question about standard deviation

[Tommy Angelo]

I have a question about standard deviation. If one guy plays smart-loose for ten years and ends up even, and another guys plays tight-tight for ten years and ends up way up, which of them has the lower standard deviation? The guy who ended up exactly where he started after ten years, but took a highly deviant road to get there? Or the guy who steadily deviated to the plus, year after year?

Thanks,

Tommy

[Chr]

standard dev. has nothing to do with where you end after 10 years, it's about how big swings (positive and negative) you encounter in small time samples, like a session or an hour or pr. 100 hands or what ever timeframe you want to use.

So, I don't think your question can be answered, since we don't know their swings in those 10 years. But from what you seems to indicate, it is the loose player who generally had the bigger swings and therefore he has the higher std. dev.

[BeerMoney]

Population parameters mu and sigma are independent.

[KidPokerX]

In order to solve this you need to know how far the swings deviate from the mean (average). In the S.D. formula:

S = root E(y-y*)2/n-1, where y*= sum of all observed values from y (the diff. swings) / total sample size

Since you are working with the mean, your end results don't fall into consideration.

[Justin A]

The loose player has a higher standard deviation.

Look at it this way. For the tight player, his "normal" occurance is to win. Deviations are from this norm, not from even. So if his results graph looks like a very straight upward sloping line, his standard deviation will be very low.

On the other hand, the loose player will likely have a high standard deviation, regardless of his results. You'll see a lot of mountains and valleys on his results graph which correlates to a high standard deviation.

[KidPokerX]

this is wrong.
Since we do not know each players' swings, we do not have enough information to tell. Look at the SD formula.

[bobbyi]

Standard deviation means deviation from the "mean". If you aren't a breakeven player, then your mean isn't breaking even, so being "exactly where you started" isn't a sign of low standard deviation.

As a clarifying example, if you won exactly $50 every hour that you played poker, then your standard deviation would be zero.

[Tommy Angelo]

"Standard deviation means deviation from the "mean". If you aren't a breakeven player, then your mean isn't breaking even, so being "exactly where you started" isn't a sign of low standard deviation.

As a clarifying example, if you won exactly $50 every hour that you played poker, then your standard deviation would be zero."

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T h a n k Y o u ! !

This moment is the first time in my life I understood what standard deviation meant in words. I guess it's the time-unit thing that always throws me.

If someone asked you, "What is standard deviation a measurement of?" how would you reply?

Before your post, I would have said, "It is a measurement of fluctuation per unit time, and you can make up whatever time unit you want. You can do it by the hour, or by the 100 hands, which are the usual ways. In this case, you are measuring fluctuation over the smallest time scales. Or you could use years or decades as a time unit I suppose, which no one ever does, which has always been the root cause of my puzzlement over this matter."

Now, instead, I would just say, "Go ask bobbyi."

Tommy

[NLSoldier]

[ QUOTE ]
If someone asked you, "What is standard deviation a measurement of?" how would you reply?

[/ QUOTE ]

Variance (from the mean).

[stigmata]

To answer the original question, I think it's established that SD is correlated with loose-tightness (amongst other things) more than WR. Therefore, without any other information, we would expect the looser player to have a higher SD, irrespective of their winrates.

The way I understand it, playing more hands increases variance, which I think Justin A was trying to get at.