Loser Counts for Squeezing Winners
Consider two different situations:
xxx
x
x
--
Kx xxx
AKQ xx
-- --
-- --
AQx
J
A
--
The lead of the free winner -- the ace of diamonds -- squeezes a winner out of Lefty. Then Lefty is thrown in, in hearts, to eventually lead a spade.
The key thing to note is a loser count of three. The loser count for a stripe squeeze and throw-in is usually two. Consider in contrast this position:
xxx
x
x
--
Kx xxx
AKx 10x
-- --
-- --
AQx
J
A
--
Now Lefty can pitch a high heart. The throw-in doesn't work, because Lefty has saved an exit card to partner's hand.
The point is this. If you are squeezing only winners, there is no loser count for the thrown in. When you have to squeeze out exit cards, then the loser count has to be right. In this case, it is a loser count of two.
xxx
x
x
--
Kx xxx
AKx 10x
-- --
-- --
AQ
J
AK
--
Now the throw-in works.
The Stepping Stone Squeeze
The same thing occurs for the stepping stone squeeze. Here is a stepping stone squeeze that is just squeezing winners:
Ax
x
xx
--
QJ --
A x
AKQ J109xx
-- --
K
K
xx
A
This stepping stone squeeze works fine, even without the propor loser count (of two). But giving Lefty an exit card ruins the stepping-stone squeeze, because the count is wrong.
Ax
x
xx
--
QJ --
A x
AKx J109xx
-- --
K
K
xx
A
Lefty saves a winner. But giving the squeeze the proper count.
Ax
x
xx
--
QJ --
A x
AKx J109xx
-- --
K
K
--
AKQ
Now the three club winners squeeze all of the diamonds from Lefty.
Winner Squeeze
As noted in the discussion of winner squeezes, the winner count can be anything from 2 to 11 if only winners are being squeezed. The loser count must be 2 in order to squeeze out exit cards.
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