As you recall from chapter 3, a double squeeze may occur when each opponent guards against one suit alone, and a third suit is guarded by both opponents (what we called a “shared threat”. We now turn our attention to the situation where one opponent has guards in all three suits, and his partner has guards in only two of these suits. The first corrective action, and perhaps the most common of these additional squeezes (at least in my experience) is the compound squeeze. If you were to study only one of these squeezes, the one to study would the compound squeeze, as in my experience, this one shows up more often than all the others. And for another, as a defender, you have some chances to break up compound squeezes if you and your partner manage to reconize the pending possible squeeze and try to make declarer's line of either more difficult, or better yet, impossible.
Compound squeeze
As you recall from chapter 3, a double squeeze occurs when each opponent guards against one suit alone, and a third suit is guarded by both opponents (what we called a “shared threat”). Clyde Love called the suit singlularly guarded the "basic threat". I don't like this term, but it is essential for compound squeeze that you do have one suit solely guarded. If I had to name it, I would call it the lone threat. To illustrate the Let’s examine one such hand played (MAY 9th 2004) on the BBO.
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| West | North | East | South |
| p | p | 2♣ | |
| p | 2♠ | p | 3♦ |
| p | 3♥ | p | 4♣ |
| p | 4♦ | p | 4NT |
| p | 5♦ | p | 5NT |
| p | 6♦ | p | 7NT |
| p | p | p |
South decide to go for all the masterpoints by playing 7NT. The play in 7@spades would be trivial (ruff a heart in south).
Opening lead was the D4 and you try, hopefully, the diamond TEN, but east covers and you win the ACE. This is followed by two more top diamonds. On the third diamond, WEST discards a club.
Let's examine BLUE. We have 12 winners 5♠ (hopefully), 2♥, 3♦, 2♣. East alone guards against the diamond threat. If either opponent holds five clubs originally a squeeze is sure. If West holds five clubs, he is busy in clubs, EAST is busy in diamonds, and hearts is the shared threat and it has two winners. The double squeeze would work. If East holds five clubs, then you have a diamond-club simple squeeze on him. From the club discard, it looks like if anyone has five clubs it is West. The heart threat (small heart in dummy), cleary both opponents will protect against. But what if diamonds are split 4-3 (as here)? Now both opponents protect against hearts and clubs.
You test clubs, by cashing the ACE (and dicarding club Ten, just in case (Ten and eight are equals, a later squeeze type will show why such unblocking discards can be important, on the current hand, it makes no difference). Then you cash three spades ending in dummy in this position, EAST pitching two hearts.
Let's examine this ending closely.
Basic Compound Squeeze Ending
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B: *defective
L: one
U: ♦5, ♥x
E: primary, ♣A, secondary ♥Ace
Shared threat: **defective
The B threat for simple squeeze is defective in that no opponent is solely responsible for two threats
**The shared-threat for double squeeze is also defective (both opponents protect against ♣ threat and against the ♥ threat)
However, watch what happens to EAST when you cash the ♠Ten. His three options are:
Discard diamond nine, which gives you your 13th trick immediately, so he will not do that
Discard a club, which makes his partner soley responsible for the club guard, turning this into a double squeeze, with hearts the shared-suit, the R-suit is clubs, the L-suit is diamonds. North now must give up hearts but it doesn't matter. You play next spade (east and south throws club), but this squeezes the heart protection from west. Now the ♣ACE is the squeeze card on EAST, he can't keep three hearts and the diamond nine.
Or East can throw a heart. Now West alone is left guarding ♥, east diamonds, and the shared-suit is clubs. A simpler squeeze could not be imagined. Cash spade, cash hearts. On last heart, East has to keep diamond, so he comes to one club, south can then pitch his diamond. North has to keep the 13th heart, so he comes to one club. The club king-nine win the last two tricks.
So what happened? If you have been following the sequence of thoughts for a squeeze ending, when you find loser = 1, and you have an upper theat (this case diamonds), you looks to see if EAST alone can guard against a threat in another suit (for a simple squeeze, if entries are right). But it is easy to determine that East probably does not guard hearts alone, and probably not clubs either (west with 3S and 2D would not discard a club from a doubleton with six hearts). So the simple squeeze "B" requirement is defective. When "B" is defectve, you turn next to see if there is only one "Shared-threat", such that one opponent alone guards against threats in one suit, and both share only one suit. Here you see instead of one 'shared-threat", there is likely two shared-suits (♣ and ♥).
This dual "shared-threat" along with one isolated threat (which lies in the UPPER hand), however, provides you with an opportunity for a new type of squeeze: a compound squeeze. This example nicely shows how this squeeze works. On the next to last free winner, the hand protecting against the lone threat is squeezed in three suits. If he gives up the lone guard there is your extra trick. If he gives up his guard in either of the two suits in which he shares responsibility, he converts the hand to a potential double squeeze, as in this example.
What are the general requirements for a compound squeeze?
Loser must equal one,
The lone threat must lie in the upper hand,
Each of the dual shared-threat suits must have a winner/entry in its own suit
This hand makes compound squeeze look easy, and sometimes they are. But some hands that meet these general requirements fail. The problem deals with after the hand with three suits stopped gives up a suit, you have to have a double squeeze that works, and remember some double squeezes failed if you didn;t have two winners in the "B" suit, or if you cashed your free winners in the wrong order.
What have we learned?
If you can isolate only one guard to one hand, then a compound squeeze might work
The lone threat must be in the upper hand for compound squeeze, and
Each of the dual shared threat suits must have a winner/entry.
