coin is fair?

[gaijinuronin]

Is online poker rigged or not?

Hard to say because we need aggregated hand histories to do serious research.

But I can do some math against specific situations.


Let's say I have hands H1,H2,....Hn where you are against one opponent and one of you two (or both) moves ALL IN.
Let's name your equities E1(H1),E2(H2).....when an ALL IN situation occurs.

Let's define a function that returns the size of the pot when you WIN and 0 when you loose.
R(i)={ 0, if you loose
POTSIZE(i), if you are the winner
}


R(1)+R(2)+R(3)..................
and
E1(H1)*POTSIZE1+E2(H2)*POTSIZE2+...............

SHOULD CONVERGE!!!




Feel free to download and test it!



It works only for NL Holdem and with Postgresql based PT for the moment.

I think I need help to interprate the result from a statistic point of view.

Is it possible to reduce somehow the above problem to a classic coin is fair ?

...and then apply "Estimator of true probability" ?

At the end my programs shows something like this:

"Analyzed games: 230
!!!! aggregate should_win= 8205.752
aggregate actually_win=7618.891
Average pot:57.372"

[T50_Omaha8]

[ QUOTE ]
Analyzed games: 230

[/ QUOTE ] I think this might have something to do with the statistical discrepancy. [img]/images/graemlins/smile.gif[/img]

In all seriousness, based on my ghetto central limit theorem check this is a quite a few standard deviations away from the mean (around 11). But we need more info to be at all certain.

For example: if one pot was, say, $1000, and the others were quite small, having your winnings diverge this much from your expected winnings would be much more probable--likely, in fact. Assuming ALL your pots were exactly $57 causes the standard deviation of your winnings/hand to plummet, making anomalies appear much less probable.

I'm no statistician, however, so I'll leave it to the pros to say for sure.

[AaronBrown]

First of all, I think E depends on both your hand and the other player's hand. If you just compute the equity given your hand, you are confusing poker skill with rigging. Also, I think E is the probability of winning the pot, rather than your pot equity (E*Potsize = Pot Equity).

I suggest adding one more piece of information, the sum of:

E(i)*[1 - E(i)]*Potsize(i)

The square root of this sum will give you the standard deviation of your winnings, assuming the game is fair. If your result deviates from expectation by many standard deviations, there is evidence the game is not fair.

[gaijinuronin]

[ QUOTE ]
First of all, I think E depends on both your hand and the other player's hand. If you just compute the equity given your hand, you are confusing poker skill with rigging. Also, I think E is the probability of winning the pot, rather than your pot equity (E*Potsize = Pot Equity).


[/ QUOTE ]

If choose E to be the probability of winning the pot then I don't deal with pot splits.
E is the equity(includes pot splits stuff as far as i know) of my hand against the opponent hand when one of us moves ALL IN.If there are community cards when one of us moves ALL IN I take them into consideration otherwise I don't (Preflop ALL IN).

Anyway I need to redefine R(i) to deal properly with pot splits:
R(i)={ 0, if you loose
POTSIZE(i), if you are the winner
POTSIZE(i)/2,if pot split
}

I think it is ok now.

[ QUOTE ]

I suggest adding one more piece of information, the sum of:

E(i)*[1 - E(i)]*Potsize(i)

The square root of this sum will give you the standard deviation of your winnings, assuming the game is fair. If your result deviates from expectation by many standard deviations, there is evidence the game is not fair.

[/ QUOTE ]

Thanks for info.
I will start to read about "Standard deviation" and include this stuff in the next release.Hopefully today.

I'm also thinking about creating a site to aggregate the results of the program analyzes.

[gaijinuronin]

I think I have some problems reducing this problem to a "standard deviation " problem
(http://en.wikipedia.org/wiki/Standard_deviation)

THe problem derives from the fact that everything is "weighted" with Pot sizes.

Could I use some kind of "weighted mean" here?
Like:
mean of X = (E(1)*Potsize(1)+E(2)*Potsize(2)+....) / (N)
and considering X(i)=R(i) where

R(i)={ 0, if you loose
POTSIZE(i), if you are the winner
POTSIZE(i)/2,if pot split
}

[AaronBrown]

[ QUOTE ]
If choose E to be the probability of winning the pot then I don't deal with pot splits.

