jfy18

When playing duplicate bridge while vulnerable, there are bonuses to successfully bidding game (ie. 500 instead of 300) but also greater costs to being set. Bearing this in mind, are there any reasons why one should be (either) more or less aggressive when bidding vulnerable, given that there are both extra costs and benefits for so doing? I can see that if one is vulnerable and the opponents are not, one should expect opponents to be more likely to attempt to make a 'save' against one (because going down will cost them comparatively little, and they will be able to stop one getting a high game bonus). But nonetheless I suppose a similar (perhaps equivalent) question to the above is: should bidding be any different when both pairs are vulnerable, compared to bidding when both pairs are not? JFY

Is it just I, or do others feel that the rather pejorative, aggressive tone of some of the replies is unproductive? Sceptic asked for views on this hand, rather than on (hypothetical) other hands.

Regarding the best defensive lead after winning the Q ♦ on, say, Trick 2: Let's think about the points declarer and dummy together have: between 25-27 one should think. They are very likely to make their 3NT. So the question probably is: how can the defence find their fourth trick? Since South has 8 points, this also therefore gives North 1-3 points in addition to his A♠. I agree that playing a ♠ now jeopardises a plausible source of a trick later on, if North has A6 ♠. I am not so sure why playing a ♥ necessarily sets up a trick in dummy. Consider: i) East may have doubleton ♥; ii) Even if East has 3 ♥, if two of them are K and Q then they will make 4 [H] tricks whatever we lead now, so in that case we would lose nothing by playing one. My conclusion, however, is that the ♦ suite is now probably a lost cause. North does not have enough room, points-wise, for the A and is more likely to have an honour in another suit (since the MAXIMUM other honour North can have is a K). Therefore, leading a ♦ back is unlikely to give anything away, since the only time it does give something away is when North has the K (which is unlikely). Does this seem reasonable?

Re. Flame's helpful comments: In that case, the situation in the example given becomes clearer in one sense (ie. that the hands East can have are limited to 5332, 4432 or 4333) but less clear in another sense (ie. because the opening trick has no count information from North). So what do we know: i) N and E between them have 4♠ 6♥ 8♦ 8 ♣. ii) On the second trick, North's low card shows he has an even number of ♦ according to Fluffy who set the problem. iii) Therefore, East must also have an even number of ♦. iv) East has no more than 3 cards in either major suit. So, what are the combinations East can have that are either 5332, 4432 or 4333? In terms of (♠, ♥, ♦, ♣) with even ♦ I make East's possible holdings: (2,3,4,4) (3,2,4,4) (3,3,2,5) But how can we now work out which one of these it is? Perhaps you could let us know your line of thinking, Fluffy; especially if I have missed something important. Cheers, Jfy

Well, let's think about it. At the start of the hand, there are 4♠ 6♥ 6♦ and 8♣ outstanding in the 2 hands we cannot see, adding to 26 cards. After the first two tricks, since East's 1NT bid shows at least 2 of each suit, those numbers will have been reduced to 2♠ 6♥ 4♦ and 8♣. North's opening discard shows an odd number of ♠ and his second discard shows an even number of ♦. {By the way, is it just I or do others think the usual convention is the other way around ie. that playing low signifies an odd number of cards in the suit?? Your comments are most welcome here, especially, Moderator}. What else do we know? i) By the Stayman bidding, East has less than 4 cards in each of ♥ and ♠ Clues in, let's first think about ♠. East must start with an odd number (since North has an odd number and they add to 4), which must be 3 ♠ since by their 1NT bid they cannot have less than 2. Therefore, North started with 1 ♠. Continue by thinking about East. Other than 3 ♠, they have 10 cards, including an even number of ♦. So East's possible ( ♥,♦,♣ ) combinations are: (2,6,2) (2,4,4) (3,4,3) (2,2,6) (3,2,5) Once we know which, we can easily work out North's hand; so one of the above 5 options must be the full answer. I cannot work out which can be eliminated, however; I suppose we could (somewhat arbitrarily) eliminate the hands with 6 in a suit on the grounds that East would not have bid 1NT with such a hand. Even if we do this, however, there are still 3 possibilities, all of which seem plausible. Any ideas?