question about standard deviation

[Justin A]

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

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

[Justin A]

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

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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]

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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).


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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.

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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.

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

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

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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]

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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?

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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]

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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?

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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.