ACES ON BRIDGE

“What would life be without arithmetic, but a scene of horrors?”

-- Sydney Smith

Today’s deal saw a battle between declarer and the defenders in a team game. One table was playing strong no-trump, and the other was playing weak, but three no-trump was reached pretty quickly in each case.

At both tables, West led the spade three to East’s ace, and East returned the spade 10. Both declarers had no choice but to take the trick, and could see that they had nine tricks if clubs broke 3-3, so they correctly went after that suit first by leading the queen, then playing to the ace.

When West accurately discarded a heart on the third club, one declarer simply gave up the ghost and tried for a miracle in diamonds. Down he went, but the second declarer found the best shot of exiting with a spade at trick six.

Let us first suppose that West happily cashes his spades, as happened at the table. On the last one, East had to keep diamonds and clubs guarded, and now the heart ace and king squeezed East in the minors.

It looks as if it might have be a mistake for West to cash his spades, but suppose instead he had gotten off lead with a heart at once. Declarer simply ducks a heart to East (as West has already had to discard a heart), and now the 13th heart is the game-going trick.

So can three no-trump be defeated? Yes: Say that instead of cashing his last spade and setting up the timing for the squeeze, West exits with the heart nine. Now declarer can’t exert any effective pressure.

ANSWER:

The opponents have appeared to come to a stop on an auction where your RHO is surely weak with short clubs, and thus the opponents have found a fit. How dangerous is it to risk a takeout double? I couldn’t say, but I’m going to find out!