Friday, August 21, 2009

Failed beer squeeze, solution

From yesterday's post:

A 10 8 5
A 2
7 4 3 2
A J 3
 
Q 2
K 10 7 4
A 6 5
K Q 5 4


SouthWest North East
1 ♣ Dbl RdblPass
Pass 1 1 ♠Pass
1 NT Pass 3 NTAll Pass


Q led, RHO plays the K and another diamond.
Say you duck twice and LHO shifts to Q. How do you play? Clubs are 1=5.


Assuming LHO has KJ9x/QJxx/QJTx/x, you can guarantee 10 tricks by winning A and running clubs to come down to:

A 10 8
2
7 4
 
Q 2
K 10 7
A



You have 5 in the bank and want 5 of the last 6 with only 3 tops. Luckily, the 3-loser progressive squeeze comes to the rescue.

LHO needs to pitch from KJ/J9x/Tx/- in this position. If he pitches:

a) A spade? This give up 2 spades immediately, and if you cash A first you'll also have a red suit squeeze to take the rest. Ouch!

b) A heart? Play 10 to set up a trick there. LHO can't lead spades without giving up a 2nd trick, so must put you back in hand with a red card, now cashing all your red winners effects a spade-diamond squeeze.

c) A diamond? That gives you an extra diamond, then A, ♠A, 7 spade exit end-plays LHO for an extra heart trick.

I guess this is my 2nd non-standard progressive squeeze in the last 2 weeks, so I'll try to write about some generalizations next week.

2 comments:

  1. It looks like you can win the heart Ace and play the diamond ace. Next cash three clubs ending in dummy and exit with the beer card to endplay West.

    I didn't comment when I looked at the problem yesterday, but this seemed likely to work (albeit not as elegant as your solution).

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  2. That gets you to 9 tricks, but if West attacks dummy's last entry with the SK you can't get to 10.

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