Posted 2011-February-01, 14:33
Agree with Rainer that ♣finesse is totally obvious (and leads to 12 tricks 99% of the time via the upcoming squeeze).
Long version:
In practice if West doesn't have both honors he will have x xxxxx Q KJT987 for his discards (discarding 78♣ is easy).
In theory we lose if West has x Qxxxxx Q JT987.
Totally unlikely ?... Hmmmmmm.... if West comes from Jupiter he could make the following reasoning in 10 steps :
1- South has KQx+♠ (no 2nd round finesse) ;
2- South doesn't have KQJx (he would have left the double) ;
3- East knew that "small spade" costs nothing : he has 7 Spades ;
4- South has ♠KQx exactly
5- West "sees" 5+13+5 = 23 HCP in W-N-E hands respectively. 17HCP are missing.
6- South has ♥Ace by counting HCPs.
7- If South has ♣K, South has 11 tricks from the top (2+2+4+3), and 12 thanks to an obvious Squeeze ;
8- So West hopes that pard has the ♣K, and South upgraded a 14HCP count because he had 5♦ ;
9- So West hopes that south has the actual hand KQx Ax AJT9x xxx
10- West thinks like Rainer, and discards 2 clubs from JT9xx to induce the OBVIOUS ♣finesse.
Aspirin please.
FD