Join Bridge Winners
How to play 6 and win or lose the Bermuda Bowl

With one segment to play Poland were leading the Netherlands by 31 imp's. First ten boards provided little scope for a major swing. Still, when the eleventh board hit the table Polish advantage was halved, although de Wijs - Muller couldn't know that their teammates have scored a thin non-vulnerable game. With six to play, de Wijs was dealt AJT98 AT732 3 76 when his partner opened the bidding with a strong club. The bidding has continued as shown on the diagram below. 

North
AJ1098
A10732
3
76
South
K6
Q64
AKJ107
A52
W
N
E
S
1
X
1
3
P
P
3
P
4
P
4
P
4
P
4
P
4NT
P
5
P
6
?

The slam here is so poor that the Polish pair at the other table, bidding mostly naturally, won't even think about it. The swing of 11 imp's was already there; the declarer could concentrate on finding the best line for his contract. On the club lead, to be expected after a club-showing double of 1, the club loser was discarded on the diamond tops. Unexpectedly, the diamond queen has appeared on the left. This has simplified declarer's task enormously. Diamond jack and ten were good for two tricks, the twelfth would come from the spade finesse through LHO short in diamonds.

There are two sensible plans hopefully yielding four heart tricks under the assumption of the outstanding hearts divided 3-2.

Plan A: play a heart toward the ace, and continue with a small heart toward the queen.

Plan B: play a heart toward the ten, and continue with the ace from the top.

At the outset, the two plans are equally good: A fails against Kx and KJx on the right, B fails against against xx and Kxx. However, once the 5 has appeared at the table, the likelihood of the success for plan B depends on the frequency with which the RHO drops the lowest from two small with or without an honour. Thus, plan B fails in 4 cases against a defender discarding up the line, but is waterproof against a defender inserting the highest first. With the RHO inclined to drop the highest from two small and the lowest from honour third, plan A will fail against K5 and KJ5, plan B only against K95 and K85. In the given circumstances, where the RHO is known to hold 9 cards in the minors, the LHO 6 in the minors plus the presumptive Q, the latter combination is less probable than the former as Kx on the right is twice as likely as Kxx. The whole deal looked as follows.

West
7
K85
98654
KQJ8
North
AJ1098
A10732
3
76
East
Q5432
J9
Q2
10943
South
K6
Q64
AKJ107
A52
D

Bringing the slam home would swing the outcome of the 2019 BB. Still, it makes no sense to criticize de Wijs' decision to play the 10 without asking about his assumptions about Narkiewicz's carding habits. Should he know? 

46 Comments
Getting Comments... loading...
.

Bottom Home Top