Finding the safe exit card

A couple of nights ago I found myself in a predicament on this defensive hand. Partner led the ♣2 (playing 4^th best) and NS were playing standard American so 1NT was not forcing.
 

West ♠ Qxx ♥ Qx ♦ KTxx ♣ JTxx
	South ♠ AKxxx ♥ Kxx ♦ AJ9 ♣ Kx

East	South	West	North
Pass	1♠	Pass	1NT
Pass	2NT	Pass	3NT
Pass	Pass	Pass

Declarer took the ♣K on the board and played a club to her Q before passing the ♥J to my ♥Q. Without really thinking I played 3^rd highest and then 4^th highest in clubs. In retrospect, I probably should have unblocked but was a bit concerned that partner led from 3 small and I wouldn’t want to be in a position to not get any club trick.

Anyway, I was in with the ♥Q at trick 3 and now I could see a potential endplay and having to break a pointed suit so I played the ♣T, hoping to deceive partner into thinking declarer has the ♣J and therefore not lead the 4^th round of clubs when she gets in with her presumed ♥A. No such luck. Declarer won the ♣A, won the ♥K, and played a heart to partner’s ace. Back came a club to me. Dummy has shed a diamond and a spade and declarer a diamond.

The bidding marks declarer with 8 or 9 hcp, possibly 10 and possibly only 7, and I know of 7 of those points already with 3 more hcp outstanding (♠J and ♦Q). The bidding also marks declarer with less than 3 spades and there’s a good chance she had a singleton spade. Dummy is left with ♠AKxx  ♥ -  ♦AJ  ♣- and I have ♠Qxx  ♥-  ♦KTx  ♣-. If only partner were on lead to play a diamond, all would be easy. I could score the ♦K and the ♠Q to set this contract.

Regardless of the spade situation, a diamond gives up one trick whenever declarer has the ♦Q (I’ll get only my ♠Q for our 4^th defensive trick) but is the winning play whenever partner has the Q, holding declarer to probably only 3 of the last 6 tricks and giving us 6 total tricks.

A spade gives up all of the last 6 tricks when declarer has ♠Jx (she gets the ♠J and that gives her an entry to cash the ♥T). When declarer has ♠xx or ♠Tx, any spade lead spade safely preserves at least 1 trick for the defense, plus a 2^nd one if partner has the ♦Q. When declarer has a singleton spade, partner may be able to score 2 spade tricks and I’ll still get my ♦K. We don’t want to lead a spade and get thrown back in with the ♠Q, only to have to lead away from the ♦K at trick 11, so maybe it’s better to play the ♠Q. Yes, if you decide to lead a spade, the ♠Q is the one to lead. It has the added bonus of winning when declarer has singleton ♠J, it doesn’t matter when declarer has ♠Jx (a low spade makes all 4 of them good) or a small stiff, and it wins when declarer has ♠xx (you can’t get endplayed again).

Both suits offer a chance to set the contract by 2 tricks while only a spade lead gives declarer a chance to make an overtrick. Does that make a diamond lead right? I don’t think so. It’s equally likely for declarer to have 8 or 9 hcp but there are a couple of things that tip the scales slightly toward leading a spade: 1) Partner chose to lead from 4 small. Would she lead that over ♦Qx or ♦Qxx? Absolutely. Would she lead that over xx, xxx, or xxxx? Probably but not necessarily. 2) Declarer did not try to establish the spade suit earlier so her spades are probably weak and therefore Jx is an unlikely spade holding.

Unfortunately, on this night declarer had ♠Jx and no ♦Q and made 4 for a tie for 6.5 out of 8 matchpoints.