A Christmas Puzzle

[lamford]

Sherlock Holmes arrived at the Baker Street Bridge Club after they resumed face-to-face bridge recently. “What do you make of this travelling scoresheet, Watson?”, he asked.

 

“Very strange,” replied the learned doctor. “EW made 7S, 7H or 7NT at every table except one where NS made 6NT. It must have been mis-boarded on the last one, I assume.” “What were the NS hands?”

 

“No it was not mis-boarded, Watson,” “But did you not notice that when South made 6NT, East-West were a retired colonel playing with his disabled wife?"

 

“Good gracious, how did you deduce that?” came the response.

 

[hv=pc=n&s=st432ht432dtck987&n=s5h5dj9832cj65432]133|200[/hv]

“Well, it was clearly a 5-table Howell movement, with 18 boards, and pair 10 was stationary, allocated to Colonel Cathcart and his wife Lady Cathcart who would have been unable to move. Also the Colonel always scores, even when he is East or West, using a distinctive purple ink, identical to his invitation to me to speak at the Rotary Club Christmas Dinner. Finally, Lady Cathcart, who has arthritis, uses one of those much criticised cardholders which have a habit of collapsing. Clearly her hand fell forward during the auction and she then had 13 penalty cards. When she became on lead, the declarer could direct her play at each trick. And I heard the TD, Professor Pedant, discussing the ruling as we arrived.”

 

“But surely that made no material difference?” pressed Watson rather dimly. “South could not have made 6NT”

 

“Not so, South could claim at trick one on the actual layout.” Holmes replied “Which is for you to work out, Watson.” He concluded. “And note that while all West’s cards are penalty cards, they are authorised to all players under Law 50E1 of the new 1917 Laws of Duplicate Bridge, so East is not constrained at all, unless he gains the lead. And I find that West can only make 7NT on a club lead, which I think makes the EW hands unique.

 

What were the exact EW cards which allowed South to make 6NT against best defence?

 

[With thanks to Nigel Guthrie, Cyberyeti, Julian Pottage and Paul Barden for cooks and corrections to my earlier feeble efforts. I hope the above is now unique! I thought I had found a way to eliminate the requirement that you could only make 7NT on a club lead, but gszes showed that was not the case. I also noted that in the main line, you have to cash the sixth club before the jack of diamonds, or you squeeze yourself in a non-material way. You are forced to decide on Weat's discard prematurely.

[mikeh]

Solved. Edit: not actually solved. I missed a point

[nige1]

Well done Paul :) Merry Christmas :)

[lamford]

Well done Paul :) Merry Christmas :)

Merry Xmas to you. Hope your health is bearing up and that you have had the vaccine.

[nige1]

Merry Xmas to you. Hope your health is bearing up and that you have had the vaccine.

:) Thank you :) Same to you :) and everybody :)

[Cyberyeti]

I couldn't solve it unless all 14 of her cards fell forward, then I might have a solution :)

 

Also what are the rules if you notice at that point partner has put 14 on the table and you only have 12 ? or declarer does ?

[lamford]

I couldn't solve it unless all 14 of her cards fell forward, then I might have a solution :)

 

Also what are the rules if you notice at that point partner has put 14 on the table and you only have 12 ? or declarer does ?

No. She had only 13 cards. Mikeh solved it, and I think Nigel Guthrie did too, although the latter had cooked earlier efforts, so he would know the theme. I think he called it an octagonal penalty card squeeze.

[lamford]

Solved. Edit: not actually solved. I missed a point

Ah ... ok. I had a small error, a typo, which might be the point we both missed. Corrected now.

[Cyberyeti]

Ah ... ok. I had a small error, a typo, which might be the point we both missed. Corrected now.

 

OK, I noted you needed the ♦7 to solve it with 13 cards, it's a lot easier now.

 

Solved.

 

I let the questioon stand, what happens if 14 penalty cards are exposed during the auction, and does it matter who notices ?

[lamford]

OK, I noted you needed the ♦7 to solve it with 13 cards, it's a lot easier now.

 

Solved.

 

I let the questioon stand, what happens if 14 penalty cards are exposed during the auction, and does it matter who notices ?

Yes there are Laws about an excess card. Law 13 is the main one. One of the cards had to be the excess one and it would have been seen. The other 13 would remain as penalty cards, I think, and the 14th given to the player whose card it was, or put in the correct board if it was not part of this board.

[Cyberyeti]

Yes there are Laws about an excess card. Law 13 is the main one. One of the cards had to be the excess one and it would have been seen. The other 13 would remain as penalty cards, I think, and the 14th given to the player whose card it was, or put in the correct board if it was not part of this board.

 

but does it become a penalty card for them if it's the other defender ? this was the key point of my 14 card solution without ♦7, which required E to be forced to discard a diamond on the clubs so the suit cashed.

