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Slightly harder puzzle


Lord Molyb

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South bid and made 6 in a 3-3 fit after a bidding misunderstanding. Afterwards he remarked that there were was at least one other slam that was unmakeable.

Construct a deal that fits the above, in addition to:

-Minimizing the N/S HCP

-Maximizing the number of unmakeable slams

 

For example:

[hv=pc=n&s=s65hakqtdaq43ckqj&w=skjh9876d987ca987&n=saq7432h2djt2ct32&e=st98hj543dk65c654]399|300[/hv]

 

N/S have 28 HCP and can make 6 (even on the A lead followed by another). In fact, every slam makes, except 6. But, surely, you can do better!

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Here's another one.

 

 

[hv=pc=n&s=s432hdaq65432c432&w=skj9h876dt987c765&n=saqthqjt5432dcakq&e=s765hak9dkjcjt98]399|300[/hv]

 

6 makes. No matter the lead, you have entries to ruff two hearts (QJ, east covers) and finesse one spade cheaply. Then you draw AKQ and run hearts until east ruffs, but he has to lead a diamond or spade, giving you another entry into the north hand to run the rest of the hearts. Note that you have to have a double stopper in diamonds.

No other slam makes!

 

 

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