Main   Puzzle 8S13 by Richard Pavlicek  

Diamond Stack

With my new-found wealth in space travel, I have reopened the Jwaneng mine. PavCo Diamonds is back in production! Our catalog features 50 new creations, like the breakfast cereal bowl pictured above, and a diamond studded hat rack with an ivory base. Low overhead, courtesy of PavCo Cayman Bank, means unbeatable prices! PavCo Diamonds should put De Beers out of business in a month (five weeks tops) as its customers flock to our offshore warehouse. The diamond craze also inspired a bridge curiosity, which I will present as a puzzle.

Click for a list of successful solvers

How many times have you found a suit stacked behind you, either as trumps or a side suit that cannot be established? Witness the egregious diamond stack on the following deal, which not only renders an excellent grand impossible but jeopardizes six as well.

6 NT South S A 5 4 3
H A K 6 5 4
D 2
C A Q 4
Both Vul   West

Pass
Pass
All Pass
North
1 H
2 S
4 NT
East
Pass
Pass
Pass
South
2 D
3 NT
6 NT
S J 10
H 9 8
D 10 8 7 6 5 4
C J 9 5
Table S Q 9 8 7 6
H Q J 10 7
D
C 10 8 7 6
Lead: S J S K 2
H 3 2
D A K Q J 9 3
C K 3 2

After winning the S A in dummy, declarer leads a diamond for what appears to be an early claimer. Oops! East’s discard changes that picture. Diamonds cannot be established, but declarer can still succeed by elimination technique. Many sequences will suffice, but best is to cash three top diamonds, H A-K, C A-Q, C K and S K; then exit with a low diamond, endplaying West. This would also work if West had three spades and two clubs, as the C K would force a spade pitch lest diamonds be established.

Given the D 2 in North, South’s D A-K-Q-J-9-3 (rank sum 62) is the weakest possible holding to succeed against the 6-0 break offside. If any card is reduced, say D A-K-Q-10-9-3 or D A-K-Q-J-8-3, South fails. Of course D K-Q-J-10-9-8 would also succeed, but that’s stronger (sum 63). Note that D K-Q-J-10-9-7 (sum 62) would fail, even if not led until six cards remain, as West holds up his ace four times.

In the above scenario declarer needed five diamond tricks. More often he will need fewer, which introduces this three-part puzzle:

With a 6-6 suit division, what are the minimal holdings to win (A) four, (B) three and (C) two tricks?

Enter South’s diamond holding (any six cards excluding the D 2) to win the number of tricks required (West will get what remains). Assume diamonds will not be played until six cards remain, and South must lead first. To succeed, South’s holding in each part must be within one pip of minimal. Successful solvers will be ranked by the sum of all South cards (lowest is better); and to celebrate the fall of De Beers, ties will be broken by the most South beer cards. I’ll drink to that!


A. 
 
Win 4 tricks    South  D
B.
 
Win 3 tricksSouth  D
C.Win 2 tricksSouth  D



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