value tends towards the mean. You can see this reflected in the plots, which shift to the right as the block size gets larger.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
[ QUOTE ]
i can't speak on math, but i know that people tend to play a lot when they're down and tend to leave when they're up

[/ QUOTE ]

Astro, that shouldn't matter since even if Josh leaves, his next session gets grouped in the stats (if I understand correctly). This would only matter if Josh tilts.

[/ QUOTE ]

It should matter. If you quit early to lock up a win, and "play through your downswings," then you're going to have a smaller winrate, and the center of your distribution will be more to the left than it could be.

Re: Are Winrates Normally Distributed?
 

This would suggest that winrates are not normally distributed, which would mean you are more likely to run good but the bad runs will be worse.


This sentence seems backwards to me. If it is skewed to the left, there are more instances of bad, but the few big wins make up for it.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
i can't speak on math, but i know that people tend to play a lot when they're down and tend to leave when they're up

[/ QUOTE ]

Astro, that shouldn't matter since even if Josh leaves, his next session gets grouped in the stats (if I understand correctly). This would only matter if Josh tilts.

[/ QUOTE ]

It should matter. If you quit early to lock up a win, and "play through your downswings," then you're going to have a smaller winrate, and the center of your distribution will be more to the left than it could be.

[/ QUOTE ]

If Josh wins his first hand and quits immediately, then plays again later, then will the hands from his second session be put in the same block as his first hand?

If so, then it doesn't matter if Josh locks up his wins.

If Josh's one hand forms its own block, then that's different -- but I don't think that's what is happening.

Re: Are Winrates Normally Distributed?
 

This is a cool idea.

I don't know about the conclusions, though, as the distributions look pretty normal to me. The variance is pretty high, so even with a lot of observations its not that surprising to see things look kind of choppy.

Another suggestion to redo the graphs using the same number of "bins" (divisions for bars) in each histogram. You have a lot more bins in the first couple, which makes things look slower to converge on a normal than they probably are.

I think something might be wrong with your BB/100 histogram. It doesn't look like there are 1500 observations there.

Re: Are Winrates Normally Distributed?
 

For the reason PTjvs states the winrate for one hand should not be normally distributed. There is an interesting fact though, that is the distribution of groups of a samples from a non-normal distribution aproach normal as the size of the group increases (Central Limit Theorem). Eg. winrate/1000 will be more normal than winrate /10.

Re: Are Winrates Normally Distributed?
 

http://forumserver.twoplustwo.com/showth...rue#Post3134879

Also there is no reason that I know of to believe that winrates are normally distributed. The normal distribution is a hammer but not every problem is a nail.

Edit: Also I forgot to say in the linked post that a bad table or a bad seat, rather than bad play, can be "Mr. Hyde".

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Astro, that shouldn't matter since even if Josh leaves, his next session gets grouped in the stats (if I understand correctly). This would only matter if Josh tilts.

[/ QUOTE ]

It should matter. If you quit early to lock up a win, and "play through your downswings," then you're going to have a smaller winrate, and the center of your distribution will be more to the left than it could be.

[/ QUOTE ]

If Josh wins his first hand and quits immediately, then plays again later, then will the hands from his second session be put in the same block as his first hand?

If so, then it doesn't matter if Josh locks up his wins.

If Josh's one hand forms its own block, then that's different -- but I don't think that's what is happening.

[/ QUOTE ]

Ok you're right about that. However, what I think Astro was getting at and I know I was, is that people play longer when they are losing. Over a large sample, this is going to mean that you play more hands when you: have a worse image, a tougher table, less confidence, etc. If you put more hours in with a lesser expectation, you move your curve to the left. If you practice excellent game selection without regard to your immediate results (aka don't tilt - just like you said) then your point stands. Many people do not do this.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
This would suggest that winrates are not normally distributed, which would mean you are more likely to run good but the bad runs will be worse.


This sentence seems backwards to me. If it is skewed to the left, there are more instances of bad, but the few big wins make up for it.

