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.

[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.

[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.