Main Column 7C83 by Richard Pavlicek
Todays deal occurred in the first round of the Vanderbilt Knockout Teams, held last month in Buffalo, N.Y. As West I was a bystander in the bidding as our opponents bid to six diamonds an excellent contract, but placed in jeopardy when I chose to lead a spade: queen, king, ace.
|6 South|| Q 3|
A K J
K Q 9 6 5 4
| 9 6 4|
9 8 6 5 4 3 2
| K J 8 5 2|
K 8 3
10 7 3
|Lead: 4|| A 10 7|
A J 10 9 7 6 2
As readily seen, declarer has 17 top tricks seven diamonds (with the finesse), six clubs, three hearts (with the queen falling) and one spade which is four more than he could ever need. Then whats the problem? Even a beginner could win 12 of those tricks!
Not so fast. Declarer is not privy to all four hands; he must choose the line of play which offers the best chance of success based on the information he has. Several reasonable plays exist: (1) Cross to dummy with a heart and take the diamond finesse. (2) Cash the diamond ace then, if the king does not fall, take the heart finesse. (3) Cash the diamond ace, two top hearts, then run the clubs (except lead the heart jack first when Easts queen drops).
Which is best? If the contract were seven diamonds, Line 1 is a standout. With nine diamonds, the finesse is by far the best chance of avoiding a loser in that suit. But the contract is only six diamonds. Lets calculate the chances.
Line 1 is basically 50-50 if East has the diamond king you succeed; otherwise you fail. Line 2 is about 56 percent it works if the diamond king is singleton (12.5 percent) or if West has the heart queen (half of the remaining 87.5 percent). Line 3 is a complex analysis, but intuitively it must be under 50 percent, so Ill write it off.
The bridge gods were cruel. Line 1 or 3 would work; but the proper play, Line 2, fails in fact declarer was defeated two tricks. This brings out the curious fact that a good player would make seven diamonds, but go down in six.
© 4-10-1988 Richard Pavlicek