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A Christmas Cracker

#1 User is offline   lamford 

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Posted 2024-December-12, 16:10


You did well to correct to 7NT, RR, gloated HH,
South, as you would not have made 7H.
Indeed any lead would have beaten you,
whereas 7NT was cold on any lead played by me.
The jack of spades was led. How do you bring this home?
I prefer to give the lawmakers credit for stating things for a reason - barmar
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#2 User is offline   Lovera 

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Posted 2024-December-21, 13:14

View Postlamford, on 2024-December-12, 16:10, said:


You did well to correct to 7NT, RR, gloated HH,
South, as you would not have made 7H.
Indeed any lead would have beaten you,
whereas 7NT was cold on any lead played by me.
The jack of spades was led. How do you bring this home?


Hi all (and many wishes for Xmas). It's strange that anyone has replied to you as usually happened. Probably is involved a type of squeeze (i'm watching and for menaces) eventually progressive too.
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#3 User is offline   apollo1201 

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Posted 2024-December-26, 03:18

Xmas is over! May we get the (funny) reveal as a gift? Thanks 🎄🎄🎄🎄🎄🎄🎄🎄🎄🎄🎄
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#4 User is offline   Lovera 

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Posted 2024-December-26, 09:50

I have the idea that we have to see more in-deep and to do any sacrifice.
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#5 User is offline   Lovera 

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Posted 2024-December-29, 10:50

I think I found the solution. Looking at the hand we have 10 tricks in high cards and to think of a repeated or progressive squeeze it is necessary to have another winner which can be the 10 if QJ are "dropped". If things are like this then W can have 3S-2H-4D-4C necessarily having to have control of in S.Therefore the Q should be a winner. We also assign to W K and J in . Having then seven winners between and , collecting the high cards in the minors and returning to the hand at when we play the last W will be squeezed in those suits and will have to give up the 13th trick.
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#6 User is offline   smerriman 

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Posted 2024-December-30, 19:02

View PostLovera, on 2024-December-29, 10:50, said:

I think I found the solution. Looking at the hand we have 10 tricks in high cards and to think of a repeated or progressive squeeze it is necessary to have another winner which can be the 10 if QJ are "dropped". If things are like this then W can have 3S-2H-4D-4C necessarily having to have control of in S.Therefore the Q should be a winner. We also assign to W K and J in . Having then seven winners between and , collecting the high cards in the minors and returning to the hand at when we play the last W will be squeezed in those suits and will have to give up the 13th trick.

I can't follow this at all. If West has 3244 how does the Q become a winner on any other lead?
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#7 User is offline   smerriman 

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Posted 2024-December-30, 19:57

OK, I got it.

Spoiler

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