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

South	West	North	East
1 ♣	Dbl	Rdbl	Pass
Pass	1 ♦	1 ♠	Pass
1 NT	Pass	3 NT	All 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
♥
♦	7
♣
♠	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...