 A 10
 A J 10 3
 K J 8 7 5
 A 2 |
||
| Q 6 5 3 2
7 5 4
10 9 2
Q 8 |
K J 9
K Q 8
Q 6 3
9 6 5 3 |
|
| 8 7 4
9 6 2
A 4
K J 10 7 4 |
| West | North | East | South |
| Pass | 1 |
Pass | 1NT |
| Pass | 3NT | Pass | Pass |
| Pass |
3
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 three of spades. Both declarers played dummy’s ten of spades at trick one. Both Easts took the trick with the king of spades and returned the jack of spades to dummy’s ace, suggesting that the spade suit was originally 5-3.
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 ace of clubs and then 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-king of one of the minors and, if the relevant queen did not appear, using the fallback position of working on the other minor for five tricks. As a doubleton queen would only yield one extra trick in diamonds compared to three in clubs, declarer cashed the ace and king of clubs.
When the queen of clubs 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-infour chance of success. The second declarer’s combination play raised his chance of making nine tricks to nearly one-in-three.
