BBO user Kenrexford posted an interesting hand on the BridgeBase Forum page a few days back. It was..
In the problem, as stated, you won the spade ACE at trick one, and discarded a diamond. That is ok, although you might choose to ruff the first lead and save the spade ACE, it will not be important for the rest of the discussion.
You have 8 heart tricks -- assuming you can find the queen (in the problem, you drop Qx offside), 1D, 1S, and 2C. That comes to 12 tricks. You can always play WEST to have QJx(x) of clubs (double finessee) which is just less than 25% (somewhere around 24%).
What squeeze chances are there? West could have 6S and 6C (ruff spade at trick two) and be squeezed, but we assume bidding eliminates that possibility. Someone could have 6D and the QJ of clubs, again a very long shot.
So as you look for squeeze chances, it seems that BLUE will be violated because of lack of a double threat. In fact, it is not clear any single threat is guarded by only one opponent (the long diamond in dummy can not be a used to isolate a threat against opponent with three diamonds because you have to lose a trick to turn it into a threat).
What squeeze chances are there? Well both black threats HAVE TO BE in north *you have no spades and the thrird club is in dummy), so for any three suit squeeze to work, the diamond threat will have to be south's remaining "x" after one is thrown on the spade ACE.
This begins to look like a hexagonal squeeze, where both opponents stop all three suits. But they have to give something up when you run hearts. You can imagine an ending where south retains 2H, 2D, and 2C, and dummy 2S, 1D, 3C. In this six card ending, no opponent can keep 3C and 2 cards in both additional side suits. So you will have isolated a threat against one opponent, as typically found in a compound squeeze.
Can a compound squeeze work here? The requirements for a compound squeeze is that whenever the opponent abandon's a suit, the remaining one of the remaining suits can serve as a "both" threat (guarded by both opponents). The problem here is that such a "both" suit requires an entry in its own suit. Here, if they make spades the "both suit", there is no double squeeze, because there is no entry in the spade suit.
When the both threat suit lacks an entry, a possible remedy is a guard squeeze. Can this hexagonal squeeze resolve to a compound guard squeeze? One requirement for guard squeeze is that one threat be partially finessable (unblock clubs and that suit is). Another is the side suit have a winner (diamond Ace) makes that true. However, if EAST keeps spades and clubs (behind dummy), west can keep diamonds and there is not squeeze of any type.
So against perfect defense (EAST knows to keep quards against threats in front of him), the squeeze must fail. Suggesting a straightforward club double finessee. But can there be any reason to forgo the finessee? Well if EAST has five diamonds, the quard squeeze has to work, but 5-1 diamond split with EAST having four plus is less likely than QJ of clubs onside (4-1, 5-0 splits are 13.2 + 1.9 or 15.1%).
However, you need not be in a hurry to take the double finesse in clubs. You can apply the pressure by running a bunch of hearts. On the hand in question, EAST did have five diamonds. Which becomes obvious by the discards. Now a double quard squeeze shows up when the ending becomes something like:
Here you lead the club Ten to the ACE, ruff a spade, and play out the last trump. East will have to keep both diamonds and the club jack, so he discards his last spade. West, can comfortably discard his next to last spade. Now you cash the diamond ACE, and WEST is squeezed in the black suits. A spade and you throw a club from dummy, a club and dummy discards the spade.
Note, if you can not READ the diamond position, the odds favor the double finessee. Once you cound down WEST with loads of clubs and spades, and EAST shows up with the diamonds, the odds switch to the double guard squeeze. EW should try to make it a little more difficult, but they really have very little room.. EAST has to keep at least two clubs. If east is able to keep three clubs and had 4+ spades, nothing will work here, but his five card diamond holding forces him to give something up.
