NL Bots on Full Tilt

[DWarrior]

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- I argue that the reason all of the data is so similar is that he is playing on all of them. Think about this: Take 400k hands and split them up into four 100k chunks. Won't the data for each of these 100k hands be very similar to each other?

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But VPIP, which should obviously be identical across the board, is statistically different between the 4 accounts listed.

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How about this: At some time t0, he decided he wanted to adjust something in the strategy. He has four accounts with x1, x2, x3, anx x4 hands each. But say they're all different # of hands. The stats for each of those will be different based on how many hands he played at his strategy before t0 and the number of hands he plays with the new strategy after t0.

Example:

Set1: 110
Set2: 111100

Both have average of .67.

Change strategy to play all hands. Add 1 to each

Set1: 110 1
Set2: 111100 1

Average for 1: 0.75
Average for 2: 0.71

Same strategy, different averages, hmm. Maybe theres is a lesson here? [img]/images/graemlins/wink.gif[/img]

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A valid point.

However, two of the accounts have basically the same number of hands (105k hands and 112k hands), so we can assume they were datamined at the same time. How come their VPIPs are so significantly different then? (13.64% and 14.08%) That difference is over 4 SDs

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has it been confirmed that the true deviation is over 4SDs? earlier there were like 4 formulas people were trying to use. if true this fact needs much more attention as it has the greatest chance of clearing them

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The reason there were 4 different formulas is because I believe 2 people used wrong ones. Also the two I gave were for slightly different questions.

Here is the test to show that VPIP of 1forthethumb (14376/105366 or 13.64) differs from that of full_tilting(15840/112514 or 14.07):

p1 is VPIP of 1for, p2 is VPIP for full_tilting, p is their combined vpip (weighted average of the two). n1 and n2 are the sample sizes.

(p1-p2)/sqrt( p*(1-p)*(1/n1 + 1/n2) )

This comes out to -2.93, so I overestimated it (the other equation where I got 4 was for a different question). The probability of these results differing by chance is 0.0017.

In case you're wondering, this is Two Sample Hypothesis Test of the difference of two Population Proportions.