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.

Free Weekly Instant Tournament - July 5 - Board 7

Board 7
Both sides vulnerable

♠ Q J 7  ♥ A 9 8 4  ♦ A Q 6  ♣ A J 7

I open with one club in first seat. Partner bids one heart and RHO doubles.

Before the days of supports redoubles, John and I would bid two notrump with this hand, an artificial bid showing four-card heart support and values to raise to the three level.

The theory was, if you had a balanced 18 or 19-count, you would redouble, since the opponents might be in trouble. Since this meant we had no need for a natural two notrump bid, we adopted the methods played by responder after a direct take-out double: Two notrump was a value raise to the three-level, and a jump raise was pre-emptive. "Pre-emptive" in this context meant a normal single raise with four trumps. So two hearts, by elimination, showed a single raise with only three trumps.

The advent of support redoubles changed all that. Redouble now showed three-card support, and a single raise showed four-card support. Two notrump was back to being natural, since you couldn't redouble any more.

The old-fashioned approach allowed you to penalize the opponents on occasion. And the pre-emptive raise to three sometimes made the opponents' life difficult. Of course, it sometimes made your life difficult when everyone passed and you found yourself too high. Most players today prefer the modern approach, where you can show your minimum with four-card support at the two level.

Back to deal at hand. I bid three hearts, and partner carries on to four. RHO leads the king of spades.

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

West	North	East	South
Robot	Phillip	Robot	Robot
	1 ♣	Pass	1 ♥
Double	3 ♥	Pass	4 ♥
(All pass)

If the opponent can take a spade ruff, I need the rest, which probably means finding both the diamond king and the club queen onside. If they can't take a spade ruff, I can afford to lose one minor-suit trick. So I need the diamond king onside, or, if that fails, I need to guess whether to finesse against the club queen or the diamond jack.

East plays the spade three; I play the four. West continues with the spade ace, and East plays the nine. The robots don't signal on opening leads, so nobody knows whether East is ruffing the next spade or not. At least that's true if I play the spade ten. Since East probably wouldn't play the spade nine if he had the ten, the ten is the card I'm known to hold. If East has a doubleton spade and I don't play the ten, I've given the show away.

I play the ten, and West continues with the spade deuce. East follows with the five. Now all I have to do is draw trump and win one of my three possible minor-suit finesses. Unfortunately, I can try only two of them.

No. I take that back. Actually, I don't need any of the finesses to work. I can draw trump and take a diamond finesse--either lead a diamond to the queen or float the ten. If East wins, he must give me a trick in one of the minors or lead a spade and give me a ruff-sluff.

Since West has at least eight cards in the minors, leaving East at most six, the right way to play trumps is to cash the ace and finesse against East. I cash the heart ace. West drops the queen. Well, I was right he had a singleton. I cash the king and jack of hearts. West discards the club nine and spade eight. This is the current position:

	NORTH Phillip ♠ -- ♥ 9 ♦ A Q 6 ♣ A J 7
	SOUTH Robot ♠ -- ♥ 10 7 ♦ 10 9 7 ♣ K 6

If my goal is to make my contract, it makes no difference whether I finesse the diamond queen or float the ten. But if I finesse the queen successfully, I can try the club finesse for an overtrick. So I lead the seven of diamonds to the queen. West takes the king and returns a diamond. I have the rest.

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

Making four is worth 54%.

Which method of handling sandwich seat take-out doubles is better? The old-fashioned approach, where redouble shows a two-notrump rebid, or the modern approach, where redouble shows a three-card raise?

As I said earlier, the old-fashioned approach will sometimes propel you to the three level when you don't want to be there. But that's more of an issue with spades than with hearts. Getting to the three-level with eight spades is overbidding total tricks. Getting to the three-level with eight hearts is OK if the opponents have a spade fit.

Of course, no one says you have to have only eight trumps. Sometimes responder has a five-card suit and you belong at the three-level anyway. And sometimes, even if you don't belong there, the opponents take you off the hook. Aggressive bidding puts a lot of pressure on the opponents and can gain even if it's a loser double-dummy. 

I suspect what's theoretically best is to play is the old-fashioned method when responder bids hearts and the modern method when he bids spades. I don't know anyone who plays that way, however. And I doubt I'll encounter anyone who does.

Free Weekly Instant Tournament - July 5 - Board 6

Board 6
Opponents vulnerable

♠ A K 8 4 3  ♥ Q 8 3 2  ♦ Q 6  ♣ A 9

I open with one spade in second seat. Partner bids two clubs; I rebid two hearts.

Partner bids three diamonds, which is explained as "five+ clubs and four+ diamonds." That's not what the bid should mean. With a diamond suit, responder should bid two notrump. Three diamonds should be a catch-all, denying the ability to bid anything else; that is, no three-card spade support, no four-card heart support, no sixth club, and only a partial diamond stopper. Something like:

♠ Q x  ♥ A K x  ♦ J x x  ♣ K J x x x

Playing the robots' methods, you have no bid with this hand. 

Why should three diamonds promise a partial diamond stopper? Why not no diamond stopper at all? Simply for reasons of frequency. With values to force to game, you are more likely to have a partial stopper than not. So if it promises a partial stopper, the hand for which you have no bid at all (e.g., the above hand without the diamond jack) is rare. If you hold such a hand, you simply have to tell whatever lie you think is least damaging.

You have an easier time of it if you play that a preference to opener's first suit shows specifically a doubleton. Then you can bid two spades with the above hand--whether or not you have the diamond jack. Playing this way, you must make a jump preference with three-card support.

