THE ACES ON BRIDGE

ANSWER: It is very tempting to bid two no-trump, going for the big payout if you are right; but here you have a viable alternative, which is to bid two spades. The point is that with such a misfit and no real club stopper, your chances of making three no-trump are very slim. Also, if partner has extras, he can still bid on if he wants. “I reason, earth is short, And anguish absolute. And many hurt; But what of that?” -- Emily Dickinson .....................

In today’s deal it would have been easy enough to defeat three no-trump after a top heart lead. (Declarer wins the opening lead and runs five rounds of diamonds, but West simply keeps his club 10 till the last diamond (to prevent declarer from finessing in the suit), then discards his spade on the top clubs. He keeps four winning hearts and the spade ace. Mission accomplished.

However, you have done well to reach five diamonds instead, on the lead of a top heart. You could rely on the spade ace being onside, but before you do that, try something else. The key to the deal is to try to establish clubs to allow you to discard your spade losers without letting East on lead. To that end, you win dummy’s heart ace perforce and take the diamond queen, all following. Next, you lead a club, and when East plays low, you put in the eight.

West wins the club 10 and does best to punch dummy with a heart. You ruff, draw the second round of trumps, cash the club ace and king, and ruff a second heart. Now you ruff a club to set up the suit and ruff another heart with dummy’s last trump. At this point, the fifth club has been established and takes care of one of your two spade losers.

The avoidance play in clubs requires West to hold one of the three remaining top clubs. You also need either diamonds or clubs to break evenly.