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All comments by Michal Czerwonko
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An important correction. There is a proper procedure designed for small samples, where small sample is defined as the lowest frequency in the table of 5 or less. The test applied by Henk was asymptotic, which was not valid since the lowest observed frequency was 2. Unfortunately, two exact tests, chi2 and Fisher produce higher p-values for the table provided by Henk, respectively of 0.1134 and 0.0900. Since I wrote a small code, which performs the test, I'll be glad to apply to new frequency tables. The code is in SAS. I provide the code should anyone have access to SAS and would like to replicate the results or derive new results:

/*beginning of code*/

data bz_table;
input suit5 signal count;
datalines;
0 0 14
1 0 9
0 1 2
1 1 6
;
proc sort data=bz_table;
by descending suit5 descending signal;
run;

proc freq data=bz_table order=data;
tables suit5*signal / chisq;
exact pchi;
weight count;
title ‘BZ Test’;
run;

/*end of code*/

Binary variable “suit5” is set equal to 1 if the five-card suit was present, to 0 otherwise. Binary variable “signal” is set equal to 1 if the signal was observed, to 0 otherwise. Variable “count” provides the number of occurrences of a given event.
Oct. 18, 2015
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In the poll I bid 7 but upon reflection I realized that pulling to 5 already promised a reasonable offensive hand. 6 is enough.
Oct. 18, 2015
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I don't agree with your computation of the expected frequency of the signal. It is based on the actual observations in the sample you test, which approach is strictly invalid. The contingency table you use to test your hypothesis requires the expected values as entries. Your observed frequency in the sample you test is not equal to the expected frequency. It will also be better to split the total sums of your columns with expected frequencies along the actual probabilities of 5-card suit, i.e. 0.443 and 1-0.443

To estimate this expected frequency of a signal, you need to look for a larger set of say, N hands and get the overall frequency of signal n. In the best case this larger set will not contain the test set. Then for your test sample of the size N1, you get expected frequency in each column as (n/N)*N1*p(5card) and (n/N)*N1*p(no5card)

I apologize for not doing any hard work myself. I have hectic time with revising two papers for a journal, polishing a submission of a different paper and bringing another to circulation.
Oct. 18, 2015
Michal Czerwonko edited this comment Oct. 18, 2015
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That's not the case. Generally p-values lower than 5% are considered to soundly reject the null hypothesis. It is possible that small sample size is the problem, but this problem may work either way. For instance, if for another set of hands of similar size you again obtain a similar p-value of 0.08, it would indicate the p-value for the joint sample roughly of 0.08^2<1%; of course, a similar p-value should obtain for the test on the joint sample.
Oct. 18, 2015
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“The sole problem in communication is the illusion the communication has happened”
Oct. 17, 2015
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Peg Kaplan
“So knowingly playing on a team with cheaters is OK if you're not getting paid? Hmmm; I sure don't think so!”

What has it to do for Pete's sake with “The problem Jessica raised is whether Boye cashed his checks while knowingly being on the team with cheaters.”
Oct. 17, 2015
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“So knowingly playing on a team with cheaters is OK if you're not getting paid? Hmmm; I sure don't think so!” The most polite comment I may add to this statement is that it makes no reference to any statement I've made
Oct. 17, 2015
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The problem Jessica raised is whether Boye cashed his checks while knowingly being on the team with cheaters. The best Boyle's argument was ‘I don’t go to a bar as often as Jessica, so how could I know the FS talk.' I don't think it's a perfectly serious response.
Oct. 17, 2015
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My question was for what exactly video data you look for 5/4-finger gesture
Oct. 17, 2015
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If you develop a bidding system, you'd like your bids to pertain to situations which appear with reasonable frequency; for instance, no system is good to bid 8-5 distributions. I hypothesize that i. ‘good’ cheating should occur often; ii. spreading your (almost) entire hand is less suspect than showing one finger. Having said that, I still think that a bit more work is required to put the case ‘beyond reasonable doubt.’
Oct. 17, 2015
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Kit, Are you speaking about the videos submitted by Nicolas or about all the BZ evidence? As far as I've been advocating skepticism throughout the whole BZ debate, I'm inclined to think that the analysis of the time when a useful information for defense may be provided is critical.
Oct. 17, 2015
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A quick question: did you use one degree of freedom? (d.f.=(Nrows-1)x(Ncols-1)
Oct. 17, 2015
Michal Czerwonko edited this comment Oct. 17, 2015
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Sirkant, These are expected frequencies you use to built a contingency table. Then the null hypotheses is that observed frequencies are equal to the expected ones. The differences between the expected and the observed you compare to the critical values of chi-square distribution
Oct. 17, 2015
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The suit of the lead is provided, so it identifies the code
Oct. 17, 2015
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I read an official statement by PBU that suspected pairs (without specific mention of BZ) are presently investigated by a committee designated by EBL with the investigation results to arrive in November
Oct. 17, 2015
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“Odprysk” means a splinter, here in this sense that the scandal has a Polish part. The general meaning is correct
Oct. 17, 2015
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Arek, chi^2 rather than x^2. You may look at Nicolas Hammond data provided in his available article if you'd like to built a contingency table and test formally very unusual BZ hand gestures
Oct. 16, 2015
Michal Czerwonko edited this comment Oct. 16, 2015
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I watched for the first time BZ videos and must agree that the finger extending gesture is extremely unusual. Great piece of detective work. BTW, why to be so unoriginal and copycat Reese-Shapiro?
Oct. 16, 2015
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Based on the hand played by Ken Cohen against BZ, I suspect Ken peeked at his LHO hand. How otherwise to explain a coincidence of an aggressive preempt and a problem hand to the left of the preemptive action? I'm 100% certain that the coincidence of the actions described by Ken Cohen in his story establishes him as a cheat with a very high probability.
Oct. 16, 2015
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Self reply: 4 may also be bid with many weak hands which contain no fit
Oct. 15, 2015
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