Std q; true win rate?
 

As one plays more hand, is it intuitive that your std of your winrate will decrease? Or just more accurately resemble your true variance?

I cant figure this out.

Also, is std a good measurement of "running good," or is this simply votality of how I'm playing?

thanks.

Re: Std q; true win rate?
 

The standard deviation will approach its true value as n increases. If it were decreasing, as you said, then eventually you would have a standard deviation approaching 0, which doesn't make much sense here.

Re: Std q; true win rate?
 

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The standard deviation will approach its true value as n increases. If it were decreasing, as you said, then eventually you would have a standard deviation approaching 0, which doesn't make much sense here.

[/ QUOTE ]

The standard deviation of the win RATE (called the standard error) does approach zero as the number of hours N approaches infinity. It is the true standard deviation of the of the winnings for 1 hour divided by sqrt(N), and the fact that it goes to zero reflects the fact that that the average win approaches the true win rate as N becomes large. This is not to be confused with standard deviation of the winnings for 1 hour which is what we normally refer to as the standard deviation, and this approaches its true value as N becomes large.