[/ QUOTE ]

Yeah you're right. I think Josh got it mixed up.

In an extreme sense, it's like we're usually treading water with a few really good runs in between that makes our results better.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]

In an extreme sense, it's like we're usually treading water with a few really good runs in between that makes our results better.

[/ QUOTE ]

LOL, this is funny because this quote seems to very accurately describe my experience at poker. Nice big bursts between periods of losing or breaking even.

Re: Are Winrates Normally Distributed?
 

Hi Josh,

I'll take a stab at addressing a few points.

The fundamental Random Variable in poker is the amount of money you win on one hand. This random variable has a distribution, which is certainly not Guassian.
First off, it's a discrete random variable. The mean is your winrate per hand. The max value values it can take are +12BB and -12BB (on Party Poker). The most probable event is 0, since you fold most hands.
Other frequently occurring values are -0.50BB and -0.25BB since these are the values you lose when you fold your blinds, and maybe -1.5BB since this is how much you lose when you raise pre-flop, completely blank the flop, bet the flop, and get raised.

So we get a sense of what the probability mass function of this random variable looks like: It's centered at your winrate (say .02bb) but its peak value is at 0. Then it has smaller peaks at popularly occuring values, such as -0.50BB, -0.25BB, etc. It is, obviously, not a normal distribution.

The Central Limit Theorem tells us that if we ADD together enough of these strange random variables, the sum, regarded as a random variable, must start looking more and more Guassian.

In your charts, when you group together a string of hands, you are adding all the random variable in each group, and this sum should starting looking Guassian the larger the group is (BB/1000 should look more Guassian than BB/10). With a 150k hand sample, I don't think you have enough hands to get a graph that shows this, since if you went to, say, BB/1000, you would only have 150 sample points. But I'm pretty sure that at some point, it would look like a nice bell-shaped curve.

Edit: You can start to see at BB/50 how the graph is looking more Guassian. Below BB/50 you have the nice feature that you have many smaple points. BUT each sample point is not yet being taken from a very Guassian distribution. Above BB/50 (BB/100 and up), you have the nice feature that the samples are being taken from a pretty Guassian distribution, BUT you don't have enough samples to draw the curve. If your DB was much larger, I think you would see the BB/100 look much closer to Guassian than the BB/50.

-v

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
[ QUOTE ]

In an extreme sense, it's like we're usually treading water with a few really good runs in between that makes our results better.

[/ QUOTE ]

LOL, this is funny because this quote seems to very accurately describe my experience at poker. Nice big bursts between periods of losing or breaking even.

[/ QUOTE ]

This has been my experience as well. Since I moved up to 15/30 in May and later 20/40 in September, I have never had a losing month, but I did make about half my money in one 30 day span in which I ran insanely well, and as a result, played a ton of hands.

What the other poster said about taking random hands and combining them to make a sample is appropriate. As tilt proof as all of us think we are (or aren't), it is still perhaps not an accurate statement to call each group of 100 hands independent. Combining hands from different sessions to form samples would be a much better indicator of overall play in my opinion.

Re: Are Winrates Normally Distributed?
 

There are two easy numbers that reflect the "normalness" of the distribution. Skewness measures the bias towards one side of the mean, and kurtosis measures the fatness of the tails of the distribution. In excel, you can easily get these numbers using the functions skew() and kurt(), with the arguments for the functions being simply the winrate series. Can you do that, and post or PM me the results, for each of the block sizes?

Re: Are Winrates Normally Distributed?
 

I liked this better when you tried to explain it drunk off your ass at Craftsteak.

Re: Are Winrates Normally Distributed?
 

Isn't your theoretical max winrate (or win) for a hand 12BB x N(number of players)? You can only lose 12 bets but you can certainly win far more.

Re: Are Winrates Normally Distributed?
 

Yes, you are right. Max loss is -12BB and Max win is 9*12BB = 108BB - rake.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
Hi Josh,

I'll take a stab at addressing a few points.

