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Assumption: cheater plays with 90% VPIP.

In the 25 hands analysed by Adanthar (the first ones listed in the cliff notes) the cheater folds whenever another player - and only when another player - has AA, KK, QQ, JJ. There is another hand analysed, but it involves a re-raise squeeze, which is a slightly wider range, and we can ignore it for this analysis.


So - what are the chances that a player who plays 90% of hands will randomly fold the 4 precise hands that someone else has AA, KK, QQ, or JJ, and none others?

I think the mathematics can be written out like this:

.1^4 * .9^21

ie, it is .1 * .1 * .1 * .1 * .9* .9* .9* .9* .9* .9* .9 etc.
^--these are the four premium hands ^--these are the rest

chucking that into excel provides the following answer:

0.00109419% chance of occuring.

in other words, a 1 in 100,000 chance of occuring randomly.

hopefully this methodology is right. can someone who is smarter than me (ie, almost anyone) confirm or correct me?

does this then mean that we can say with 99.99890581% certainty that the cheater was cheating in these hands? the more i think about it, i think not, but i feel that there is a calculation somewhere here that would give us the probability that this was cheating - can someone who knows mathematics/statistics well chime in?

incidentally, the figures vary a bit depending on what the cheater's VPIP is. if the VPIP is .95, then there is a 0.00021285% and a 99.99978715% chance respectively. if the VPIP is .8, 0.00147574% and 99.99852426%)

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Well.... no.

With a VPIP of 90%, there is a 1/10 that Villain folds.

It's a little more complicated to figure out how often one of his opponents has AA-JJ, but 10% is somewhere in the neighborhood, and will make the calculations easier, and won't much effect the unlikeliness of the result.

Four possible outcomes on each hand:
Villain folds, an opponent has AA-JJ: (.1 * .1) = .01
Villain folds, an opponent doesn't have AA-JJ: (.1 * .9) = .09
Villain doesn't fold, an opponent has AA-JJ: (.9 * .1) = .09
Villain doesn't fold, an opponent doesn't have AA-JJ: (.9 * .9) = .81

Note the sum of the probabilities is 1.

Now for 25 independent events, we just multiply, so you end up with:

(.01^4) * (.81^21) = 1E-10 = 1 in 10 billion. [img]/images/graemlins/shocked.gif[/img]