[nige1]

[hv=pc=n&w=SAKQJHAKQJDAKQCAQ&n=S5H5DJT943CJ98765&s=ST432HT432D2CK432&d=S&a=6NPPP]400|300| :) This seems to work...:) [/hv][hv=pc=n&w=SAKQJHAKQJ&n=S5H5DJT943CJ&s=ST432HT432&e=S98H98D8765&a=-NN&p=CJ]400|300|:) ♣J executes a pentagon squeeze :)

- triple-squeezes East - and

- misère squeezes West. If East has abandoned a major, then West is forced to discard the same major.[/hv][hv=pc=n&w=SAKHAK&n=S5H5D43&s=ST4HT4&e=S98H98&a=-NN&p=D4]400|300| :) If West abandons ♦, then North's penultimate ♦ executes a double squeeze :)

- East must abandon a major - and

- West is forced to discard the major that East gave up.

+++++++++++++++++++++++++++++++

Edited to make West hand unique[/hv]

Edited December 25, 2020 by nige1

[smerriman]

Oh, I had been thinking about this for quite a while and hadn't figured it out. The edited version with the 7 of diamonds makes things considerably simpler. I do wonder what the intended solution was without it, though. Oh right, typo, so it wasn't intended.

[Cyberyeti]

The solution I had to the rewritten original problem required no squeeze.

[lamford]

The solution I had to the rewritten original problem required no squeeze.

What was that? Did it meet the requirement that 7NT only made on a club lead?

[lamford]

[hv=pc=n&w=SAKQJHAKQJDAKQCAQ&n=S5H5DJT943CT98765&s=ST432HT432D2CK432&d=S&a=6NPPP]300|300|This seems to work...:) [/hv][hv=pc=n&w=SAKQJHAKQJ&n=S5H5DJT943CT&s=ST432HT432&e=S98H98D8765d=S&a=-NN]300|300|:) ♣T triple=squeezes East [/hv]

The queen and jack of clubs can be interchanged ... You can give East a singleton ten of clubs though, and I think then the triple squeeze is the only line.

[Cyberyeti]

What was that? Did it meet the requirement that 7NT only made on a club lead?

 

Yup, W has A, AKQJxxx, AKQ, AQ leads ♣AQ, you run the clubs pitching 3 diamonds and ♥A, then 5 diamonds pitching hearts from the top, then score 2 heart tricks.

[lamford]

Yup, W has A, AKQJxxx, AKQ, AQ leads ♣AQ, you run the clubs pitching 3 diamonds and ♥A, then 5 diamonds pitching hearts from the top, then score 2 heart tricks.

Yes you are right, as you would be with the majors transposed.

 

[hv=pc=n&s=st432ht432dtck987&w=sakqjhakqjdakqcaq&n=s5h5dj9832cj65432&e=s9876h9876d7654ct]399|300[/hv]

 

That seems to work. Essentially that is Nigel Guthrie's construction. But with a unique EW hand as shown. Can you cook that, Cyberyeti?

 

I don't think you can make now against your hand(s).

[Cyberyeti]

Yes you are right, as you would be with the majors transposed.

 

[hv=pc=n&s=st432ht432dtcK987&n=s5h5dj9832cJ65432]133|200[/hv]

 

That seems to work, and I now think that West has to be ♠AKQJ ♥AKQJ ♦AKQ ♣AQ. Essentially that is Nigel Guthrie's construction. But with a unique EW hand. Can you cook that, Cyberyeti?

 

I don't think you can make now against your hand(s).

 

Nope, which is why I gave the potential 14 card construction.

[lamford]

Nope, which is why I gave the potential 14 card construction.

Great. So have amended the original hand. Hope it is now sound!

[gszes]

Late to the party as usual ASSUMING I have not totally misread the situation

 

 

[hv=pc=n&s=st432ht432dtcq987&w=s987hakqjdakq7caj&n=s5h5dj9832ct65432&e=sakqj6h9876d654ck]399|300[/hv]

Trick 1 club A

trick 2 club J*

trick 3/4/5/6 clubs forcing west to pitch the AKQ7 of diamonds

trick 7/8/9/10 diamonds forcing west to pitch AKQJ of hearts this makes west hand irrelevant and we arrive at the following[hv=pc=n&s=stht4dc&w=shdc&n=s5h5d2c&e=sah98dc]399|300[/hv]

trick 11 east is done on the lead of the dia 2 south pitching the suit opposite whatever east discards

 

 

 

My solution is totally boring compared to Nige1

 

If you block the club suit you deserve to go down

 

HAPPY HOLIDAYS and STAY WELL

[smerriman]

Late to the party as usual ASSUMING I have not totally misread the situation

This doesn't satisfy the constraint of E/W only being able to make 7NT on a club lead (they make it on all leads).

[lamford]

This doesn't satisfy the constraint of E/W only being able to make 7NT on a club lead (they make it on all leads).

To be fair, I had removed the condition in the rubric, but have now reinstated it and restored the previous layout. Also, the penultimate version was cooked similarly by West having AKQJxx A AKQx AJ. Now 6NT is cold, but 7NT is beaten by a club lead, taking out the entry to dummy prematurely!

[nige1]

[hv=pc=n&s=S5432HT432d432cKJ&w=SAKQHAKQJDAKJ9CAQ&n=S6H5H5dQTCT98765432&d=S&a=6NPPP]400|300|

Another construction based on Gszes solution, which satisfies the conditions that EW

- Can make a grand-slam in 3 suits.

- Need a ♣ lead to make 7NT.[/hv]

[nige1]

Compared with my earlier suggested formulation of the problem, the OP now has an extra diamond trick, which might make solutions easier.