| Dlr: West |
♠ A 10 |
|
| Vul: Both |
♥ A J 10 3 |
|
|
♦ K J 8 7 5 |
|
|
♣ A 2 |
|
|
|
|
|
|
|
|
♠ 8 7 4 |
|
|
♥ 9 6 2 |
|
|
♦ A 4 |
|
|
♣ K J 10 7 4 |
|
| West |
North |
East |
South |
| Pass |
1♦ |
Pass |
1NT |
| Pass |
3NT |
All Pass |
|
Both North players in a team game made the same aggressive drive to what turned out to be quite a dodgy game. The lead was the same at both tables, a fourth-highest ♠3. Both declarers played dummy’s ♠10 at trick one. Both Easts took the trick with ♠K and returned the ♠J, suggesting that the spade suit was originally 5-3. How should declarer proceed at trick three?
Solution
At the first table, declarer decided that his best chance for nine tricks lay in clubs rather than diamonds. At trick three, he cashed the ♣A and led a club to the jack. West took this with the queen and then cashed three spade tricks to defeat the contract.
The declarer at the other table was a more experienced operator. He knew that in this type of situation it was normal to cash the ace and king of one of the minors and, if the relevant queen did not appear, use the fallback position of working on the other minor for five tricks. As a doubleton queen would yield only one extra trick in diamonds compared to three in clubs, declarer cashed the ♣A and ♣K.
When the ♣Q fell, he had nine tricks — the major-suit aces, the two top diamonds and five clubs. What are the odds of the plays chosen by the declarers? On the assumption that the spades are indeed 5-3, the first line has a little less than a one-in-four chance of success. The second declarer’s combination play raised his chance of making nine tricks to nearly one-in-three. The full deal:
| Dlr: West |
♠ A 10 |
|
| Vul: Both |
♥ A J 10 3 |
|
|
♦ K J 8 7 5 |
|
|
♣ A 2 |
|
| ♠ Q 6 5 3 2 |
|
♠ K J 9 |
| ♥ 7 5 4 |
|
♥ K Q 8 |
| ♦ 10 9 2 |
|
♦ Q 6 3 |
| ♣ Q 8 |
|
♣ 9 6 5 3 |
|
♠ 8 7 4 |
|
|
♥ 9 6 2 |
|
|
♦ A 4 |
|
|
♣ K J 10 7 4 |
|
