ACES ON BRIDGE

A thought is often original, though you have uttered it a hundred times.

— Oliver Wendell Holmes Try your hand as South at this declarer play problem.

After partner’s game-forcing response of two diamonds, your jump rebid shows a solid or semisolid suit, setting spades as trumps. It should also imply a little more than a dead minimum in high-card terms. I believe you should probably not make the call without either a control or least two queens on the side.

Partner’s four diamonds is therefore (at least initially) a cue bid rather than natural. When you sign off in four spades, he comes again with five hearts, another cue bid. With a club control in hand, you must take a shot at six spades.

On the lead of the heart queen, you can count 11 tricks and could be forgiven for thinking you must rely on the club finesse for another. However, you can bolster your chances by allowing for a 3-3 heart split too.You would then be able to establish dummy’s fourth heart for a discard.

You cannot afford to lose a heart trick in the process, so you must throw a heart on a diamond and then ruff a heart. If they do not break evenly, you can fall back on a club to the king, thus combining your chances.

Win the heart lead and draw two rounds of trumps with the ace and king. Then cash the diamond ace-king, discarding a heart. Next come the heart king and a heart ruff. When the hearts break, you use the spade nine as the entry to throw your club loser on the established heart. Had the hearts not broken, you would have fallen back on the club finesse, of course.

ANSWER: Lead the club 10. It sounds like the opponents are missing a club control, and with spades controlled to your right, there are a limited number of tricks that a spade lead might develop. By contrast, a club lead might set the game at the very outset. I think I would lead a club even if I had the spade king instead of the jack.