Bridge Problem 9


8 4
10 4 3
A 10 6 2
Q 7 6 3
K Q J 9 5
J 9 6 2
8 7 5
7 6 2
K 8 7 5
K 3
J 10 8 2
A 10 3
Q J 9 4
A K 5 4
The contract is 3 NT. K was led.

At the table South held up A twice. Next South played Q, taken by East with the king. East continued with a small heart. South decided against the heart finesse, since the contract will certainly go down if West should win the trick. A good decision! Unfortunately the clubs broke 4-1, therefore declarer had to be contented with 8 tricks: 3 in either of the low suits + the aces of hearts and spades.
How can the contract be made all the same?
For solution go to the bottom of the page[IMAGE]


Holding up A was correct, but before taking the diamond finesse declarer should have tested the clubs. After A declarer should cash Q + K. The 4-1 club break means that only 8 tricks are available if the diamond finesse fails. If Q is taken with the king and a small heart is played, a successful heart finesse is the only way to obtain 9 tricks. Consequently South plays Q after East's small heart and this gives South nine tricks.
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