Playing Both Opponents for the Queen

[rod]

This one is from yesterday's club game in Coffs Harbour. I played with Hema and of course this is matchpoints. Board 11, nobody vulnerable, I dealt as South and opened 2NT. Hema raised to 3NT and West led her fourth-best club.

♠ A109
♥ 965
♦ 105
♣ KQ654

♠ Q4
♥ 1083
♦ 842
♣ J9872

♠ 87632
♥ AK74
♦ 763
♣ 10

♠ KJ5
♥ QJ2
♦ AKQJ9
♣ A3

Taking 10 tricks would be a poor result. They have two top heart tricks but I can make 12 if the clubs are 3-3. I win the Ace and lead another club to the King with East's spade discard revealing the 5-1 split.

Playing for 11 tricks I lead a heart from dummy. East wins the King, returns a diamond and I play another heart which East ducks!

♠ A109
♥ 9
♦ 10
♣ K65

♠ Q4
♥ 10
♦ 84
♣ J98

♠ 8763
♥ A7
♦ 76
♣ --

♠ KJ5
♥ Q
♦ AKQ9
♣ --

Now I can make 12 tricks by locating the spade Queen.

Oh wait, I don't have to locate it!

Playing 4 more rounds of diamonds leaves this:

♠ A10
♥ --
♦ --
♣ K6

♠ Q4
♥ --
♦ --
♣ J9

♠ 876
♥ A
♦ --
♣ --

♠ KJ5
♥ Q
♦ --
♣ --

West had to keep 2 clubs so is down to 2 spades. Cashing dummy's ♠A and ♣K then forces East to discard a spade. My ♠K is now guaranteed to drop the Q, no matter who has it.

This is called a non-simultaneous double squeeze and would have worked even if I held a small spade instead of the J.