Thursday, August 20, 2009

A failed beer squeeze

A few years ago I tried to construct a beer squeeze: a squeeze where the opponents can successfully prevent you from gaining a trick only at the "cost" of allowing you to win the last trick with the 7. But, there was a cook: declarer can in fact always succeed in gaining the extra trick. Give it a try:

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.

A little color below, full answer tomorrow.

The auction and early defense were constructed to make this a double dummy problem -- LHO must be 4441 with all the remaining missing honors. My original intent was that you duck the Q (to rectify the count), and then LHO would continue a diamond. You could cash A and run 4 clubs to reach this position:

A 10 8
Q 2
K 10

If LHO started with KJ9x/QJxx/QJTx/x this is a non-progressive triple squeeze: unguarding hearts would allow the squeeze to repeat, and unguarding spades would give up 2 tricks instantly, but ungaurding diamonds just gives up one trick and the squeeze doesn't repeat. But it would allow declarer to, say, play K, heart pitching 2 spades, then win ♠A and 7 and earn a beer.

But, rolling back to the Q play at trick 3, declarer has a line for 10 tricks against the same layout. Stay tuned...

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