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All comments by Okan Zabunoglu
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In critical cases like this, the director, himself/herself, had better carry it to the appeal if (s)he is not really sure of his/her own decision. Well(!), in this case, it seems that the director did not even give a ruling. So, what I am saying is just in vain.
April 6, 2016
Okan Zabunoglu edited this comment April 6, 2016
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5/3 = 1.67
April 6, 2016
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Reflective analysis. I was referring to the English version, though.
April 4, 2016
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It is up to you what to appreciate!
April 4, 2016
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I'd expect a saying which involves a ‘horse’ :)
This one may not completely reflect the following (for instance):
Compare the sadness when you lost it to the happiness when you found it…
April 4, 2016
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Some people do not find it funny because they are concerned. It is interesting to me that for the same reason, i.e., being concerned, I found it funny.

We have an old saying in Turkish (although my translation may not give the exact sense): “Losing your donkey first and then finding it is the best way to appreciate it.”

NOTE: If someone can tell me its equivalent in English, thank her/him in advance.
April 4, 2016
Okan Zabunoglu edited this comment April 4, 2016
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My comment is general, or for the original poster. Did what I said mean that I thought or felt that the joke is bad/immature? Please don't let me be misunderstood!
April 2, 2016
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Because I see BW as a multicultural site, I cannot help saying this. What may be a joke (for a conspicuous day) in a country can be a routine occurrence in another country. This (for some) makes it harder to catch the joke but also yields a more enjoyable effect.
April 2, 2016
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I was going to ask the same question, and am still asking:
“Is there anything that members can do to support the administration?”
April 2, 2016
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To answer your question: “Most of what time?”, I was about to say that simple C(8,4)/C(8,2) does not work most of the time because it is also needed to consider how the other suit is distributed and so on…

But then I read the rest of the sentence carefully and looked through your previous comments and found this “…there are equally many ways to distribute the other 4 hearts 1-3 or 3-1.”

Now I think you know all these. Frankly, I was doubtful about your estimations since you had started with 11:3 and then ignored the effect of placing Q in East hand (upon Kit Woolsey's comment).

I guess we both know that the answer to “Most of what time?” is when the C values related to one of the suits are the same and a ratio (not the percentage of any specific distribution) is needed (so, two of the C’s cancel out).

If I had grasped your approach, I would not have answered your question “or was I glib?” as “yes, I think so (in a positive sense).” It is a smart (quick) way of getting a ratio of two distributions correctly.

In this hand (with 5 and 7 vacant spaces in South and East, and 8 s and 4 s to be dealt), for ratio of 6142/6322 in South, the C(4,1) and C(4,3) of s will cancel out; and the result is 2.5.

Using your approach, if we give East Q only, now we will have 6 spaces in East (5 in South) and 11 cards (4 s and 7 s) to deal. The ratio of 6142/6322 is 7/3 = 1.67. And when East is known to have both QJ, 6142/6322 = 1.0.

As a final and different example, if we were after, say, the ratio 6142/6412 in South, with 5 and 7 spaces and 12 cards (8 s and 4 s) to deal; we would have:

6142/6412 = C(4,1)C(8,4)/C(4,4)C(8,1)=(4x70)/8 = 35

I believe we agree on these.

NOTE: As for DEALER, I don’t want to extend this thread further, I will send you an example input and output related to one of the above cases.
March 31, 2016
Okan Zabunoglu edited this comment March 31, 2016
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Replacing a with a is different from replacing a with a , because a is free to move between East and South (which leaves 7 vacant spaces in East); whereas a is not free to move, it has to go to East, now East will have 6 vacant spaces. We cannot stipulate anything else.

As for your other remarks, I will convey my thoughts, but not tonight. It is 11:00 pm here.
March 30, 2016
Okan Zabunoglu edited this comment March 30, 2016
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I have to look up “glib”…!
Yes, I think you were glib (in a positive sense). Yet, the result of your glib calculation (i.e., 5:2) looks pretty close, according to my rough estimation.

“Is this kind of computation equivalent to some empty spaces computation?”

I don't think so. At first glance, say that we have 8+8=16 empty spaces, now all distributions (8-0, 7-1, etc.) are available (4-4 = 38 %). But if we have 5+5=10 spaces, now only 5-3, 4-4 and 3-5 are there (now, 4-4 = 55 %). And if there are 4+4=8 spaces, 4-4 = 100 %. And it is not only this, there are further issues.

Simple C(8,4) and C(8,2) should not work most of the time. It seems to give a good result (5:2) here; I am not sure why, need to work it out.

Hand calculation is pretty complicated and not worth it (IMO). If you are interested, you may take a look at this:
http://bridgewinners.com/article/view/how-to-play-this-suit/?cj=279061#c279061 (EDITed)

NOTE: When necessary, I use the software DEALER, which generates many (10 000 000, if I remember correctly) hands and gives the ratio of the number of the hands specified (in a short program written by the user) to the total number. If I find time, I will run the Dealer for the case at hand here.
March 30, 2016
Okan Zabunoglu edited this comment March 30, 2016
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I have just noticed that I overlooked the obvious in my last paragraph right above.