The fundamental Random Variable in poker is the amount of money you win on one hand. This random variable has a distribution, which is certainly not Guassian.
First off, it's a discrete random variable. The mean is your winrate per hand. The max value values it can take are +12BB and -12BB (on Party Poker). The most probable event is 0, since you fold most hands.
Other frequently occurring values are -0.50BB and -0.25BB since these are the values you lose when you fold your blinds, and maybe -1.5BB since this is how much you lose when you raise pre-flop, completely blank the flop, bet the flop, and get raised.

So we get a sense of what the probability mass function of this random variable looks like: It's centered at your winrate (say .02bb) but its peak value is at 0. Then it has smaller peaks at popularly occuring values, such as -0.50BB, -0.25BB, etc. It is, obviously, not a normal distribution.

The Central Limit Theorem tells us that if we ADD together enough of these strange random variables, the sum, regarded as a random variable, must start looking more and more Guassian.

In your charts, when you group together a string of hands, you are adding all the random variable in each group, and this sum should starting looking Guassian the larger the group is (BB/1000 should look more Guassian than BB/10). With a 150k hand sample, I don't think you have enough hands to get a graph that shows this, since if you went to, say, BB/1000, you would only have 150 sample points. But I'm pretty sure that at some point, it would look like a nice bell-shaped curve.

Edit: You can start to see at BB/50 how the graph is looking more Guassian. Below BB/50 you have the nice feature that you have many smaple points. BUT each sample point is not yet being taken from a very Guassian distribution. Above BB/50 (BB/100 and up), you have the nice feature that the samples are being taken from a pretty Guassian distribution, BUT you don't have enough samples to draw the curve. If your DB was much larger, I think you would see the BB/100 look much closer to Guassian than the BB/50.

-v

[/ QUOTE ]

Ok you seem to know a lot about stats. When I first started looking into this I did so because I was wondering if the confidence interval calcs we've done in the past are accurate when dealing with BB/100. So if you have a winrate of x bb/100 after 100k hands or whatever, then we do a calc for a 95% or 99% or whatever confidence level we choose to find out where are true winrate most likely falls. Likewise we can do the same for level of confidence that winrate > x.

So my question to you, is how accurate are these confidence intervals when dealing with a statistic that is not distributed normally?

Re: Are Winrates Normally Distributed?
 

If the distribution is not normal but not much different from the normal distribution, then in practical terms I don't think it would make much difference. Even if it were not strictly accurate, it would be good enough and probably not worth doing all the extra work to get an accurate confidence interval.

I am still not convinced that win rates are not normally distributed around the mean.

If you want to base you confidence interval on the actual distribution, you might look into bootstrapping.

Re: Are Winrates Normally Distributed?
 

you pictures are huge, even on 1600*1200 in makes reading this thread no fun. If u can, resize them before u post next time.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
you pictures are huge, even on 1600*1200 in makes reading this thread no fun. If u can, resize them before u post next time.

[/ QUOTE ]

He can resize them right now through photobucket.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
If the distribution is not normal but not much different from the normal distribution, then in practical terms I don't think it would make much difference. Even if it were not strictly accurate, it would be good enough and probably not worth doing all the extra work to get an accurate confidence interval.

I am still not convinced that win rates are not normally distributed around the mean.

If you want to base you confidence interval on the actual distribution, you might look into bootstrapping.

