Posted 2012-February-24, 09:50
Very interesting Hand.
I think the following works (didn't read what the others wrote, so pardon me if there is a repeat). (EDIT: I see Hanoi and Fred have a similar line).
Win the club, spade to K, cash heart A, Diamond K,Q and then A. Then play clubs throwing 3 hearts. Play heart K throwing the diamond (if not a winner already).
Basically, we have cashed our winners in clubs, diamonds and hearts, one spade for 10 tricks and are left with QT9 in North opposite Axx in hand.
At this point, I claim that we have enough information to play spades correctly.
This is how I tried to prove it without a (more)cumbersome case analysis:
There are 21 cards of clubs+diamonds+hearts outstanding, and so at least one of the opponents must have 10 or more cards in those three. That opponent cannot be the one who holds four spades.
Now, if we know how the clubs, diamonds and heart suits split, we will have a perfect count of the hand and know exactly which (second) spade honour to cash(first being the K or Q in North) and who to finesse.
Thus, let us assume for the moment that there is no singleton heart about, and each has at least two hearts. We play the heart AK, so we can confirm that.
The diamond and club suits are such that by playing 3 rounds of diamonds and 4 rounds of clubs, we will exactly know how they split. Now there are 14 clubs and diamonds outstanding. So, if any opponent has at least 8 or more of the minors, then that person has at least 10 cards in minors+hearts, and so we can guess spades.
So assume each person has exactly 7 cards in the minors.
Now consider the guy with 4 or more diamonds (if they split 3-3, the T will be a winner). Since the only problem case is when he has exactly 7 cards in the minors, he has 3 or less clubs.
On the last club, he cannot un-guard the diamond stopper (we haven't cashed diamonds yet) and is forced to throw a heart (if he throws a spade, we are done).
Now when we cash the heart K on trick 10, if he throws a diamond, then we know how the hearts split, and have an exact count.
So we can assume he follows to the heart K. Thus he will have at least 3 hearts and exactly 7 cards in the minors so no more than 3 spades. We know how to play spades in that case.