Bridge Problem 7
South/NS
|
 A Q
 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 K
A Q 10 9 4
A K
| |
| South | West | North | East |
| 1 | Pass | 1 | Pass |
| 3 NT | Pass | Pass | Pass |
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.
Problem:
How can the contract be made?
For solution go to the bottom of the page![[IMAGE]](beneden.gif)
Solution:
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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