[/ QUOTE ]

one thing thats important to consider is the nature of the process that drives the win rate of a given player or a pool of players.

in discrete time, its easier to deal with but when we move to continuous time, the driving force could be a set of stochastic processes which COULD nullify any inferences made from using the current normal distribution as a base for analysis.

basically, if random processes drive parts of winrate (one process could be how a given person does x,y, or z and have it based on randomness or even have real life like jumps-like the poker graphs show- by making those processes brownian motions that accumulate quadratic variation at rate 1 per unit time) then we wont see the distribution as normal or even a good approximation unless ALL processes meet a few criteria:

-they all have to individually be random and not deterministic (though they can change over time, so long as its random)

-their drift/diffusion (mean/variance) must be adapted to the SAME information that drives the whole system

-they are jointly normally distributed

NOTE: these are some seroiusly strict conditions ... especially the last one.

if these are met then the distribution of the results of the process may approximate a normal distrubution with some confidence.

either way, studies have shown that almost all biological/psychological phenomenon are normally distributed or very easily and readily approximated by a normal distribution. since the win rate is driven by largely biological phenomenon, it would seem as if on a large enough scale, the results of the win rate observations would converge to a normal distribution as well.

the whole thing is interesting and i like thinking about it but im not good enough at all types of higher level math to write out a proof of this...

well, its bedtime.

Barron

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
Hi Josh,

I'll take a stab at addressing a few points.

The fundamental Random Variable in poker is the amount of money you win on one hand. This random variable has a distribution, which is certainly not Guassian.
First off, it's a discrete random variable. The mean is your winrate per hand. The max value values it can take are +12BB and -12BB (on Party Poker). The most probable event is 0, since you fold most hands.
Other frequently occurring values are -0.50BB and -0.25BB since these are the values you lose when you fold your blinds, and maybe -1.5BB since this is how much you lose when you raise pre-flop, completely blank the flop, bet the flop, and get raised.

So we get a sense of what the probability mass function of this random variable looks like: It's centered at your winrate (say .02bb) but its peak value is at 0. Then it has smaller peaks at popularly occuring values, such as -0.50BB, -0.25BB, etc. It is, obviously, not a normal distribution.

The Central Limit Theorem tells us that if we ADD together enough of these strange random variables, the sum, regarded as a random variable, must start looking more and more Guassian.

In your charts, when you group together a string of hands, you are adding all the random variable in each group, and this sum should starting looking Guassian the larger the group is (BB/1000 should look more Guassian than BB/10). With a 150k hand sample, I don't think you have enough hands to get a graph that shows this, since if you went to, say, BB/1000, you would only have 150 sample points. But I'm pretty sure that at some point, it would look like a nice bell-shaped curve.

Edit: You can start to see at BB/50 how the graph is looking more Guassian. Below BB/50 you have the nice feature that you have many smaple points. BUT each sample point is not yet being taken from a very Guassian distribution. Above BB/50 (BB/100 and up), you have the nice feature that the samples are being taken from a pretty Guassian distribution, BUT you don't have enough samples to draw the curve. If your DB was much larger, I think you would see the BB/100 look much closer to Guassian than the BB/50.

-v

[/ QUOTE ]

this is very well said, and based on my understanding of statistics, exactly correct.


the assumption here is that the win/loss per hand is a random variable, which it strictly speaking is not, as dcifr mentioned. the distribution of the random variable will change based on game conditions, improvements in your play, tilt, who you're playing against, the number of dumps the guy in seat 6 has taken that day, etc. but these things shouldn't change the random variable so much that you can't approximate win rate per N hands where N is large as a normal distribution fairly accurately (the last point in particular has very little effect.)

Re: Are Winrates Normally Distributed?
 

Hi Justin,

The confidence interval calcs that people do only apply to a Normal distribution with mean and variance equal to the mean and variance of our special BB/100 distribution.
Since our BB/100 distribution is not normal, these first two "moments" are not enough to figure out exactly what our confidence intervals are. So, our calculations introduce an error. How big is the error?

Let's go to the extreme and say that we were interested in BB/1 (BB/hand). Our mean, let's say, is 0.0200 (this would mean we had a winrate of 2.00 BB/100). And let's say our standard deviation is 1.5 (which would result in an SD/100 of 15). This means that a one sigma event would put us between -1.48 and 1.52 BB, and a five sigma event (which happens less than once in a million trials on average) would be between -7.48 BB and 7.52 BB. Clearly this is way off. We win (and even sometimes lose) more than 7.5 BB well over once in every million hands. So our confidence interval calcs for BB/1 are way off because the real poker distribution is nowhere near Normal and can NOT be approximated well using just its first two moments (mean and variance).