[/ QUOTE ]
You may have missed my point, I'm not sure. I wasn't talking about the possibility of pot splits, I was saying that you have to compute E knowing both hands. For example, say you go all-in with KK when the board is KQJ3. There are 45,540 combinations of cards that could make up the other player's hand and the river card. 42,095 (92%) of them win for you. But that doesn't mean you have a 92% chance of winning. If the other player holds AT, you have only 10 outs, for 23% equity. If you set E = 92%, you're going to thing you're unlucky, because the other player will usually have better than a random hand. Only if you know what the other player has, and compute E on that basis, will your results be valid.

The variation of pot size is not a problem. The formula I gave shows the standard deviation of winnings. You could instead treat every pot size as the same, in all three of the profit calculation, expected profit calculation and standard deviation calculation. If you thought the site was rigged ignoring pot size, this would be a better test. But if you thought the site was more likely to cheat you on the biggest pots, then you should do your initial computation.

[gaijinuronin]

[ QUOTE ]

Only if you know what the other player has, and
compute E on that basis, will your results be valid.


[/ QUOTE ]
This is what I'm doing in my program. I take into
consideration only the ALL IN hands that went to
showdown.This way I can compute the exact equity(my
hand vs his hand) at the moment ALL in
happens(preflop/flop/turn).


I've applied the formula you gave me:
standard deviation = sqrt( sum after i (E(i)*[1 -
E(i)]*Potsize(i)))
k = abs(total expectation - actual result ) /
standard deviation

When I run my program for smaller pots (< 80BB) I
have the following result:
-------------------------
Analyzed games: 577
aggregate should_win= 5518.49
aggregate actually_win=5578.96
Average pot:15.52
k=1.55
----------------------

When I run the program for bigger pots (>80BB) I get the following results:
---------------------------
Analyzed games: 230
aggregate should_win= 8309.21 aggregate actually_win=7703.83
Average pot:57.37
k=13.86
-----------------------------------

Can I use on this data the Chebyshev's inequality implication?
"At least (1 − 1/(k*k)) · 100% of the values are within k standard deviations from the mean."

For small pots(k=1.55) I have 20%
For big pots(k=13.86) I have 99.995%

I didn't touch any math problems for a long time so sorry if there are any obvious errors in my judgment.

Thanks,

[T50_Omaha8]

I'm going to try to recover SD of expected winnings and apply the central limit theorem. Chebyshev's inequality is for a general random variable and thus tends to be highly inaccurate. Plus I forgot the damned thing and it showed up on the probability actuarial exam I just took. [img]/images/graemlins/blush.gif[/img]

Okay, so you have added the following identity:

k = (E$-A$)/SD

From my math you have

13.86 = (8309.21-7703.83)/SD
SD = 605.38/13.86
SD = 43.68

Now we can apply the central limit theorem, noting that Phi(z) denotes the cumulative distribution function of the standard normal distribution, and also noting all equals signs approximate:
P((sum(A$(i)) - n(E$))/(SD*sqrt(n)) < (7703.83 - 8309.21)/(43.68*sqrt(230)))
= Phi((7703.83 - 8309.21)/(43.68*sqrt(230)))
= Phi(-605.38/(43.68*15.17))
= Phi(-605.38/662.44)
= Phi(-.914)
= 1-Phi(.914)
= 1-.819
= .181

18.1%

My guess is you're in the unlucky fifth of poker players. I'm always in that fifth as well, so welcome to the club. [img]/images/graemlins/smile.gif[/img]

[gaijinuronin]

What if 80% of the winning poker players ( 5%-20% from total) are in the "unlucky fifth of poker players" ?

[gaijinuronin]

I've found some interesting stuff like
"weighted standard deviation"
http://www-minos.phyast.pitt.edu/disdocs/weightsd.pdf
and
"weighted mean"
http://www-minos.phyast.pitt.edu/disdocs/weigmean.pdf

Could it be possible to apply them with

W(i)=E(i)
and X(i) to be the outcome of an ALL in:
X(i) = {
0, if you loose
POTSIZE(i), if you are the winner
POTSIZE(i)/2,if pot split
}

Hmm I think I'm playing a guessing game so pls ignore this post.