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.

Free Weekly Instant Tournament - August 30 - Board 1

 

Board 1
Neither side vulnerable

♠ 8 6  ♥ 7 3  ♦ K Q J 10 8 2  ♣ A K 4

Partner opens with one heart. RHO passes. I bid two diamonds, game-forcing, and partner bids two hearts. This bid does not promise a sixth heart in the robots' methods.

I would like to bid three clubs to enable partner to bid three notrump with a spade stopper and no club stopper. But the robots take these bids seriously. If partner raises clubs, there may be no way to avoid playing this hand in clubs. I'm sure he will interpret a four-diamond bid by me as a cue-bid. I'm not sure what he will make of four hearts, but I don't want to find out. To avoid partner's tunnel vision, I'm stuck with bidding three diamonds. 

I bid three diamonds, and partner bids three notrump. I pass, and RHO leads the spade deuce.

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

West	North	East	South
Robot	Phillip	Robot	Robot
			1 ♥
Pass	2 ♦	Pass	2 ♥
Pass	3 ♦	Pass	3 NT
(All pass)

Partner had a tough decision over three diamonds. He could either show his sixth heart, possibly giving me a headache if I have a singleton and only one black suit stopper, or he could show both his stoppers with three notrump. This would have been an easier auction if two hearts had promised six. 

At one time, I didn't care for that treatment. One thing I dislike about Eastern Science Fiction is that it can be difficult to show extra values. So I liked using two hearts as a catch-all, allowing two spades or three clubs to show extras. But I'm beginning to think I was wrong about that. I keep seeing deals where I wished opener's rebid of his suit showed six.

Since West led the spade deuce, he should have three or four spades. So I'm down one or two after I drive the diamond ace. Fortunately, others will have the same problem, so this should be a normal result.

East plays the spade jack, and I take the king. I lead the seven of diamonds--four--king--nine. It's weird they would duck this trick, since I have two side dummy entries. Whoever ducked must know hearts aren't running. But table presence is useless against robots, so I have no idea who ducked. 

Incidentally, Ron Andersen once told me the best way to read which hand ducked. It's not tempo, which can be unreliable. The player who ducked is the player who, after seeing his partner's card, glanced at dummy's spots.

I continue with the queen of diamonds--six--three--ace. West had the diamond ace, so, as we decided earlier, he must hold a heart honor. Although I can't imagine it's going to help to know that. I'm just practicing.

West shifts to the nine of spades, and East takes the ace. I don't think East would have falsecarded at trick one, so he can't have the ten. West must have made one of the robots' weird leads of the nine from ten-nine. I don't know why they do that. I know who has the ten. So why not tell partner?

East continues with the queen of spades, and West unblocks his ten. East continues with the spade three. The three? Not the seven? So West has 10972 and they've blocked the suit? Yes. West takes his spade seven and shifts to a club. Making three.

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

The winning defense should not have been difficult to find. If West shifts to the spade ten. East can duck and the rest is easy. In fact, with ten-nine-seven West should lead the ten at trick one.

Four hearts makes. So in a real field, this would be a poor result despite the bad defense. In this field, plus 400 is worth 57%.

Eight players did reach four hearts, but only because they were unaware that opener's two-heart bid did not promise six and forgot to read the tooltip. They raised two hearts to four. Fortunately, five of them went down on creative lines of play. Otherwise this would have been a below-average result.

Four hearts is the wrong bid, by the way, even if partner's rebid promises six hearts. A jump to four hearts should show good heart support. This isn't some "fast arrival" situation; we are still looking for the right strain.

Free Weekly Instant Tournament - July 5 - Board 8

Board 8
Neither side vulnerable

♠ A 4  ♥ A 10 6  ♦ A J 7 4 2  ♣ Q 10 7

LHO opens with one spade. There are two passes to me. 

I double, and partner bids one notrump. This shows five to ten HCP according to the tooltip. Eight to eleven makes more sense. You don't want to have to jump to two notrump with eleven opposite a balancing double. But I have no say in the robots' methods.

Given the bid shows five to ten, we could have 25 HCP combined, in which case we belong in game. But it's unlikely partner has a complete maximum, so it isn't worth getting to the two-level hoping he does. Besides, we've wrong-sided it. Opener might be hard-pressed to avoid giving away a trick on the opening lead. Responder will have an easier time.

I pass, and RHO leads the eight of spades.

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

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

It looks normal to duck this trick to set up my spade queen. But if diamonds run, I can always lead toward the queen later. If they don't run, I'd like to concede a diamond trick before the defense establishes the heart suit.

I play the spade ace--six--three. I lead the deuce of diamonds from dummy--ten--king--three. Now the eight of diamonds. West plays the five.

Let's assume for the time being that East has a doubleton diamond. If he has three, it makes no difference what I do. We can consider his holding a singleton later. What is my correct play if East has two diamonds? A priori, the odds are three to two in favor of finessing. But there are special considerations here that may change that.

Some, noting East's play of the ten, might reason this way: East has Q10 or 109. By restricted choice, Q10 is twice as likely, so it's two to one to go up. That's faulty reasoning, however. This is not a restricted choice situation. If we know nothing about the opponent's high cards, the right play is to finesse. (More on this in the post mortem.)

