Bridge Problem 7


10 8 6 4 3
J 7 5
6 4 2
J 10 7 6 4
7 5
K 3 2
Q 9 8
K 2
Q J 9 2
8 6
J 10 7 5 3
9 8 5 3
A Q 10 9 4
3 NT PassPassPass
The lead is 6
At both tables during a team match the players reached 3 NT. At the one table the contract went one down, at the other the contract was made. At the first table declarer took the spades finesse and East won with the king. A spade was returned, which removed declarer's stopper. Therefore when West won the diamond trick with the king, declarer lost three more spade tricks and the contract was one down.


How can the contract be made?
For solution go to the bottom of the page[IMAGE]


Declarer realised the following:
If East has K or if the spades are divided 4-3, the contract will always be made. Therefore....declarer concentrated on a manner of play with a 5-2 spade split, with West in possession of K. However, if West would have K J 10 x x and K, he was likely to have bid 1 being not vulnerable. With any other 5-2 split, going up with the ace would be the right procedure. For if East had doubleton J 10, the 9 would become a stopper. If East held the king, the spades would be blocked, if declarer takes trick one with the ace.
So: North took the first trick with A and played J for West's king. The spade which was returned, was taken by East's king , but since West had no fast entry, North was able to collect nine tricks.
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