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All comments by Henk Boer
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From the 116 video boards where BZ defended, Balicki held a 5-card as longest suit 55 times. That is 47.41 %
Oct. 20, 2015
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Please have a look at the figures I posted here:
http://bridgewinners.com/article/view/the-videos-speak-balicki-zmudzinski/?cj=259357

Michal, can you (or anybody) check them and recalculate the appropriate test results?
Oct. 20, 2015
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Now I have observed and counted ALL video boards where BZ defended in Opatija. Again thanks to Nicolas' website.

Total nr of boards: 116
- Balicki on lead: 56
- Zmudzinski on lead: 60

- Balicki held a 5-card as longest suit: 55 times
- Balicki made the 5-gesture: 22 times

Again the hypothesis is that the 5-finger-gestures by Balicki are independent from holding a 5-card suit (as longest suit).

————— No5card —- 5card
NoSignal —- 57 ——— 37
5-Signal —- 4 ———– 18

Total ——– 61 ———– 55

The expected numbers are based on:
5-signal occurred 22 times out of 116
5Card occurred 55 times out of 116

————— No5card —— 5card
NoSignal —- 49.43 ——— 44.57
5-Signal —- 11.57 ———– 10.43

Result:
X2 = 12.89
p-value = 0.000331
Oct. 20, 2015
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For who is interested: in the example of 550 heads the p-value is 0.001565. Very fishy.
Oct. 18, 2015
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Michal, you assume I took a sample from a bigger set. But actually I observed and counted them ALL. There are 60 boards where Zmudzinski is on lead. So in this case N = N1 = 60.

Regarding p(5card) and p(no5card): the overall theorical p might be 0.443, but it may depend on whether you declare or not and on the bidding style of the player. Just like I used the whole set to determine the probability of a 5-signal (13/60) I used the whole set to determine the probability of a 5card (30/60).
Oct. 18, 2015
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Here are the numbers based on all boards where Zmudzinski is on lead. The total nr of boards is 60.
The numbers are based on my own observations.

————— No5card —- 5card
NoSignal —- 28 ——— 19
5-Signal —- 2 ———– 11

Total ——– 30 ———– 30

The expected numbers are based on the fact that the 5-signal occurred 13 times out of 60:

————— No5card —— 5card
NoSignal —- 23.5 ——— 23.5
5-Signal —- 6.5 ———– 6.5

Result:
X2 = 7.95
p-value = 0.00480
Oct. 18, 2015
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Srikant, you are right. I updated the expected numbers in the post above. And I removed my conclusion. Others may comment on that.
Oct. 18, 2015
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@ Michal: yes, 1 df
Oct. 18, 2015
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Arkadiusz suggested to do a Chi^2 test. And so I did. I tested the hypothesis whether the 5-finger-gestures by Balicki are indipendent from holding a 5-card suit.

I first selected all 116 video records from Hammond's famous excel sheet at http://www.bridgescoreplus.com and then made a selection with the following criteria:
- Only the boards where Zmudzinski is on lead
- Only the records where BZ did not bid

Total nr of boards: 31

Then I did my own observation on these 31 boards to determine whether B signalled a 5 card or not. I regarded a 4 card signal as ‘No signal’. In one occasion the 5-signal was very late. I still counted this as ‘5-Signal’.
The result was the following table:

———— No5card —- 5card
NoSignal —- 14 ———– 9
5-Signal —- 2 ———– 6

Total ——– 16 ———– 15

I would have expected the following count if there was no relation between the signal and the hand:

EDIT: I changed the expected results, since indeed the 5-signal only occurs 8 times out of 31.

———— No5card —- 5card
NoSignal —- 11.87 ——- 11.13
5-Signal —- 4.13 ——– 3.87

Result of the calculations:
X2 = 3.06
p-value = 0.08035

I removed my first conclusion since the values are now different. Is this significant enough?
Oct. 17, 2015
Henk Boer edited this comment Oct. 18, 2015
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BZ61: edge;edge;small;medium;medium;medium;large;medium
BZ62: edge;edge;small;medium;medium;small;medium

Edit: nrs corrected
Oct. 4, 2015
Henk Boer edited this comment Oct. 4, 2015
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BZ61, BZ62
Oct. 4, 2015
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