To keep in practice (and to get article material), I sometimes play online. Take my cards at favorable vulnerability (IMP scoring):
J43
AJ964
A32
Q5.
LHO opens 1, partner passes and RHO responds 1. You overcall 1, LHO passes and partner raises to 4. RHO doubles and you buy it there. I wondered what partner could have whereby he couldn't come in the first time, but could now bid game. The K was led and I saw:
Vul:East-West Dlr: West | 10876 10752 J AK103 | |
Q52 KQ KQ10 J9762 | AK9 83 987654 84 | |
J43 AJ964 A32 Q5 |
I suppose partner was a bit aggressive, but having escaped a spade lead, there were chances. What do you do after winning the A?
Let's say you trump a diamond (LHO plays the 10) and lead a low heart. RHO plays low and since KQx is unlikely (RHO must have decent spades since LHO didn't lead from a spade sequence--so he can't have enough points to also hold 5 HCP in hearts), you play the ace, dropping the queen. Now what?
Everyone follows on the Q and the moment of truth is coming. If clubs are 4-3, you can continue with two high clubs to throw a spade. But, clubs are probably 5-2.
Why? If LHO has only 4 clubs, what is his shape? He has at most two hearts (since the queen fell). He could be 4=2=3=4 in that order, but that is the only shape where your third club will live. Meanwhile, if he has only 3 spades or only 1 heart, he is just about sure to have 5 clubs. Accordingly, the odds favor finessing dummy's 10. Even if clubs are an unlikely 4-3, you will live if the 1 opener has the jack. This was the Real Deal:
Vul:East-West Dlr: West | 10876 10752 J AK103 | |
Q52 KQ KQ10 J9762 | AK9 83 987654 84 | |
J43 AJ964 A32 Q5 |
As was with the odds, finessing the 10 was the winning play. Next comes a high club, trumped and overruffed. You ruff a diamond and finally get rid of a spade on the last high club.
Notice what happens without finessing the 10. You would play Q, A, K and RHO would trump. Now, there is no way to discard a spade and declarer loses 3 spades and a trump trick for down one. +590 was worth a gain of 11 IMPs.