PARD (dummy) - K J - Q 8 6 |
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ME K 10 8 10 10 |
Q 9 4 2
Q 6 2
A K Q ? x
7
But where's the
J? There can only be
two possible explanations.
Q.
If my hand were, say:
K 10 x x
x x x
x x x
A x x
J.
J, which holds. Lefty has given
me a chance to make 3
, and I wonder
if there is any way to maneuver both of:
PARD (dummy) - K - Q 8 6 | |||
LEFTY Q Q ? x - |
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RIGHTY - 5 ? J 9 | |
ME K 10 - 10 10 |
K and pitch a
diamond, but the followup club head will promote lefty's
Q. Where else could the diamond
loser go, though? Fortunately, there is a chance to shed
of this loser if righty holds the
J.
Since this is my only possible line to gain a trick, I'm
going to take it.
I lead the
K from dummy and ruff(!)
with the
10. Now I cash the
K, pulling lefty's last trump and
I have caught righty in a backward squeeze! He resigns
by pitching the
9, which makes dummy
good. Making 3
was worth 100% of the matchpoints (not to mention bragging
rights). I couldn't have asked for friendlier
opponents than these to enjoy the remainder of the matchpoints.
K instead of the
J, just in case righty was playing
with me. I could have blown my chance in the end position
by this impulsive play.
K
certainly should have cost, but it didn't. This was a
simple case of failure to count the hand.
Axxx opposite
K10xx is to
first play the
A and then lead towards the remaining
K10x, playing the
10
if second hand plays low. It is important to note
that this would not have been correct on this hand. Suppose
righty played low to the second round of spades and lefty
won the
10 with the
J. Another high diamond would
again endplay dummy, still with 2 diamond losers in hand
and a high trump out. The safety play would have been correct if there
was still an entry to declarer's hand.
AK was a matchpoint
play to get all of our deserved uptricks. At IMPS, a
more secure route to 9 tricks would be to win the
J in hand with the
K and try to ruff out as many
diamonds as possible.