kfay Posted September 16, 2009 Report Share Posted September 16, 2009 [hv=d=e&v=b&n=sq54haq52dkqj9c86&e=sa10hj109763d65ck93]266|200|Scoring: IMPP-(1♣)-P-(1♦)P-(1NT)-P-(3NT)All Pass[/hv] Partner leads the 8 of spades. You win with the ACE (declarer playing the 9) and... Quote Link to comment Share on other sites More sharing options...
ArtK78 Posted September 16, 2009 Report Share Posted September 16, 2009 ♣9. Partner may have AQ7x of clubs, and it will be necessary to lead the 9 to take 4 tricks in the suit. Hopefully partner can work out to return a small club upon winning declarer's 10 or J with his Q on the first round. I almost led the ♣K at trick 2, but that would make it impossible to take 4 club tricks if partner has AQ7x. If partner holds AQJx or AQTx of clubs, my shift to the 9 may make it hard for him to continue the suit. But I think it is unlikely that declarer's clubs are that weak. From the bidding, the lead and the play to trick one, declarer could hold this hand: KJ9KxxAxxJTxx Quote Link to comment Share on other sites More sharing options...
quiddity Posted September 16, 2009 Report Share Posted September 16, 2009 A slightly harder problem: what if declarer plays a low spade instead of the 9 to the first trick? Quote Link to comment Share on other sites More sharing options...
pooltuna Posted September 16, 2009 Report Share Posted September 16, 2009 [hv=d=e&v=b&n=sq54haq52dkqj9c86&e=sa10hj109763d65ck93]266|200|Scoring: IMPP-(1♣)-P-(1♦)P-(1NT)-P-(3NT)All Pass[/hv] Partner leads the 8 of spades. You win with the ACE (declarer playing the 9) and... fire back the ♣9 hopefully partner will have enough brains to return the suit. Quote Link to comment Share on other sites More sharing options...
ArtK78 Posted September 16, 2009 Report Share Posted September 16, 2009 A slightly harder problem: what if declarer plays a low spade instead of the 9 to the first trick? If declarer plays a low spade at trick 1 then partner's 8 could be fourth-best from KJ98(x). Having won the ♠A at trick one, it is almost impossible for there to be a successful defense at this point, as declarer must have the ♣A and either the ♣Q or ♣J. But the little chance that remains is that declarer's clubs are no better than AJ5x. You need to switch to the ♣9 in the hope that partner has QT7x of clubs. Declarer's ninth trick is the ♠Q in dummy - you need to establish 3 club tricks before declarer gets to play a spade towards dummy's ♠Q. Quote Link to comment Share on other sites More sharing options...
nigel_k Posted September 16, 2009 Report Share Posted September 16, 2009 Would partner lead his highest or second-highest from small cards? Quote Link to comment Share on other sites More sharing options...
kfay Posted September 16, 2009 Author Report Share Posted September 16, 2009 Would partner lead his highest or second-highest from small cards? As high as he could afford, which is often 2nd highest. Quote Link to comment Share on other sites More sharing options...
phil_20686 Posted September 17, 2009 Report Share Posted September 17, 2009 My analysis was flawed. 9 clubs must be the only chance. It would be worth knowing if south would open 1d with 4-4 int he minors, or if partner would lead from an interior sequence. Im inclined to beleive that south is 3235, in which case there is no way to beat this i think. Quote Link to comment Share on other sites More sharing options...
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