Winner Squeeze

The Two-suited Winner Squeeze, Part II: When the Aces are in Different Hands

So far, the assumption has been that the two aces (or controls) are in different hands. What if they are in different hands? This was the hand I gave for when the aces were in the same hand.

	KQx
	KQx
	Kxx
	xxxx

9876           AJ10x
987            AJ10xx
J1098x         x
x              xxx

	xx
	xx
	AQJx
	AKQJ10

What if the aces were divided? That one of them is onside creates a second way to play the contract, but let us ignore that. The last club simultaneously squeezes both opponents. One opponent has to save 3 spades and one has to save three hearts. So the situation is this, when the aces are with the long suits:

      KQ
      KQ
      --
      --
Axx        x
x          Axx
--         --
--         --
      xx
      xx

Now either suit can be safely led. Essentially, a key loser has been squeezed out of each hand. (Putting the aces in the short suits changes nothing.)

If Righty had kept two spades and Lefty kept three, then it would not be safe to lead spades. It would take good defense, but Lefty could duck the first round. Then Righty could get to Lefty's winning spades.

But that means also that the defense retained 5 spades and only 3 hearts. So you could safely lead hearts first.

So the two-suited winner squeeze is remarkable easy to run whenever two suits can be cleared and the count is right and the aces are split. You simply run all of your winners coming down to your two KQs. If the opponents have saved 4 of each suit, you may lead either suit to get two tricks. If the opponents have saved 5 of one suit and 3 of the other, you need to lead the suit they have saved 3 of.

That has some implications for defense when the two aces are in the same hand. The defender with two aces must save three of one suit and one of the other, or else there is no hope. The other defender must save at least 3 of the second suit, and preferable 4. If not then declarer can safely lead the suit in which only 3 cards are out.

From actual play

     K10x
     Kx
     QJ10xxx
     xx

AJx         xxxx
xxxxx       J10x
9xx         K
Ax          J109xx

     Qxx
     AQx
     Axx
     KQxx

This occurred on 11/21/07. The opening lead was a small heart against 3NT. Declarer won in hand, led the queen of diamonds from board, which Righty quickly covered, hoping partner would have 9xxx. Declare now has 9 tricks and a sure play for 10. One possibility is the spade finesse against the jack. A second is to lead a heart to the board, lead a club, and hope the ace of clubs is onside. If it is, then a club trick can be won safely, then spades can be attacked. This, however, does not work because there is not necessarily a route back to the good heart. One could hope that the ace of spades is also onside.

The third possibility is to run a two-suited winner squeeze hoping the aces are in the same hand. Declarer in fact rand the diamonds.

This hand has the complication of the heart suit. If, after running diamonds, the situation is this:

     K10x
     K
     --
     xx

Axx      
--      
--        
Axx        

     Qx
     Ax
     --
     KQ

Then declarer has to cash the king of hearts to effect the two-suited winner squeeze. There is a spare winner in hand that cannot be cashed, but fortunately the squeeze works on the next-to-last free winner.

However, if the situation is this:

     K10x
     K
     --
     xx

Ax       
xxx      
--        
A         

     Qx
     Ax
     --
     KQ

then declarer cannot afford to lead the king of hearts and instead must lead a black suit. In fact, defender saved hearts and declarer won the king of hearts before attacking a black suit.

If the aces had been split, then the winner squeeze is not so automatic. First, there is the technical problem of cashing the hearts, now that they have been blocked. There is also a problem that for the winner squeeze to work in this situation, there cannot be any remaining hearts.

Another hand from play

The date is three days later. How common is this squeeze that I would find it in two consecutive sessions of bridge?

      3
      K953
      AKQ72
      1065
A1054         J987
A62           J1087
54            1086
KJ73          Q8
       KQ62
       Q4
       J93
       A942

The opening lead was a small club against 3NT. Declarer let the queen win, a mistake, Lefty won the second club and returned a high club, which was a second mistake balancing the first. So that is how the play went.

Again, one could hope that the ace of spades was onside. Then a spade could be safely established, then a heart could be established. This works because the jack of diamonds is an entry to declarer's hand. If the jack of diamonds is exchanged with a small diamond, then declarer still has 5 diamond tricks. But declarer does not necessarily have an entry to the spades, either if Righty wins the first trick with the ace or if Lefty has the ace and does not clear the suit.

Again, the two-suited winner squeeze works whenever the same hand has both aces, which is the case here.

      x
      Kxx
      --
      --

      KQ
      Qx

If the hand with both aces saves 3 spades, it is simple to establish a spade and then a heart trick. If the hand with both aces saves 3 hearts, then you can lead a spade to set up your spade. Now that hand has to lead away from the ace of hearts, allowing your queen to be an entry.

When the aces are split, the two-suited winner squeeze can fail. First, Lefty has to save 3 hearts to the ace, while Righty has the ace of spades. Then if you attack spades, Righty can win the ace and lead a heart through. You have no entry to the good spade. If you attack hearts by leading to the queen, Lefty cannot afford to duck. So Lefty has to win and play on hearts. That means Righty also had to save three hearts in order to beat the hand.

Next: A Winner Squeeze on Entries