EricK, on Aug 5 2006, 02:02 AM, said:
I would have called it a "triple squeeze without the count."
Multi-loser squeezes are often called "squeeze without the count". As your comment points out, a triple squeeze typically has two losers (triple squeezes can also have only one loser, but then the victim is really mincemeat). So while non-triple squeezes with multiple losers are often called "squeeze without the count", and it might be logical to extend that definition to triple squeezes with more than two losers, this blurs some fine distinctions between hands.
Let's consider "squeeze without count" hands. What types are there? I can think of at least four general types off the top of my head.
1.) A developmental threat. For a developmental threat, by losing a trick in a suit, we develop an extra winner. On this hand there are two developmental threats (diamonds and spades). Development squeeze work when loser count is one greater than the "correct count". So for a two suit squeeze, the loser count has to be "2", for a three suit squeeze, the count has to be "3". Sometimes the squeeze will repeat to gain two tricks when in the context of a triple squeeze and losers count is "3" (such triple squeezes that work where loser count is higher than 3 is rare, unless... see next type).
2) Strip endplay - is where by losing a trick to an opponent you receive a favorable return that gains a trick (the "squeeze" here generally removes excess winners and/or cards of exit). The loser count can be any... Even wiht five losers, this type of play can gain one trick.
3) Losing Card Squeeze. Here the "squeeze card" just happens to be a trick won by the victims partners. In a triple squeeze setting, the loser total has to be "3" and when you lose a trick to your victims partners, he can not have another winner to cash, obviously.
4) Guard squeeze plus developmental threat: This is a type of three suit squeeze without a count where in addition to the a developmental threat, one option the opposition has is to expose his partner to an immediate finesse. Like the case with a developmental threat, this on works with a loser count that is too high by one.
I suspect there are more combinations we can think of if we try. But if you lump all of these into one category "triple squeeze without the count", it will be hard to recongize them at the table, and you do not convey the nature of the squeeze play very accurately. You need some context in which to examine the nature of your threats early in the hand in order to manipulate the play to the corresponding ending. Also, it turns out, for instance, that a triple squeeze with a developmental threat has the "correct count" when loser count is "3", as it is a different squeeze than a normal triple squeeze, just as a two suit squeeze with a developmental threat or a losing card squeeze has a "correct count" when loser count is "2". That is, the correct count for these "squeezes without the count" is one more than normal. While a strip squeeze the loser count can be anything.... by classifying them correctly, you also can try to "correct" the count to what is needed (see the duck of the first club, to go from four losers to three on this hand, as an example).
The unique this about the hand shown is that there are two developmental threats, and when setting up the developed threat, it doesn't ahve to be the victim of the squeeze that wins the trick. Give West KQJ of spades rather than QJx and it is a normal triple squeeze with a single developmental threat. But here, West had to keep the QJ of spades to prevent declarer from developing a spade trick. This is sort of a the developmental squeeze equivalent to a guard squeeze (rather than protect partenr from a finessee, you have to protect him from being exposed to a developmental play in spades). Note, swap the spade Queen with a small spade in his partners hand, and the squeeze fails. Why? Because you have just turned the triple squeeze into a double squeeze but the loser count is off by two for a simple squeeze with a developmental squeeze.
Help
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South on lead needs 4
more tricks to make 4♥.