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.

[Justin A]

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

[/ QUOTE ]

I'd reply that it's a measure of the luck factor in poker.

[Justin A]

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

[/ QUOTE ]

It's not wrong. I'm going off of experience and common sense that tells me a looser player is going to have greater swings and a higher standard deviation.

[KidPokerX]

But you do not know the mean for both players. What if Player A plays micro-limit penny poker and comes up $5 over the course of 100,000 hands and Player B plays $3,000/$6,000 HE and losses $12,000 over 100,000 hands. Since you do not know the average (mean) gain/loss for either player, you cannot determine who experienced greater swings (on a % basis). Your common sense is telling you one thing, but the hard evidence is not there. Player B may have a smaller deviation than Player A - you just don't know.

If you still wish to argue this point I would like you to include the S.D. formula.

[Justin A]

[ QUOTE ]
But you do not know the mean for both players. What if Player A plays micro-limit penny poker and comes up $5 over the course of 100,000 hands and Player B plays $3,000/$6,000 HE and losses $12,000 over 100,000 hands. Since you do not know the average (mean) gain/loss for either player, you cannot determine who experienced greater swings (on a % basis).


[/ QUOTE ]

You're kidding right? The mean for the loose player is 0, and the mean for the tight player is some positive number, which will most likely be his total winnings in big bets divided by hours played if that's the units we choose.

[ QUOTE ]
Your common sense is telling you one thing, but the hard evidence is not there. Player B may have a smaller deviation than Player A - you just don't know.

[/ QUOTE ]

This is ridiculous. Tight players play less hands and that's going to give them smaller swings per hour.

[ QUOTE ]

If you still wish to argue this point I would like you to include the S.D. formula.

[/ QUOTE ]

If you insist. s = [(1/n-1) * E (x - X)^2] ^ 1/2, where s is S.D., E means the sum of, x is each value in our sample, and X is the mean. Guess what, the tight player plays less hands, which means his x - X for each hour played will be much smaller on average than the loose player.

Please feel free to argue further, but I'm right about this.

[Str8Fish]

What can you do with your standard deviation once you know it? Are there any cool formulas you can plug your standard deviation into to tell you something about your game?

[bobbyi]

[ QUOTE ]
What can you do with your standard deviation once you know it? Are there any cool formulas you can plug your standard deviation into to tell you something about your game?

[/ QUOTE ]
Using your standard deviation and your win rate, you can answer questions like "how large a bankroll do I need to have a less than one percent chance of going broke?" and "how likely is it that I will break even or worse over the course of 10k hands (or over 100 hours if you are tracking that way, etc.)?" I don't have the forumulas for these things off the top of my head, unfortunately.

The problem is that thse require an accurate sense of your winrate as well as your s.d. and win rates take a long time to converge (s.d. doesn't take long to converge), so you need to make an educated guess unless you have played for thousands of hours. How long does it take for your winrate to converge? The answer depends on your standard deviation...

[Justin A]

[ QUOTE ]
What can you do with your standard deviation once you know it? Are there any cool formulas you can plug your standard deviation into to tell you something about your game?

[/ QUOTE ]

Yeah, standard deviations are useful in figuring out a confidence interval for your winrate. You can find your winrate and standard deviation in poker tracker, and use those to figure out a range that your true winrate most likely lies in. Here's a simple example:

Suppose you've played 100,000 hands with a winrate of 1.8 bb/100, and a standard deviation of 15 bb's per 100 hands. You have to find your standard error, which is simply std deviation divided by the square root of the number of samples. 100,000 hands converts to 1,000 sets of 100 hands.

So, S.E = 15/(1000^1/2) = .474

Now lets say we want to find a 95% confidence interval for our winrate. 95% corresponds to a "Z" value of 1.96. The Z value basically corresponds to number of standard deviations from the mean. All we do is multiply this value by our standard error of .474, and we get .929.

So 95% of the time, our true winrate will be between .871 bb/100 and 2.729 bb/100.

I realize this is probably confusing because it's really hard to explain it and type it out coherently. Let me know if you have any questions.