So where does that leave us with BB/100? How much error do we introduce in our confidence interval calculations by assuming BB/100 is Guassian? Good question. I'm not sure. I think you will still see remnants of the longer flat tail on the positive side and the shorter, steeper tail on the negative side (a by-product of the fact that you can win a lot more in one hand than you can lose). But I suspect it will be close enough to normal to not worry about it.

Re: Are Winrates Normally Distributed?
 

vkh, great posts, and thanks to Justing and Josh for bringing this topic up.

I think this...

[ QUOTE ]

So where does that leave us with BB/100? How much error do we introduce in our confidence interval calculations by assuming BB/100 is Guassian? Good question. I'm not sure. I think you will still see remnants of the longer flat tail on the positive side and the shorter, steeper tail on the negative side (a by-product of the fact that you can win a lot more in one hand than you can lose). But I suspect it will be close enough to normal to not worry about it.

[/ QUOTE ]

is very worth studying. BB/100 and the associated confidence intervals are important to a lot of people. Finding out how off our confidence intervals have been, and exactly what we should use as x for bb/x with a sample size y, are very worthy pursuits. Even if you suspect its close enough to normal not to worry about, Josh's graph with (only?) 1500 datapoints suggests otherwise.

Re: Are Winrates Normally Distributed?
 

If I am not mistaken, according to the central limit theorem they certainly must be normally distributed.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
It looks pretty close to normal here, but I think the way you did the sampling is not quite right.

You need to draw random samples from the total group of hands. Chopping them up into blocks is easier, but not appropriate. The way you have done it, we might find a different results using a different database.

[/ QUOTE ]

You probably want to draw largish random blocks of hands to randomize position as much as possible between the samples.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
[ QUOTE ]
It looks pretty close to normal here, but I think the way you did the sampling is not quite right.

You need to draw random samples from the total group of hands. Chopping them up into blocks is easier, but not appropriate. The way you have done it, we might find a different results using a different database.

[/ QUOTE ]

You probably want to draw largish random blocks of hands to randomize position as much as possible between the samples.

[/ QUOTE ]

OK, I figured out a way to do this in SPSS. I thought the data file would crash my computer but it doesn't

I have a data file with the amount I won/lost for 164,724 hands at 15/30. My win rate over these hands is a pitiful 1.13BB/100.

How large should the samples be and how many should I pull? I was thinking of selecting 10,000 samples of 1000 hands each.

Then I can I plot them and get the skewness, kurtosis, etc.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
It looks pretty close to normal here, but I think the way you did the sampling is not quite right.

You need to draw random samples from the total group of hands. Chopping them up into blocks is easier, but not appropriate. The way you have done it, we might find a different results using a different database.

[/ QUOTE ]

You probably want to draw largish random blocks of hands to randomize position as much as possible between the samples.

[/ QUOTE ]

OK, I figured out a way to do this in SPSS. I thought the data file would crash my computer but it doesn't

I have a data file with the amount I won/lost for 164,724 hands at 15/30. My win rate over these hands is a pitiful 1.13BB/100.

How large should the samples be and how many should I pull? I was thinking of selecting 10,000 samples of 1000 hands each.

Then I can I plot them and get the skewness, kurtosis, etc.

[/ QUOTE ]

I just had a grim college statistics flashback when you mentioned SPSS. And I am now scared of this thread. You guys are so much smarter than me. [img]/images/graemlins/frown.gif[/img]

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
If I am not mistaken, according to the central limit theorem they certainly must be normally distributed.

[/ QUOTE ]

In order the central limit theorom to be applicable, the sample's variate's must be:

1) independent
and
2) distributed arbitrarily.

I think arguments can be made in support of both being true or false.

