question about standard deviation

[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.