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
7 4
Q 2
K 10 7

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.


  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).

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