A K J 10 3 Q 5 A J 10 J 8 7
	7 2 A 8 7 4 K Q 7 A K 6 3

W	N	E	S
		P	1NT
P	2	P	2
P	4NT	P	6NT
P	P	P

We are a point short of 33 and there is duplication in diamonds. We have only eight sure tricks. Even counting four pretty likely tricks in spades gets us only to ten.

Well, here’s hoping the spades come in—and therein lies the point. It's tempting to cash the A first, “in case East has the singleton Queen.” But there are two problems with this theory. First of all, while playing Ace first does save a trick if East has the singleton Q, West would then have 98654 and would get a spade trick in the end, so you’d be down anyway, which won’t get you any matchpoints when you’ve stretched to slam. But even more to the point, playing the Ace first guards against the danger of East’s singleton Q only by exposing you to a much greater danger: West might have Qxxx. In that case, you need to finesse through West twice to pick up the suit, and cashing the A first will leave you able to finesse only once. Queen-fourth with West is much likelier than singleton Queen with East, so that’s the danger to guard against.

So I win the K and take a first-round spade finesse through West, and I am happy to see that East plays low. Now I’m up to 11 tricks, provided spades aren’t worse than 4-2 (and East isn’t diabolically holding off with the Q), with possibilities for an indirect finesse, a good club break, or perhaps a squeeze for 12. I might as well lose a trick to help my squeeze chances, so I play back to the A and then send a low club toward dummy’s Jack. No luck; East produces the Queen. But after I take East’s diamond return, cashing the K reveals a 3-3 club break. Now things look good. I cash my last club and finesse in spades again. East plays low a second time, so I have 12 tricks. 1440 turns out to be worth all the matchpoints. There are several 660s and a couple of -100s, so I think some players did play the A on the first round of spades.

The lesson here is that you cash the Ace to guard against Queen singleton when a suit is divided something like this:

K J 8 5

A 7 4 3

but when you have AKJ10x in one hand opposite xx(x), the first round finesse is better.

Here's the full deal:

	A K J 10 3 Q 5 A J 10 J 8 7
Q 8 6 5 10 6 3 2 6 4 10 5 4		9 4 K J 9 9 8 5 3 2 Q 9 2
	7 2 A 8 7 4 K Q 7 A K 6 3

P.S.: If East turned out to have the 10 instead of the 2, I would still have made the hand with a squeeze: Assuming a diamond return when East is in with the Q, I win, cash my last club, and run the spades. On the last spade East must either bare the K or discard his last club. Either way I get the last two tricks.

[Alexandria Regional, 6-30-2005]