gwnn, on Nov 3 2009, 11:32 AM, said:
LHO probably doesn't have 6 spades. He probably has either 4 or 5. If he has 5 then there is a 2 card discrepancy between slots and if he has 4 there is none. I read Frances' post here a while ago that you should deviate from '8 ever 9 never' only if there is a known 2 card difference which there is none here. So I'd play for the drop.
No this reasoning is not right. You need to evaluate your "probably doesn't", and your "probably has" properly, otherwise the result of your estimation will be random.
Even if we don't know the leading agreements of oppos,
in the worst case WC for finessing (i.e. you know by agreement of the WC that LHO has KQ
♠, and NEVER the Jack), the finesse would be a strict 50% bet (at the time you have to make up your mind, you'd know KQ
♠ by west, J
♠ by East, x
♦ by West, and xx
♦ by East, so exactly the same number of cards). And... OK, if their card says "K from KQx, x from KQxx+, and Q from KQJx+", I'll change my mind (definitely not the case where I play), but then I wouldn't even cash the
♦A !
In other cases, the Finesse is the winner. Depending on what we know from the lead, here is the percentage for finessing :
- KQJ+ : 56%
- KQJx+ : 56%
- KQJxx+ : 59% (in my partnership, K would have shown KQJxx+)
Here for vacant places calculus, it's the spade honors and the situation in diamonds which is important.
Actually you could try to infer something from the non Heart lead and the spot plays in clubs, but I think it's really secondary.
Odly enough in this case, even if you knew the parity of the spade suits at trick one (which may be the case against some oppos), the %age when they play KQJx+ don't change.
Clear case for finessing a priori, but let's have a look at their CC.
IMP