Cyberyeti Posted February 17, 2018 Report Share Posted February 17, 2018 [hv=pc=n&s=sjt43h86dkt52c765&w=sq62hkq52daj6cqj9&n=s95haj974dq87cak3&e=sak87ht3d943ct842]399|300[/hv] The contract is bizarrely 2♥x by S after a MI ruling against NS, how many tricks do you think it should be ruled to make (EW actually played 2♠ at the table). Quote Link to comment Share on other sites More sharing options...
blackshoe Posted February 17, 2018 Report Share Posted February 17, 2018 How did the bidding go? Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted February 17, 2018 Author Report Share Posted February 17, 2018 How did the bidding go?N opened 1N-P-P-X-P-P-2♥(not alerted, intended as transfer, agreed as their system)-P-P-2♠. It was ruled that if 2♥ had correctly been alerted as a transfer, the auction would have gone X-P-P-P. I didn't want to get into whether S was allowed to pass 2♥x (I don't think he is, for me and I suspect for the pair at the table the failure to complete the transfer only shows 2 spades, so xx, AKx, AQxxx, xxx is plausible and could conceivably take only 3 tricks in hearts while making 8 in diamonds) and concentrate on the number of tricks which is why I didn't include it. Quote Link to comment Share on other sites More sharing options...
ahydra Posted February 17, 2018 Report Share Posted February 17, 2018 It's too late at night for a full analysis but I would guess -2, making 3H 1D 2C. West has a natural club lead and declarer will want to play trumps from hand, so will have to find a way there, but the defense can just persist with clubs (cashing SAK when required). ahydra Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted February 17, 2018 Author Report Share Posted February 17, 2018 It's too late at night for a full analysis but I would guess -2, making 3H 1D 2C. West has a natural club lead and declarer will want to play trumps from hand, so will have to find a way there, but the defense can just persist with clubs (cashing SAK when required). ahydra Declarer can always get out for -1 seeing all the cards, as the defence will be endplayed eventually, but the sort of question is which heart does declarer play if declarer plays ♣AKx and W returns a small one ? or if it comes down to just diamonds left, does declarer play small to the 10 which wins if the J is one side or small to the K which wins if it's the other (given that the A is known to be with the doubler). Quote Link to comment Share on other sites More sharing options...
blackshoe Posted February 17, 2018 Report Share Posted February 17, 2018 If not passing 2♥X is a possibility in their methods, what might happen after South doesn't pass (3♦, 3♦X?) should be included in a weighted score. I think a poll is needed here. Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted February 17, 2018 Author Report Share Posted February 17, 2018 If not passing 2♥X is a possibility in their methods, what might happen after South doesn't pass (3♦, 3♦X?) should be included in a weighted score. I think a poll is needed here. Which is exactly why I didn't want to include the auction, I was purely interested in the assessment of how many tricks 2♥X should be awarded. Quote Link to comment Share on other sites More sharing options...
BudH Posted February 17, 2018 Report Share Posted February 17, 2018 [hv=pc=n&s=sjt43h86dkt52c765&w=sq62hkq52daj6cqj9&n=s95haj974dq87cak3&e=sak87ht3d943ct842]399|300[/hv] The contract is bizarrely 2♥x by S after a MI ruling against NS, how many tricks do you think it should be ruled to make (EW actually played 2♠ at the table). I think six tricks is most likely, and the chance of only five tricks is low enough not to use a weighted average. 1 Quote Link to comment Share on other sites More sharing options...
blackshoe Posted February 18, 2018 Report Share Posted February 18, 2018 I think six tricks is most likely, and the chance of only five tricks is low enough not to use a weighted average.Works for me. Quote Link to comment Share on other sites More sharing options...
Cyberyeti Posted February 18, 2018 Author Report Share Posted February 18, 2018 I think six tricks is most likely, and the chance of only five tricks is low enough not to use a weighted average. This was what I felt, I objected somewhat to the director's award of 7. Quote Link to comment Share on other sites More sharing options...
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