My chance of being a winner?

[Rudbaeck]

Since I bought PokerTracker I've played 21,415 hands, I've won 2.63BB/100, and have a standard deviation of 16.7454BB/100.

How do I calculate the confidence interval of me being a long term winner?

[NUReedy]

You can be 95% confident your true BB/100 hands is within
[2.63 - 1.96(16.745/sqrt(21415), 2.63 + 1.96(16.745/sqrt(21415)]
OR
[2.405, 2.854]

[Rudbaeck]

Thank you. [img]/images/graemlins/smile.gif[/img]

[Rudbaeck]

Giving it some more thought, shouldn't the divisor be sqrt(214.15)?

Both win rate and sd are calculated for sets of 100, so reasonably we should count how many sets of 100 I have?

That would give a 95% confidence interval that I am within a deviation of 2.63+-1.96(16.7454/sqrt(214.15))=2.63+-2.24

[Nottom]

[ QUOTE ]
Giving it some more thought, shouldn't the divisor be sqrt(214.15)?

[/ QUOTE ]

yes

[ QUOTE ]
That would give a 95% confidence interval that I am within a deviation of 2.63+-1.96(16.7454/sqrt(214.15))=2.63+-2.24

[/ QUOTE ]

This is probably about right, now you see why people are always saying you need many many many hands to have even a reasonable estimate of your true winrate.