Semi-Real

Author: Larry Cohen
Date of publish: 01/01/2023
Level: Intermediate

This Deal comes from a 2022 ACBL "Bidding-Box" competition. Since ACBL uses the term "semi-forcing", let's call this a "semi-real-deal."

The goal was to reach six of a major on this layout:

♠ AKJ65
♥ 105
♦ J95
♣ AJ7
 
♠ Q98
♥ AKQJ97
♦ 6
♣ Q106

Suppose South reaches 6♠ and West leads the ♠K. This looks easy; declarer can throw clubs on spades and doesn't need the club finesse. 

West continues diamonds at trick two to East's ace. South ruffs and starts to draw trump, West showing out on the second round. Now what?

Suppose this is the "Real Deal:"

 

  ♠ AKJ65
♥ 105
♦ J95
♣ AJ7
 
♠ 107432
♥ 8
♦ KQ1084
♣ 94
  ♠ --
♥ 6432
♦ A732
♣ K8532
  ♠ Q98
♥ AKQJ97
♦ 6
♣ Q106
 

When West shows up with only a singleton in hearts, he is the opponent who might have spade length (it is surely against the odds for West to have been dealt 0-1 in the majors). Accordingly, declarer should finish drawing trump (keeping all of dummy's spades, of course) and then start spades with the ♠Q. If all follow, claim. If the layout shown in the diagram exists, no problem. Lead the ♠9 for a marked finesse and there are still 5 spade tricks. The trap was to start spades by leading low to dummy (the instinctive play).

This is all semi-fantasy, because if the layout shown existed, East-West have screwed up. No doubt, dummy would have bid spades along the way. East would (should) have made a Lightner double. That would ask for the lead of dummy's first-bid suit. West would lead a highish spade (suit preference) and East would ruff. Likely he'd cash the ♠A for down one, but an intrepid diamond underlead would produce a second undertrick (nobody has such nerves). 

Even after the ♠K lead, any black-suit shift at trick two would still defeat the contract (because on a club shift, declarer would win the ace and West could later cover the second spade to block the suit).