IF IT WALKS LIKE ....

by Jos van Kan.

Playing Imps you hold as west in 4th seat, all vulnerable  96  K9765  8765  K3 and partner opens in second seat with 1. RHO overcalls 1 and you have just enough for a negative double. This earns little respect from LHO, who jumps to 3, partner passes and RHO goes to game. So the bidding has been:
 

west north east south
-- pass 1  1
 X 3 pass 4 
pass pass pass
You lead K and the dummy comes down with
 
Q7543
  JT2
AK3 
JT 
Partner encourages with 8, you continue Clubs and partner wins with Q as declarer follows to both rounds (with 4 and 6). Partner now surprisingly switches to T, won in dummy with the Q and declarer continues with a second round of trumps on which partner tosses a , as declarer wins the Ace. Declarer continues with Q from hand.
  1. Who has A? Who has Q?
  2. How many Clubs does partner have at least?
  3. Who has 9?
  4. How many tricks does declarer have? <br> a) if partner has 6 clubs <br> b) if partner has 5 clubs
  5. How do you defend?
  1. At this point there are 3 picture cards whose location is unknown: A, Q and J. Partner, who has opened the bidding only has shown up with AQ of clubs and hence certainly has A. He probably also has Q unless he has extraordinary distribution. He did not bid over 3, so that is not very likely. He may or may not have J.
  2. This is a tricky one. Maybe you'd be inclined to say at least three, but then you have missed an important inference. He had a singleton , remember? And a player who has a singleton  (or ) and opens 1  has five or more s. Huh?? Yes. He either has some 5 card or more or exactly 1-4-4-4 distribution. But that distribution is usually opened 1. (unless the  quality is abysmal, but we have already established that he holds Q).
  3. Only of academic value, but nevertheless: declarer has 9. Partner encouraged with 8 and if he'd had the 9 he'd have played it.
  4. If partner has six Clubs declarer has three and he can come to 8 tricks: 5 , a  ruff and  AK. If partner has five Clubs declarer has four and he can come to nine tricks since he can now ruff two clubs.
  5. So what kind of problem is this anyway? Isn't declarer bound to go down, since we are always going to make two   tricks? Always? Think about it. Suppose declarer has a singleton ? If declarer's Q is singleton you must be very careful, since otherwise you give it away. Suppose you win with the K. Now you cannot prevent declarer from ruffing away partners Ace. So, you must duck the Q and let partner win his A. He'll play back a  and in the fullness of time you will make either your K or partners Q will score.
Post mortem
All this was necessary since the full hand was:
 
 
Q7543
JT2
AK3
JT
 
96 
K9765
8765
K3 
 
T
A843
Q92
AQ852
 
AKJ82
Q
JT4
9764 
 

Partner could have made the defense somewhat easier by first cashing A before switching to trump. But that would have made it a kind of dull hand. :-) This is an OKBridge hand and it was played just as I described it. W was the well known Canadian player Shelagh Paulsson. At the critical point she ducked the K without so much as pausing for thought! Well done. A question you might still ask is: "How will declarer know that he must ruff out A and not take the  finesse?" The answer to that is that declarers ALSO count. Your partner has opened the bidding and YOU have shown up with two kings already. So your partner must hold the rest of the HCP and the  finesse is out of the question.


copyright © 1997 by Jos van Kan. All rights reserved.