I also think that the nature of the player is very important, as well as the game they are plaing (limit, NL)

I think an equally interesting question is, if you were to model BB/100, what would it be a function of?

Re: Are Winrates Normally Distributed?
 

A discrete function will never become perfectly normal if the underlying discrete function is not normal. It will however become MORE normal as you increase the number of independent trials.

For example say your indivdual hand distrbution is:
- 1 90%
+ 11 10% of the time

If you run 100 trials there is still a .1^100 chance you will have won 11 * 100. (you win every trial) There is a zero chance you will have lost 11 * 100 (in fact your largest possible loss is 100).

So what you need to decide first is how close to normal do you need the aggregate results to be to be considered normal?
----------------------------------------------------------
On the changing parameters. (Good table, playing well, etc.) This will be VERY hard to empirically estimate, and will make the calculations more tedious.
----------------------------------------------------------

Some methods that may be usefull in seeing how many hands you need to approximate a normal function would be:

Get every hand you have and put it into a discrete curve with the values as percentages. There won't be that many buckets. (You can lose up to 12 BB, but probably never have, and win up to 108BB <-- That I'd like to see). Most likely your range will be something like -8BB to +30BB with an interval size of .25 BB <-- The small blind.

Now you can either
A) Run X simulations off that curve and see if the result is 'normal'

Or

B) Do a FFT on the curve with X convolutions(?) and see if the outputs are 'normal'.

If the results are still to skewed to be called 'normal' increase X.
The FFT method has the advantage it gives the 'true' distribution but it's a pain to do (I would guess you will need somewhere in the neighborhood of 5,000 convolutions(?) and you certainly won't pull this off using FFT in excel)
The simulation method if you get a 'normal' result you should probably still run it about 10 times (minimum) to make sure you didn't just get a 'normal' simulation.

Re: Are Winrates Normally Distributed?
 

[ QUOTE ]
A discrete function will never become perfectly normal if the underlying discrete function is not normal. It will however become MORE normal as you increase the number of independent trials.

For example say your indivdual hand distrbution is:
- 1 90%
+ 11 10% of the time

If you run 100 trials there is still a .1^100 chance you will have won 11 * 100. (you win every trial) There is a zero chance you will have lost 11 * 100 (in fact your largest possible loss is 100).

So what you need to decide first is how close to normal do you need the aggregate results to be to be considered normal?
----------------------------------------------------------
On the changing parameters. (Good table, playing well, etc.) This will be VERY hard to empirically estimate, and will make the calculations more tedious.
----------------------------------------------------------

Some methods that may be usefull in seeing how many hands you need to approximate a normal function would be:

Get every hand you have and put it into a discrete curve with the values as percentages. There won't be that many buckets. (You can lose up to 12 BB, but probably never have, and win up to 108BB <-- That I'd like to see). Most likely your range will be something like -8BB to +30BB with an interval size of .25 BB <-- The small blind.

Now you can either
A) Run X simulations off that curve and see if the result is 'normal'

Or

B) Do a FFT on the curve with X iterations and see if the outputs are 'normal'.

If the results are still to skewed to be called 'normal' increase X.
[The FFT method has the advantage it gives the 'true' distribution but it's a pain to do, the simulation method if you get a 'normal' result you should probably still run it about 10 times (minimum) to make sure you didn't just get a 'normal' simulation.

[/ QUOTE ]

i think a more realistic hand distribution would be much closer to approaching normality.

lose 0sb with P1
lose 1sb with P2
lose .5sb with P3
lose 2 sbs with P4
lose 3sbs with P5
lose 4sbs with P6
lose 5sbs with P7
.
.
.
lose 12bbs with Pn

then the upside:
win 0 with Pa
win
.
.
.
win 55bbs with Pm

i think that would be better, but still, there is a skew due to the limited downside and positive upside.

but we're aggregating in such a way around the MEAN of this random variable over time which might be enough to make it fairly close to gaussian.

as vks stated, "how close"? we dont know....we'd need more data than anybody here has to look at it.

Barron