Free Weekly Instant Tournament - August 30 - Board 2

 

Board 2
Our side vulnerable

♠ Q 9 7 6 5  ♥ K Q 6  ♦ Q 6 4  ♣ A 5

I open with one spade in second seat. Partner bids three spades, a limit raise.

I have only 14 total points, and even that is an over-valuation. I have a bad spade suit and two unsupported queens. By point count, then, I don't have an acceptance.

What does loser count say? I do have six losers, which is usually an acceptance opposite a limit raise. But loser count assumes partner can have a useful doubleton. If he can't, you should add a loser. In this case, no doubleton in partner's hand will cover a loser in mine. So I should count this hand as seven losers, which is not an acceptance.

I pass. LHO leads the king of diamonds.

	NORTH Robot ♠ A J 8 2 ♥ A 10 4 ♦ 9 5 ♣ Q 10 7 2
	SOUTH Phillip ♠ Q 9 7 6 5 ♥ K Q 6 ♦ Q 6 4 ♣ A 5

West	North	East	South
Robot	Robot	Robot	Phillip
		Pass	1 ♠
Pass	3 ♠	(All pass)

Did I make the right decision? Partner has 12 support points, a maximum for his limit raise, and game is only a little better than 33%. That's certainly not a game you want to be in at matchpoints. If you take away partner's jack of spades or give him king-jack instead of ace-jack, he is in the middle of his range and game has virtually no play. So I seem to have made the right decision. Still, a handful will accept. So if game does make, this board will be below average.

On West's king of diamonds lead, East plays the three and I play the four. West continues with the ace of diamonds, on which East plays the jack.

A third diamond play could be problematic. Should I play the diamond queen to prevent that? I don't need to score the diamond queen. I can always ruff my low diamond later. Or can I?

If I play a spade to the jack, cash the ace, and LHO still has the trump king, I'd like to cash winners and endplay him with the third round of trumps. I can't do that if I have to use the fourth trump to ruff a diamond. So dropping the diamond queen may be a luxury I can't afford. West didn't overcall at favorable after all. So it's unlikely he has ace-king-ten sixth of diamonds anyway.

I follow with the diamond six, and West shifts to the nine of hearts. I want to be in my hand for a spade finesse, so I play low from dummy. East plays the five. West knows East would play the king if he had it, so I win with the king, the card I'm known to hold.

I lead the five of spades--four--jack--three. Now the spade ace--ten--six--king. Oh, well. Four spades makes. I'm getting below average unless I can manage another overtrick somehow. 

This is the current position:

	NORTH Robot ♠ 8 2 ♥ A 10  ♦ -- ♣ Q 10 7 2
	SOUTH Phillip ♠ Q 9 7 ♥ Q 6 ♦ Q  ♣ A 5

The only thing I can think of is to lead the club queen. If East thinks I have,

♠ Q x x x x  ♥ K x x  ♦ Q x x  ♣ A J

then covering would allow me to take three club tricks and pitch a heart. Of course that makes no sense. If I had that hand, I'd lead a low club to the jack, then take a ruffing finesse against the king to pitch my heart.

How about this hand:

♠ Q x x x x  ♥ K x x  ♦ x x  ♣ A J 9

Covering from king fourth indeed costs a trick. In fact, ace-jack-eight of clubs makes covering from king fourth or king-nine fourth dangerous. It will cost a trick if I guess the nine. Unfortunately, I can't have that hand because it's not an opening bid. Now I wish I had dropped the diamond queen at trick two. If I had, East could have a legitimate problem and might easily go wrong.

I can't think of a good reason for East to duck the club queen. But maybe he can think of one. I can't see anyone's having a stiff king of clubs, so it's worth a shot. I lead the club queen. East covers with the king. I take the ace, and West follows with the three.

Do I have any other chances? Maybe I can pseudo-squeeze someone at trick twelve. That wouldn't work against a human. A human would reason I could just ruff a red-suit loser if I had one, so the club jack is the only card worth keeping. But robots don't draw inferences. It might work against them.

First I must cash the heart ace. No one is going to set up the club ten if dummy has an entry. So I play a heart to the ace, then run the spades and cash my red-suit winners. No one pitches the club jack. Making four.

	NORTH Robot ♠ A J 8 2 ♥ A 10 4 ♦ 9 5 ♣ Q 10 7 2
WEST Robot ♠ K 4 ♥ 9 7 3 2 ♦ A K 8 7 2 ♣ J 3		EAST Robot ♠ 10 3 ♥ J 8 5 ♦ J 10 3 ♣ K 9 8 6 4
	SOUTH Phillip ♠ Q 9 7 6 5 ♥ K Q 6 ♦ Q 6 4 ♣ A 5

Dropping the diamond queen at trick two would not have helped. With both the eight and nine of clubs, East has no reason not to cover.

A rather larger handful accepted than I anticipated: nine out of fourteen players are in game. So plus 170 is worth only 25%. And it's worth that much only because a couple of players managed only nine tricks. I can't imagine why so many accepted. Did they think three spades was forcing? I'll have to take consolation in knowing the odds were in my favor: Roughly two thirds of the time, I would be scoring 89%.

It's worth taking a look at how loser count works on this deal. I have six losers. A limit raise has eight. For the supporting hand, it usually makes more sense to think of cover cards than of losers. Eight losers is four cover cards. (You subtract the number of losers from twelve.) These four cover cards typically comprise three high cards and a doubleton. And that's exactly what partner has: two aces, a queen, and a doubleton diamond.

Loser count assumes one of the cover cards will be duplicated, so partner's four cover cards should, on balance, cover three of my six losers, producing game. In this case, however, two of partner's cover cards are duplicated: The club queen is duplicated by my doubleton, and the doubleton diamond is duplicated by my queen. That means partner covers only two of my losers and we are a trick short.

But I knew any doubleton in partner's hand would be duplication. That's why I treated my hand as seven losers rather than six. And seven losers opposite eight does not produce a game. Ruben's suggestion of adding a loser when you have two more queens than aces would work as well. But I like my method better.

Many players dislike loser count, probably because they've had bad experiences applying it mechanically. Personally, I find loser count a useful tool. But I believe thinking in terms of cover cards for dummy makes more sense than adding the losers of both hands, as the method is generally taught. To say that six plus eight, or fourteen, combined losers produces ten tricks doesn't make a lot of sense to me. To say four cover cards will probably cover three of my six losers makes more sense. In addition, thinking in terms of cover cards helps you visualize when blindly applying loser count might yield the wrong answer, as on this deal.