Posted 2005-June-02, 10:58
Clearly we can't get this right 100% of the time. The various odds calculations given are not all the same, so somebody must be wrong!
Partner's club should have been suit preference, but we've just been told it's reverse count. What's the point of count when there's another entry on the table? Anyway, as he's given count it looks like he has 3 clubs. In which case I should have ducked the club. Never mind.
The first question to address is:
- Does declarer have 9 tricks if we get this wrong?
If we trust partner, then declarer has 4 spades: declarer's play a little odd missing the K, so if declarer has the KJ then partner is discouraging; but partner would play his highest spade to discourage. If partner's 6 is his highest spade then declarer has the 7 and hence 4 spade tricks. Alternatively partner is encouraging from J76, but then I don't think declarer can afford to falsecard holding K9xx because the S9 might be a trick.
So declarer has 4 clubs and 4 spades. He must have a red ace to give him the 1NT bid, so we need to play the right red suit now.
The second question is which red suit is more likely to run?
A heart is right if partner has AJ7x or AQ7x or AQJx That makes declarer exactly 4243 with the KJ of spades, the DA, the CA and any of {HQDJ,HQDQ, HQDQJ,HJDQ,HJDQJ,DQJ} i.e. 6 different honour combinations. Plus we need partner to have the H7 which will happen slightly under 2/3 of the time the rest comes true.
A diamond is right if partner has AQJx or AQxxx. AQJx doesn't give declarer enough HCP. AQxxx gives declarer KJxx AQxx Jx Axx or KJxx AQJx Jx Axx.
Now, all these different honour holding have different probabilities but 2/3 of 6 combinations seems enough better than 2 combinations that I'm not going to work it out.
Heart (the 8, of course).
South: 3NT
Lead: S2