Two Extended Menaces (Advanced)

The situation gets more interesting when you have two extended menaces. First, I will present the boring instances.
        A876
        K5
        --
        32
 
QJ109        --
QJ109        --
--           AKQ
--           QJ1098
 
        K5
        A876
        --
        A4
The second extended menace creates yet another busy card in LHO’s hand, which means that the squeeze works even a trick earlier than when there is one extended menace. Here, the loser count is 3, yet the squeeze works on the last free winner. Declarer cashes the ace of clubs, then sets up whatever suit lefty abandons. A trick is lost at the end.
        A876
        K5
        --
        K2
 
QJ109        --
QJ109        --
--                AKQ
--              QJ1098
 
        K5
        A876
        --
        A4
I have added a club winner, so that there the loser count is only two. Now the squeeze works on the next to last free winner.

In the first case, there is still a loser to fight over. However, I cannot think of very many interesting ramifications. Lefty can’t be squeezed for the last trick, and by definition there are no winners in the other suits, so righty can’t be squeezed either. Once a trick is ducked to set up the extended menace and the menace is cashed, the loser count is one, so a throw-in would be useless. So the opps usually get the last trick. The only exception I can think of is when the throw in can occur when the trick is lost to set up the extended menace. For example:

        A1076
        K5
        --
        32
 
QJ98           --
Q1098          --
--             AKQ
--             QJ1098
 
        K54
        AJ76
        --
        A
On the last club, no matter which suit LHO abandons, LHO is thrown in to establish the extended menace in that suit and to endplay LHO in the other suit.

The more interesting situation is when there is a free winner. This free winner can be used to compensate for potential problems with the extended menace squeeze. For example:

        A876
        A432
        --
        --
 
QJ43           1098
J1075          Q98
--             KQ
--             --
 
        K5
        K6
        A
        A
I have set this up so that no simple squeeze is possible -- both lefty and righty stop both suits. However, only lefty guards the two extended menaces, so lefty is squeezed on the next-to-last minor suit minor. If declarer cashes the ace of diamonds, lefty is squeezed. But when declarer tried to lose a trick, righty can win the trick and cash a good diamond.

However, if declarer cashes the ace of clubs, then righty has no good card to set up. In this setup, so matter which suit lefty unguards, declaer will have a winner in the other suit.

Here is another one. In this case, diamonds are trump:

        A87
        A32
        --
        --
 
J104           Q2
J107           Q5
--             --
--             KQJ10
 
        6
        6
        9
        A32
More typically, the extra winner is used for communication. In this situation:
        A1092
        2
        --
        K2
 
J1043          Q
J107           Q
--             A
--             QJ109
 
        K6
        A109
        --
        A3
If LHO unguards spades, there is no entry to the long spade except for the king of clubs.

What if both suits lack entries?

        A109
        2
        --
        K2
 
J104           Q
J107           Q
--             A
--             QJ10
 
        6
        A109
        --
        A3
If you knew that lefty was going to abandon hearts, you could lead a club to the king and sill have an entry to the long heart; if you knew that lefty was going to abandon spades, you could cash your ace of clubs and retain an entry to the long spade. But you have to commit yourself in clubs before lefty has to commit with suit is discarded. So you just have to hope that lefty unguards the wrong suit.

Instead, you need this situation:

        A109
        2
        --
        K2
 
J104           Q
J107           Q
--             A
--             QJ10
 
        6
        A109
        --
        AQ
You lead the ace of clubs. That does not commit you to where the second club will be won. If you play the king of clubs under the A, you will have a club winner in your hand; if you play a small club, you will have a club winner in the dummy. Now lefty has to commit first. If lefty pitches a spade, you keep your king of clubs as an entry after the long spade is established; if lefty pitches a heart, you pitch your king of clubs and have an entry in the queen of clubs to the long heart.

In all of these examples, it has not mattered where the squeeze trick was won, because either hand could start the attack on the long suit. When only one of the hands can start the attack on both suits, the squeeze card has to come from that hand. (Or, the free winner can be used as an entry, but I am assuming here that the free winner is needed for another purpose. For example, suppose the situation is this:

        A1092
        --
        --
        K52
 
J1043          Q
J107           Q
--             A
--             QJ109
 
        K6
        A109
        --
        A3
Everything is fine -- the second club winner is needed for an entry to the spades, so the first club must be won in hand. Fortunately, you want to win the first trick in the hand, because your hand can lead either spades or diamonds. This situation would not work:
        A1092
        --
        --
        AK2
 
J1043          Q
J107           Q
--             A
--             QJ109
 
        K6
        A109
        --
        53
When you lead a club to the AK, lefty is squeezed out of guarding an extended menace. However, lefty can safely discard a heart, because dummy cannot attack hearts.

Now consider this hand

        A109
        --
        2
        KJ5
 
J104           Q
J107           Q
4              A
--             9876
 
        6
        A109
        --
        AQ10
Let's think about this. You have an two extended menace threats against lefty. You have only two losers, so the squeeze occurs on the next-to-last club winner. This is good, because you will need the last club as an entry to the long spade or heart. So the second club must be led from your hand, and you should be able to choose, depending on how you play from dummy, which hand wins the third club. However, and last but not least, the club itself must be won in hand, because only your hand can attack both threat suits.

So you have to win the first club with the ace in your hand and playing the king from the board, to leave this club situation;

J5
 
Q10
Now you can lead the highest club from your hand, which will win no matter what you play from dummy, but by your choice of plays from dummy you can determine who wins the second club. It would work just as well to lead the queen of clubs and play the jack from dummy, leaving
K5
 
AQ
It seems important to constrast this to a trump squeeze. Suppose spades are trump and this is the situation:
        K5
        32
        --
        --
 
--                10
AK                Q
AK                Q
--                Q
 
        AQ
        --
        32
        --
Except for righty's trump, you would have an easy cross-ruff of the remaining tricks. But righty has a trump, so you have to draw one rount of trump. Fortunately, on this round of trump, lefty is squeezed. You could, if you want, think of the two small hearts and two small diamonds as being extended menaces. I prefer not to. Essentially, the trump can be used to win the first round of the suit, so only the second card in the suit is a threat.

However, it is the nature of trumps that they only win in a suit if the suit is first led. In other words, if left discards a heart, you want to win the trump trick in dummy, so that you can lead and ruff a heart. If lefty discards a diamond, you want to win the trump trick in your hand, so that you can lead and ruff a diamond.

So, in this type of trump squeeze, you need to make lefty commit, then you decide which hand wind the FIRST trick in trumps. When you have two extended-menace suits, both of which might lack entries, you either do not care which hand wins the first trick, or you already know which hand should win the first trick. You need to make lefty commit and then you decide which hand will win the SECOND trick. (Another difference is that this type of trump squeeze can also work with 3 (or even more) trumps in hand, whereas the two-extended- menaces squeeze only works on the next to last free winner.
Return to Part 1: The Simple Extended-Menace Squeeze (Tutorial)

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