A Cute Bridge Problem by Paul Lukacs

Sarasota Herald-Tribune – Aug 5, 1973

In a recent issue of The Bridge World, Paul Lukacs of Tel AVIV presents this cute bridge problem:

3 2  A 10 5 3  7 5 4 2  8 6 3

A K Q 6 5 4    A K Q  A K Q 2

At rubber bridge. South is the declarer at six spades. West, on opening lead, plays the spade jack. You win and immediately take another round of spades on which East shows out.

What is your best play for your contract from here on to?

We won’t keep you in suspense on this one, but don’t read on if you want to cogitate a bit on your own.

Here’s Mr. Lukacs answer: “Don’t look askance at the second trump play. Starting clubs would be wrong if a defender short in clubs had three trumps — which is more likely than that such a defender will hold a singleton trump.

“Once trumps are known to be 4-1, South’s problem is to cash minor suit winners stripping West, then throw him in with a trump. Declarer can assume that West has at most three diamonds and at most two clubs. If West has four or more of either minor, the through in won’t work.

If West has three clubs, the hand will always be made anyway. Therefore, the five winners declarer will try to cash should be three diamonds and two clubs.

“Does the order matter? Yes, It does.

The LAST card of the series should be a diamond, because it is more likely that West started with 4-5-2-2 than 4-5-3-1. Once this decision has been made, it is a toss-up which suit should be chosen for the next-to-last card. West holdings of 4-6-2-1 and 4-6-1-2 are equally likely.

Continuing in this way, we see that Declarer, after drawing the third round of trumps, should play from either minor, then from the opposite minor, and finally a diamond. If West has not ruffed. Declarer can play a club and then a trump, and hope.”

The four hands:

	3 2  A 10 5 3  7 5 4 2  8 6 3
J 10 9 7  K 7 6 4  10 8 6  7 5		8  Q J 9 8 2  J 9 3  J 10 9 4
	A K Q 6 5 4    A K Q  A K Q 2

Esta entrada también está disponible en: Spanish