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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.

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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!

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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.

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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:

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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.

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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.

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.

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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).

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.

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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.

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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).

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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.

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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 %.

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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.

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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.

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Let me add this Rolf:

Playing against you (i.e., your team) in a third-place play-off would be the last thing I would want. (In such an incompatible mood, such a determined opponent!)

Okan Zabunoglu

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.

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

“Is there anything that members can do to support the administration?”

Okan Zabunoglu

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.

Okan Zabunoglu

As for your other remarks, I will convey my thoughts, but not tonight. It is 11:00 pm here.

Okan Zabunoglu

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.

Okan Zabunoglu

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.

Okan Zabunoglu

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.

Okan Zabunoglu

Okan Zabunoglu

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.

Okan Zabunoglu

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.

Okan Zabunoglu

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 %.

Okan Zabunoglu

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.

Okan Zabunoglu

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.

Okan Zabunoglu

Okan Zabunoglu

Okan Zabunoglu

'Irony medal' did not sound good to me (that's why I did not know where to put the y). How about a ‘tiny medal’?

Okan Zabunoglu

Playing against you (i.e., your team) in a third-place play-off would be the last thing I would want. (In such an incompatible mood, such a determined opponent!)

Okan Zabunoglu

… If there has to be a third-place play-off, at least the loser may get some relief (with an iron medal) obviously.

It ends with a y.

Now, is it okay?