NL Bots on Full Tilt

[Zele]

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WTF, I'm rusty on my statistics, but are you going to tell me that a .5% difference in VPIP over 100,000 hands is 3 SD's? [censored], post your math.

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(14%)*(1-14%)/sqrt(100000) = .038% = 1SD

Of course it's higher than that because all 100,000 decisions are not independent identically distributed.

[DWarrior]

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WTF, I'm rusty on my statistics, but are you going to tell me that a .5% difference in VPIP over 100,000 hands is 3 SD's? [censored], post your math.

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(14%)*(1-14%)/sqrt(100000) = .038% = 1SD

Of course it's higher than that because all 100,000 decisions are not independent identically distributed.

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I think the test statistic formula is .005/sqrt(.14*.86/100000) = 4.6

Assuming independence, and I'm just using a textbook formula for population proportions.

[IRuleYouHard]

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WTF, I'm rusty on my statistics, but are you going to tell me that a .5% difference in VPIP over 100,000 hands is 3 SD's? [censored], post your math.

[/ QUOTE ]

(14%)*(1-14%)/sqrt(100000) = .038% = 1SD

Of course it's higher than that because all 100,000 decisions are not independent identically distributed.

[/ QUOTE ]

I think the test statistic formula is .005/sqrt(.14*.86/100000) = 4.6

Assuming independence, and I'm just using a textbook formula for population proportions.

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We just went over this last week at work..
(sigma) = squareroot( (sumation((x-change of x)squared) divided by n-1

[DWarrior]

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WTF, I'm rusty on my statistics, but are you going to tell me that a .5% difference in VPIP over 100,000 hands is 3 SD's? [censored], post your math.

[/ QUOTE ]

(14%)*(1-14%)/sqrt(100000) = .038% = 1SD

Of course it's higher than that because all 100,000 decisions are not independent identically distributed.

[/ QUOTE ]

I think the test statistic formula is .005/sqrt(.14*.86/100000) = 4.6

Assuming independence, and I'm just using a textbook formula for population proportions.

[/ QUOTE ]
We just went over this last week at work..
(sigma) = squareroot( (sumation((x-change of x)squared) divided by n-1

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I'm not sure what this refers to.

But can't we also use the Goodness of Fit test for this?

We have 4 suspected bots in the OP's screenshot:

VPIP1: 14376/105366 = 0.136438699
VPIP2: 15840/112514 = 0.14078248
VPIP3: 11683/82577 = 0.141480073
VPIP4: 5721/41414 = 0.138141691

The total VPIP is 47620/341871 = 0.139292306

So if they are indeed botting, the expected hands played of the 4 bots should be:
14676.67
15672.33
11502.34
5768.65

respectively

Throwing it into my calculator for GOF, using 3 degrees of freedom, we get a p-value of .0108, which means there's just over 1% chance that this happened by chance if we assume the 4 players were using a similar strategy.

[dp13368]

Dwarrior, this just takes into account VPIP. I think the more relevent information here is the astounding similarities between every single stat (vpip being one of the least complex).

[DWarrior]

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Dwarrior, this just takes into account VPIP. I think the more relevent information here is the astounding similarities between every single stat (vpip being one of the least complex).

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But VPIP is the one that has the biggest sample size, and one most susceptible to botting, won't you agree? There were 3 theories being thrown: 1) two players just learning together 2) player using bots for PF and then taking over post-flop 3) bot playing almost exclusively with the player only jumping in in rare situations. Only #2 and #3 are illegal and involve botting, and the VPIP in both of these situations would be 100% determined by the bot pre-flop.

My test proves, with almost 99% certainty, that the VPIP of the four accounts in question was not the same, meaning that they did not have the bot playing them. Once you disprove this, doesn't the whole botting argument fall apart?

EDIT: I also suspect this GOF test would prove the same with every stat, but it's a bit tedious to do, and I think there is already enough evidence on this thread to prove that these guys do not run bots.

[BluffTHIS!]

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Dwarrior, this just takes into account VPIP. I think the more relevent information here is the astounding similarities between every single stat (vpip being one of the least complex).

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But VPIP is the one that has the biggest sample size, and one most susceptible to botting, won't you agree? There were 3 theories being thrown: 1) two players just learning together 2) player using bots for PF and then taking over post-flop 3) bot playing almost exclusively with the player only jumping in in rare situations. Only #2 and #3 are illegal and involve botting, and the VPIP in both of these situations would be 100% determined by the bot pre-flop.

My test proves, with almost 99% certainty, that the VPIP of the four accounts in question was not the same, meaning that they did not have the bot playing them. Once you disprove this, doesn't the whole botting argument fall apart?

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The onus is on you to prove that VPIP is both necessary *and* sufficient along to prove/disprove botting, and thus that there aren't one or more stats which are more indicative.

[IRuleYouHard]

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WTF, I'm rusty on my statistics, but are you going to tell me that a .5% difference in VPIP over 100,000 hands is 3 SD's? [censored], post your math.

[/ QUOTE ]

(14%)*(1-14%)/sqrt(100000) = .038% = 1SD

Of course it's higher than that because all 100,000 decisions are not independent identically distributed.

[/ QUOTE ]

I think the test statistic formula is .005/sqrt(.14*.86/100000) = 4.6

Assuming independence, and I'm just using a textbook formula for population proportions.

[/ QUOTE ]
We just went over this last week at work..
(sigma) = squareroot( (sumation((x-change of x)squared) divided by n-1

[/ QUOTE ]

I'm not sure what this refers to.

But can't we also use the Goodness of Fit test for this?

We have 4 suspected bots in the OP's screenshot:

VPIP1: 14376/105366 = 0.136438699
VPIP2: 15840/112514 = 0.14078248
VPIP3: 11683/82577 = 0.141480073
VPIP4: 5721/41414 = 0.138141691

The total VPIP is 47620/341871 = 0.139292306

So if they are indeed botting, the expected hands played of the 4 bots should be:
14676.67
15672.33
11502.34
5768.65

respectively

Throwing it into my calculator for GOF, using 3 degrees of freedom, we get a p-value of .0108, which means there's just over 1% chance that this happened by chance if we assume the 4 players were using a similar strategy.

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I have no idea what I am talking about... I was just excited to use that formula [img]/images/graemlins/laugh.gif[/img] [img]/images/graemlins/tongue.gif[/img]

[DWarrior]

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(sigma) = squareroot( (sumation((x-change of x)squared) divided by n-1

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I have no idea what I am talking about... I was just excited to use that formula [img]/images/graemlins/laugh.gif[/img] [img]/images/graemlins/tongue.gif[/img]

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Can you explain what (x-change of x) means? It sounds like the formula for standard deviation, except n-1 should also be square rooted.

[IRuleYouHard]

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(sigma) = squareroot( (sumation((x-change of x)squared) divided by n-1

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I have no idea what I am talking about... I was just excited to use that formula [img]/images/graemlins/laugh.gif[/img] [img]/images/graemlins/tongue.gif[/img]

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Can you explain what (x-change of x) means? It sounds like the formula for standard deviation, except n-1 should also be square rooted.

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Yes you are correct.. that is hard to type on the innanets.
x is the mean of all the data and the change of x would be how far from the mean it is. so you take for each point of data the distance from the mean and square it... then add all those up.
divide by 1 less than the number of data points you have.
take square root of that for sigma (standard deviation)