When both South and East have five vacant spaces each and four s and six s are to be placed into those, half of the time South and East will have two s and three s, and one-fourth of the time South will have one and three s (and vice versa).

Thus, South's holding (eliminating 6412 and 6052);
6142 = 25 %
6232 = 50 %
6322 = 25 %

In other words, when, as a precondition, East has both Q and J (additionally is known to have exactly one and five s), declarer is equally likely to hold 6142 and 6322. At least it seems so from where I look. If I am wrong, please tell me.

NOTE: Of course, when no card is specified in East hand, South's 6142 is expected to be significantly greater than 6322.
March 30, 2016
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Charles Brenner: “From the point of view of a defender holding J10xxx, declarer is 11:3 more likely to hold x than xxx.”

Purely from a probability angle, do you mind telling how you get 11:3?

I agree that, in general, x is more likely than xxx, but 11:3 seems too high.

Here is my reasoning.

We have the exact distribution of North hand,
and let's give West the following hand: 6 JT654 KT8 A853.
(Note that I replaced a with a , not a with a , because it should make a difference.)

So, North and West hands are specified like this.
Now; South has 5 vacant spaces (six s and two s are known), and East has 7 vacant spaces (one and five s are known). And we have 12 cards (four s and eight s) to deal into these vacant spaces.
Under these conditions, I could not guess how you got 11:3.

And when we fix both Q and J in East, now the number of vacant places in both East and South become the same (five), and ten cards (four s and six s) remain to be placed. I'd expect this to result in an appreciable difference.
March 30, 2016
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In my opinion, the play of to 9 does not have anything to do with how opponents look. I would interpret it as use of a critical information that declarer has but opponents do not (in order to make the contract).
March 29, 2016
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Right, I should have said: 6232 does not matter since declarer is supposed to double finesse anyway.

In summary,
6412: split loses
6322: may not matter, but not sure
6232: does not matter
6142: split wins.

Seeing four s and two s in the dummy, 6142 is more likely than 6412. Then, if 6322 does not matter at all, split should be OK.
March 29, 2016
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Thanks for the search. Just a note: The poll was for JTxx, thus, p=0.10 is valid for this only. E.g., for JTxxx, I don't know the result (I just assumed that it would be 50 %.) And I did not consider JTxxxx and longer, since in these cases it will be obvious to split or, in some, a lightner double may be thought of.
Multiplying 21.17 by 0.1 means that 0.1 is valid for all (except JTx and JT tight), which is not a result of the poll.
March 28, 2016
Okan Zabunoglu edited this comment March 28, 2016
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I ignored 8-0, 7-1 and 6-2 of s (between LHO and RHO).
The remaining ones (in LHO)are
JT: 0.3
JTx: 2.5
JTxx: 7.0
JTxxx: 8.4

The total available is 18.3 %.

Assuming that, with JTxx, about 90 % of the time; with JTxxx, about 50 % of the time to 9 succeeds, then it gives 0.3 + 2.5 + 7.0x0.9 + 8.4x0.5 = 13.3 %.
March 28, 2016
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Thank you all for your comments and votes.

I “abstain”ed because I was not sure what is right to do; yet, now, after combining some of the comments, I am inclined to vote for “insert J/T”.

As said by Charles Brenner, it is more likely that declarer will have more s than s. When declarer has 6-2-3-2 or 6-1-4-2, it is correct to split. As Kit Woolsey says and Nigel Kearney implies: Even when declarer has 3 hearts, splitting may be OK…

Then, what is left behind? 6-4-1-2 only. Just let it go!

I will convey the story of the hand in a separate post soon, since it is also interesting to look at the hand from the declarer's point of view.

NOTE: Mike Summers-Smith got the declarer's distribution right.
March 28, 2016
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Yuan: shift is on the second trick, before declarer gains the lead and have a chance to draw trumps.

Say, we shift to a on the second trick. Partner gives 6 (normal count), and declarer a smaller than 6. Now, we can tell that pard has either 5 or 3 s, thus declarer has either 2 or 4. This helps eliminating 6-4-1-2 distribution in declarer; 6-3-2-2 and 6-1-4-2 still remain.

If declarer produces a greater than 6 (perhaps as a false card), now pard may also have Q652, that leaves J97 to declarer; in this case, declarer has 6-2-3-2, and our play will not matter.

In summary, even finding out that pard has an odd number of s will not solve most of the problem.

As Kit Woolsey implied, shift on the second trick may cause a big loss; say, declarer has KQT9x xx Qxxx xx. I am not sure if it is worth risking that.
March 28, 2016
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