Puzzle 8S13 by Richard Pavlicek
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.
After winning the A in dummy, declarer leads a diamond for what appears to be an early claimer. Oops! Easts 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, A-K, A-Q, K and K; then exit with a low diamond, endplaying West. This would also work if West had three spades and two clubs, as the K would force a spade pitch lest diamonds be established.
Given the 2 in North, Souths 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 A-K-Q-10-9-3 or A-K-Q-J-8-3, South fails. Of course K-Q-J-10-9-8 would also succeed, but thats stronger (sum 63). Note that 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 Souths diamond holding (any six cards excluding the 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, Souths 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. Ill drink to that!
If desired, you may submit your solutions using the form below. This may be done only once, and doing so will add your name to the list of successful solvers, ranked according to the tiebreakers. You will also receive an automatic reply with a copy of your solutions and what Richard believes are the optimal solutions.
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