But we do know something about the opponent's high cards, since East opened the bidding. On average, the 16 missing high card points will be distributed 13-3. We already know four of East's high card points. Of the remaining 12, East has, on balance, nine to West's three. So East is roughly three to one to have the diamond queen. Roughly because of granularity. High card points come in clumps; they aren't distributed one at a time. Still, three-to-one is a fair approximation. So, even with the three-two split, East is a heavy favorite to hold the diamond queen. That means the percentage play is the ace.

What about the case where East has a singleton? Then my best play is to float the eight. That play is out of the question, however. It loses to any doubleton in the East hand. So my choice is between the jack and the ace. My choice is irrelevant in the diamond suit itself. I lose one trick either way. But the ace does make my life more difficult, since I have no convenient way to get to my hand to play another diamond. This makes playing the jack somewhat more attractive. But the queen is such a heavy favorite to be on my right, that I don't think it tilts the odds.

I rise with the diamond ace; East plays the nine. I continue with a diamond, and East wins the queen. It made no difference what I did in the diamond suit. West discards the five of hearts. That's probably from a five-card suit, so West is probably 2-5-2-4.

I expect a heart shift, but East cashes the spade king. West follows with the seven. Now East shifts to the three of hearts. West plays the jack. Surely West's opening lead would have been a heart if he held king-queen-jack. So East must have led low from an honor. If I'm right that West has five hearts, it was a doubleton honor. That gives West king-jack or, less likely, queen-jack fifth of hearts. Either way, East must have the club king.

It must be right to duck this trick to tighten up the position for a squeeze against East. Suppose I duck and West continues hearts. I can win in dummy and cash diamonds, coming down to this position:

	NORTH Phillip ♠ -- ♥ 10 ♦ -- ♣ Q 10 7
	SOUTH Robot ♠ Q 10 ♥ -- ♦ -- ♣ A 4

East can't afford to come down to a singleton in either black suit. And if he comes down to king doubleton of clubs and jack doubleton of spades, I can play ace and a club and force him to lead into my spade tenace.

I duck the heart jack, arriving at this position:

	NORTH Phillip ♠ -- ♥ A 10 ♦ J 7 ♣ Q 10 7
	SOUTH Robot ♠ Q 10 ♥ 8 4 ♦ -- ♣ A 4 2

West valiantly tries to get his partner off the endplay by switching to the eight of clubs. I play the ten from dummy, and East plays the jack. Did this club switch save the day? No. If I duck, the position just converts to a simple squeeze rather than a squeeze endplay. If East returns a heart, I win and cash a diamond, reaching this position:

	NORTH Phillip ♠ -- ♥ 10 ♦ 7 ♣ Q 7
	SOUTH Robot ♠ Q 10 ♥ -- ♦ -- ♣ A 4

Now the last diamond squeezes East. He must stiff one black honor or the other.

I duck the club jack. East sees the squeeze coming, so he exits with the club king to save time. Making three.

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

We would have reached this game had I been dealer. I would open with one notrump, partner would invite, and, with three aces, two tens, and a five-card suit, I would accept despite my 15-count. Whether I would make it or not without the blueprint the opponents' auction gave me is open to question.

But, as is often the case in matchpoints, you don't need to bid the game. You just need to make it. Plus 150 is worth 86%.

Back to the restricted choice problem. Why is it right to finesse if we know nothing about the opponents' high cards?

The argument for refusing the finesse is as follows: East can have Q10 or 109. If he has Q10, his choice is restricted. If he has 109, he might have played either card. So 109 is half as likely as Q10.

I know from experience that some would reason this way. And the argument is right so far as it goes: 109 indeed counts as only half a case. But the fact is, Q10 isn't a full case either. If East has 109, then West has Q53, and his play is restricted. He would always play the five and the three. But if East has Q10, then West has 953 and could choose to play any of three pairs: 95, 93, or 53. So, by restricted choice, East's Q10 counts as only a third of a case.

This means the odds in favor of the finesse are 1/2 to 1/3, or 3 to 2, the same as the a priori odds. East's play of the ten did nothing to change the odds.

This is all very complicated, but it needn't be. The way to avoid these complications is to recognize this isn't a restricted choice situation in the first place and just stick with the a priori odds. After all, if East had played the five of diamonds on the first diamond trick, no one would even think about restricted choice. Why should the ten be any different?

Some seem to think it should be different because the ten is an honor. But that's not the criterion. Here's the rule:  Apply restricted choice only when the cards involved are critical. What makes a card critical? A card is critical if the other defender would never play that card voluntarily.

To illustrate, suppose dummy's diamonds were headed by A10 rather than AJ. You cash the king, and East drops the queen. When you play a second diamond, West would never play the jack voluntarily, so the jack is a critical card and restricted choice applies. East is twice as likely to have a stiff queen rather than queen-jack doubleton.

But in the case under discussion, when you play a second diamond, there is no reason West can't play the nine if he has it. The nine is not a critical card, so restricted choice does not apply. Or, more accurately, it does apply, but it applies to both opponents, so it cancels out. That means you can just ignore it and stick with the a priori odds.

Be sure to play in this week's Free Weekly Instant Tournament on BBO, so we can start comparing in my next blog post.