Types of compound squeezes
Types of Compound Squeezes
The above are the general requirements for a compound squeeze. These requirements are necessary, but not always sufficient for a successful compound squeeze. Let's see an illustrative example, then explain why this example fails.
Failing Case for a compound squeeze
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B: *Defective
L: one
U: ♦Q, ♠2
Entries, ♥A, ♠A
Lone threat: ♦Q
On the surface, this ending looks very much like the compound squeeze in the last post. One opponent (west) has the lone guard against the diamond threat, which lies in the upper hand. There is only one loser, and both "shared-threats" have a winner/entry. All the rule given so far are satisfied, and yet the compound squeeze will fail with best defense (this doesn't mean give up, sometimes you don't get best defense). Let's try it out. When you cash the last club, West has to discard something. If the diamond, you are home, so he will discard a major. First, what if he discards a club? Then you will make. You will cash the spade ACE, then lead the last club, executing a very standard double squeeze. WEST will have to keep the diamond, so can't keep two hearts. Once he discards his heart, dummy lets go the ♦, and then the pressure turns to east. He can't keep two hearts and the spade to stop your two. Success... However, what if WEST, discards the ♥Queen? Now, no matter which way you go, you will not gain an extra trick. If you cash the ♥ACE then come to your hand to lead the last club, there is no entry to the diamond threat or heart threats in dummy. If you cash the spade ace, there will be no re-entry to your hand and no winner in the "shared-threat suit", which fails. And if you cash the last club, West throws another heart, and dummy is squeezed.. if you throw a heart or a diamond you lose a threat in dummy prematurely and if you thow a spade, you will have no entry back to south's hand.
Exactly why this ending goes wrong is beyond the scope of this "introductory" sequence of post as it requires explaining all the different types of double squeezes (something I intentionally have avoided doing). Instead, I will just remind you of the two rules I gave for double squeezes. One is that is better to have two winners in shared-threat suits rather than one. The second is that it is best to cash your "R winners" before your last squeeze card (last free winner). Some compound squeezes will fail because the defense will set up a double squeeze for you (by abandoning the approriate suit) where you need two Shared suit winners when you have only one. Others will fail because the defense will set up a the double squeeze for you where you have to cash your last "R -WINNER" before the last squeeze card, but there is no way for you to do so and keep your entry conditions satisfied. This latter case is what happened above when WEST discards a heart. On a heart discard, the shared threat becomes spades in south, so the last R winner is the heart ACE. But if you cash that, the squeeze fails because it mucks up your entry conditions.
So this example shows that compound squeezes can have some strick requirements. What types are there and are they all this complicated? There are two general types of compound squeezes, and each one has an easy layout and a difficult one. In all compound squeezes, at least one shared threat must be in a hand opposite the lone threat (or else no threat would be in the upper hand relative to one opponent). So the two endings are: Both shared threats in the same hand, and One shared threat in the same hand as the lone threat.
When both shared threats are in the same hand (paired-shared threat), the ending is very easy if the lone threat has a winner in its own suit and an entry in any of the three suits (including the lone threat). You can run all the free suit winners, then enter (if not there) the hand with the lone threat and play the lone threat winner. These ending play themselves. For instance...
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You can cash both clubs, throwing a spade from dummy. West has to abandon diamonds or spades, and the double squeeze can not fail.
When you have a paired shared threat hand without a winner in the lone theat suit, or with no winner in either of the two shared threat suits, the compound squeeze will only work if the squeeze card (last free winner) is in the same hand as the lone threat. And here, after the hand with the lone guard abandones one of the shared threats, you must cash all the winner outside the shared threat suit in the hand with the double threat, then re-enter the hand with the sole threat and cash the last free winner. You need adequate entries for this and the LAST FREE winner has to be n the hand with the sole threat. This one is difficult to manage.
The other type of compound squeeze has the shared suit threats split between the two hands. The best kind of these to have is where the shared suit opposite teh squeeze card has TWO WINNERS. These play themselfves, and this was the type of compound squeeze shown in the last post because it seems so utterly easy to see and excecute. (REMEMBER THE RULE, TWO WINNERS ARE BETTER THAN ONE FOR DOUBLE SQUEEZES AND FOR COMPOUND SQUEEZES).
When the hand with just one shared threat lacks two winners, the compound squeeze becomes very difficult to execute and easy for the defense to mess it up. This is what happened in the failing case above. Let's revist it but give each hand an extra card, this time giving south two spade winners instead of one...
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B: *Defective
L: one
U: ♦Q, ♠2
Entries, ♥A, ♠A
Lone threat: ♦Q
Now this compound squeeze works, just cash both clubs, and throw a low diamnond from dummy.
Can the compound squeeze ever work where the lone shared suit has only one winner? The answer is yes, if the following strict requirements are met. First, the hand with only the one shared-threat must hold the last free winner. On the next to last free winner, you must have entry conditions necessary to cash the last R-winner, reenter the hand with the sole threat threat and cash the last free winner. The sequence of plays can trip up the best of us, and perhaps this will require extra study on your part, but for now, here is an example.
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On the club ACE, West is under pressure in three suits. If he discards a ♦, you have 8 tricks. If he discards a ♠, this becomes an easy double squeeze. Cash ♠AK, cross to the diamond King, cash the last club, west has to keep the diamond, so he will throw a heart, dummy small diamond let go, and you lead diamond to the ACE, squeeznig east in the majors.
So West will discard a heart. Now cash ♥ACE (last right winner) then diamnond A then king, then lead the last club. West has to keep a diamond, Easat ahs to keep a heart, so no one can keep three spades
1 comment:
On the second diagram, there is a typo. East is shown having seven cards: three clubs when he should have two.
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