what is standard deviaton?

[nicky g]

I've asked a few people this question and all they've told me is either how to calculate it, which I know, or that it's "a measure of the spread", which doesn't really tell me anything. I also know how to use it for poker purposes - but not why. For instance, I understand mean deviation; it's the mean of the distances between each measurement and the mean of the measurements. That makes sense to me and I know what it's telling me. But SD doesn't. What is it telling me, for goodness sake? For that matter, what is variance telling me? I can kind of understand why adding together all of the differences between the measurements and their mean gives you an idea of the spread of the data, but why do you sqaure them first? And what does taking the square root of that actually telling you?
Sorry if this is basic.
[img]/images/graemlins/confused.gif[/img].

[Tharpab]

I also have a question about this, is the avarege deviation(not sure its the correct name, its the nome that doest need square root) useful?why sd is so used and the ad not at all?

[topspin]

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I've asked a few people this question and all they've told me is either how to calculate it, which I know, or that it's "a measure of the spread", which doesn't really tell me anything. I also know how to use it for poker purposes - but not why. For instance, I understand mean deviation; it's the mean of the distances between each measurement and the mean of the measurements. That makes sense to me and I know what it's telling me. But SD doesn't. What is it telling me, for goodness sake?

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Standard deviation is used as a useful measure of how closely your data lies around the mean. If your random variable is normally distributed (the "bell curve"), then about 95% of the time it will lie within 2 standard deviations.

For example, if you came up with a strategy for playing JJ that yielded an average profit of 4BB with a standard devation of 0.5BB, then 95% of the time when you got that hand and used your strategy, you would win between 3BB and 5BB.

You can easily get more information by digging around on Google -- e.g. here.

[FlashFunk]

I'm not sure if my details are totally correct (its been a while since my last statistics course)

But the reason you must square the differences from the mean is due to the fact you have numbers both about and below the mean. By squaring you get rid of all the negatives (the samples that were under the mean) and you have a number that actually measures the squared absolute value of the distances from the mean. If you sum all these differences up, then take the square-root of them (basically reversing the square you did to get rid of the negatives), and finally take the average of these, you will have the standard deviation.

[tubbyspencer]

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For that matter, what is variance telling me? [img]/images/graemlins/confused.gif[/img].

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From the standpoint of poker, variance tells you what your swings in bankroll are likely to be. For example, at the lower limits where people play much looser, and see more flops, your BB/hr or BB/100 may be higher; but your variance will be higher as well. At higher limits, your BB/hr or BB/100 will be lower, but with fewer folks seeing the flop, and capitalizing on garbage, the lower your variance. So the lower your bankroll swings will be.

[uuDevil]

I'll give it a shot:

Variance:

Suppose you have a group of people and you measure their heights. Suppose the average height is X ft. Is everyone in the group X ft tall? Probably not. Since they are not all the same height, there is some "spread" or "variation" or "dispersion". Well how much variation? Maybe a little, if these people are all NBA centers. Maybe a lot if there are adults and children in the group. But saying there is "a little" or "a lot" is not very precise. So we construct a way to measure the spread and call it the variance.

Why do we square the deviations from the mean?

Suppose we did not. Add up the deviations from the mean without squaring and what do we get? Zero. Always. Because deviations for values below the mean are negative and deviations above the mean are positive, when you add them up, they cancel out. This is not useful. So we square the deviations to give us positve values. Now when we add them, they don't cancel. The mean of the squared deviations is called the variance. The further away from the mean, on average, our observations are, the greater the variance.

Standard deviation-- why do we take the square root of the variance?

Well, if you look at the units associated with variance, it is square feet in the example above. If we want to express the amount of spread in our data in the same units as the data itself, we just take the square root of the variance and call it the "standard deviation." It is just as valid a measure of spread. Now we can say meaningful things like "My height is 4 standard deviations above the mean, so gimme the rock!"

[nicky g]

" [ QUOTE ]
Why do we square the deviations from the mean?

Suppose we did not. Add up the deviations from the mean without squaring and what do we get? Zero. Always. Because deviations for values below the mean are negative and deviations above the mean are positive, when you add them up, they cancel out. This is not useful. So we square the deviations to give us positve values. Now when we add them, they don't cancel. "


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Yes... but when you calculate the mean deviation, you solve this problem by simply changing all the negatives to positives. Why not just do that?

I realise variance is telling us about variation around the mean; that the bigger it is, the more variation there is. But what exactly is it telling us? Mean deviation already tells us about variance around the mean. What makes variance and standard deviation more useful?

I'm not sure I'm really getting my question across here so I'll try some other ones:
What is standard deviation telling us that mean deviation isn;t?
Why do you square the differences from the mean to get variance, when you could simply get rid of the negativce signs (as you do to calculate mean variance)?
Why will 95% of results fall within two standard deviations of the mean? Where does this magical property come from?

[donkeyradish]

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For example, if you came up with a strategy for playing JJ that yielded an average profit of 4BB with a standard devation of 0.5BB, then 95% of the time when you got that hand and used your strategy, you would win between 3BB and 5BB.


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Hmm, I measured the SD of my win rate at 0.5/1 hold'em and it was 16BB/hour with a win rate of 2BB/hour (it was a very small sample).

So that means 95% of my hours I should expect to fall somewhere between winning 34BB or losing 30BB? Not exactly a revelation [img]/images/graemlins/tongue.gif[/img]

[nicky g]

There is a BruceZ thread somewhere on a way to get much more accurate figures. I forget exactly what it is but it involves number of hours played; the bigger they are, the more accurate it is. Try a search.

[uuDevil]

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What makes variance and standard deviation more useful?

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Hopefully I don't get in too much trouble with the mathematicians....

The Magical Part

Take a set of measurements like height in my previous example and arrange them in order from smallest to biggest. Properly plotted, this "distribution" has a characteristic shape. This shape is called "bell" or "normal" or "Gaussian." Many different things in nature exhibit this same shape.

The Math Part

It would be nice if we could use math to represent what is going on with normal distributions, so we look for a mathematical function that has this same shape. Here it is:

f(x)=B*exp(-(x-u)^2/A), where A, B, and u are constants

and B=sqrt(1/(pi*A))

Look at the argument of the exponential. Does the (x-u)^2 part look familiar? We used this form in the definition of variance (where u=mean).

[Sorry if the following is not that clear.] The argument of the exponential has to be dimensionless, so A has to have the same dimensions as (x-u)^2. And what constant can we choose that is both relevant and has these dimensions? One defined in terms of the variance will work. So the variance is important because it shows up in the function that represents Normal distributions.

The Good Part

With properly defined constants, we can now use the distribution function to calculate general features of normal distributions like the fact that 95% of results fall within 2 SD of the mean, etc.

Does that help?