Winner Squeeze

Loser Count in the Winner Squeezes

Loser count is very important to almost all squeezes. Consider even the so-called "squeeze without the count". This phrase is usually used to describe what I call the "extended menace squeeze". This squeeze has a well-defined loser count, it just isn't a loser count of one.

In contrast, the winner squeeze has nothing resembling any requirement for loser count. In fact, the loser count can range from 2 to 11.

Loser Count of Eleven

	KQJx
	xxxxx
	Kxx
	x

Axx          J10xxxx
KQxx         xxx
AJxx         xxxx
xx           --

	--
	A
	KQ
	AKQJ10xxxxx

Lefty is on lead against 6NT. The lead of the ace of spades winner squeezes declarer. The loser count is 11.

Loser Count of Two

The lowest loser count is two. This loser count has special importance. With a loser count of two, you not only can squeeze out all winners from a suit, you can also squeeze out all entries to partner's winners. For example, you are in 5C with an opening spade lead.

xxx
KQx
AQxx
xxx

Axx
x
Kxx
AKQJ109

Lefty has made a takeout double. If you win the first spade trick, your only legitimate play for your contract is a 3-3 diamond break. If Lefty has four diamonds, you can make the contract by winning the second lead of spades. On the run of the clubs, Lefty will be forced to save 4 diamonds and the ace of hearts, hence no spades. (The position should not be that difficult to read.)

Some More Hands

	x
	KQxxx
	Kxx
	xxxx

Qx             KJ10xxxxxx
Axxxx          xx
J1098x         x
x              x

	A
	x
	AQxx
	AKQJ10xx

Lefty leads a spade. You have 11 tricks off the top and would like a 12th. Running all of the clubs forces Lefty to discard the entry to Righty's spades. The loser count was 2.

You are in 6 NT and Lefty leads a spade.

    109xx
    KQJ
    Qxx
    10xx
 
 
    A
    x
    AJxx
    AKQJxxx

You cross to the 10 of clubs, lead a diamond, and RHO pops the king of diamonds. Things are looking much better! But you still have only 11 tricks. Your twelfth trick could come from diamonds being 3-3, except the diamond king is probably singleton. A heart-diamond squeeze is perfect, except that you cannot rectify the count. However, if the same player has the diamonds and the ace of hearts, a winner squeeze will work. Because the loser count is the magic number -- two -- you can squeeze out all of that player's spades. You cannot afford to play another round of diamonds. This is the hand.

	x
	KQxxx
	Kxx
	xxxx

KJxxx          Q10xxxx
Ax             xxxxx
J1098x         x
x              x

	A
	x
	AQxx
	AKQJ10xx

On the run of the clubs, Lefty can first throw a heart, then a diamond, then three spades, keeping a small spade as an entry to Righty's good spades. Unfortunately for Lefty, there is still one discard to be made:

	--
	KQx
	Kxx
	--

x             Q10xxx
A             x
J1098         --
--            --

	--
	x
	AQxx
	x

The last club forces out the exit card, just as it would have forced out a spade winner. (Note also that if Lefty had started out with AK of hearts, or even AKQ, Lefty would be forced to pitch those winners, coming down to a blank A of hearts.)

Next: Squeezing Potential Winners