This is a more common approach in Acol. I know of only a handful of American experts who play this way. Perhaps not even a complete handful. Eastern science fiction players don't like to jump in forcing auctions. But I don't see the problem in this case. Do you really need to start investigating slam at the two-level? To my mind, having to bid three spades to show genuine support isn't much of a hardship. Being able to bid two spades with a doubleton and an awkward hand can be quite valuable at times. And those are precisely the times where you want to keep the auction low: when you need room to explore the right strain.

In any event, whether playing my methods or the robots' methods, I have an obvious three notrump call over three diamonds. I bid three notrump, everyone passes, and LHO leads the king of hearts.

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

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

Leading an honor in declarer's second suit is a strange choice. But I suppose no suit is especially attractive in this auction.

East plays the five; I play the deuce. West continues with the ace of hearts, and East plays the six. If I can run clubs, I can take the rest: five clubs, three diamonds, two spades, and a heart. Making five. If I must concede a trick to club queen, I make four.

The likeliest defense is a diamond shift. Let's say West shifts to a low diamond, I play the eight from dummy, and East plays low. They've given me a diamond trick, but I can't untangle it. If I cash my diamond queen, I either have to spurn the club finesse to get back to dummy or take the finesse and risk never reaching dummy again if it loses.

Fortunately, the extra diamond trick is an illusion. I don't need it. If clubs run, I have the rest. If they don't I make four. It makes no difference whether I take three diamonds or four.

How should I handle the club suit? To run the suit, I need to find one opponent with queen doubleton or third. It may appear I can't afford to finesse East for the queen, since if it loses, a second diamond kills my dummy entry before I can unblock my club ace. But, in fact, I can unblock clubs by discarding the club ace on a diamond. So, unless clubs are five-one, I can safely finesse either way.

Since I have no reason to believe one opponent is more likely to have the club queen than the other, it's a guess which way to finesse if the queen is doubleton or third. What if someone has queen fourth? In that case, I want to finesse against West. No matter which hand has queen fourth, I make four by finessing against West and conceding a trick to the club queen.

In short, finessing West for the queen works more often than finessing against East. So that's my plan if I get a diamond shift.

I don't. West makes the friendly shift of a club, and East covers the jack with the queen. Making five.

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

Plus 460 is worth 67.9%.

The club shift is a poor choice. A diamond shift should be routine. As a general rule, when defending a misfit notrump contract, the best defense is to attack declarer's communication suit; i.e., the suit that is solid or close to solid and has honors split between the two hands.

The first time I encountered this principle was this deal from the 1979 Bermuda Bowl:

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

West	North	East	South
			1 ♦
Pass	1 ♠	Pass	1 NT
Pass	2 ♣	Pass	2 ♥
Pass	3 NT	(All pass)

West led the club four. Declarer was Mike Passell, and it appears he has nine easy tricks: Five diamonds, two clubs, and two spades.

Passell won in dummy with the king, crossed to the spade ace, led a diamond to the queen, and led the spade queen, setting up his ninth trick. Unfortunately, he had Benito Garozzo on his right. Garozzo took the spade king and returned a diamond. There was now no way for Passell to take his nine tricks.

Attacking the communication suit isn't nearly so deadly on my deal. While it does scramble my communications, that doesn't really matter. The main reason it works is that it avoids picking up the club queen. But the principle is the same: Attacking declarer's entries is a better plan than helping him set up his suits. The robots would do well to learn from Garozzo.

Free Weekly Instant Tournament - July 5 - Board 5

Board 5
Our side vulnerable

♠ A K 10 9 5  ♥ A K Q J 6 4  ♦ J  ♣ K

Partner passes, and RHO opens with one diamond. I'm willing to drive this hand to game. In fact, I don't need much for slam. The spade queen and an ace suffices. A small doubleton spade and an ace might be enough opposite adequate heart support.

I could start with two diamonds, a Michaels cue-bid. But I'm not a fan of Michaels with good hands. When it's our hand, I like to keep the auction low to find out as much as I can. If I bid one heart and partner raises to two hearts, at least I know he has something. But if I bid two diamonds and partner bids two hearts, I know very little. Partner could have a hand where he would have raised an overcall. Or he could have nothing. He might even have a doubleton heart. In short, I'm better placed if I give partner a chance to raise my suit voluntarily.

One downside to overcalling is that you might buy it there. In general, I don't worry about that too much, since opponents rarely sell out at the one-level. But there is more danger with this hand than with most--not because of the high cards but because of the spade suit.

If the auction goes one diamond--one heart--pass--pass and opener is looking at a stiff spade, he might well choose to sell out. He knows his partner can't have four spades unless he's broke. So either his partner has a terrible hand, or the opponents have a nine-card or better spade fit they haven't found yet. Either way, the opponents probably have a better spot than one heart. Why give them another chance to find it?

Maybe I'm giving the robots too much credit. But the prospect of the auction's ending in one heart does worry me a little. And, while bidding two diamonds may not be best, it isn't a terrible choice. Many players wouldn't even consider a different action. 

A bit reluctantly, I bid two diamonds. LHO passes, partner bids two heart, and RHO passes.

This was just the auction I was worried about. Now I wish I had bid one heart. If I bid one heart and partner passes, I can give up on slam. I would continue with three spades over RHO's presumed balance. But if partner raises to two hearts, I can be more aggressive. A reasonable approach would be to make a "game try" of two spades, bidding Blackwood if partner accepts and settling for four hearts if he doesn't.

How do I get partner's cooperation after this start? Two spades isn't even forcing. It's simply a forward-going bid with a sixth spade. Three clubs and three diamonds are available as "game tries," but they should show fragments. Partner will expect the king of that suit to be worth something. Furthermore, they are passable. With a misfit, he might decide to pass three of a minor. Since I want to play game even opposite a misfit, I can't take that risk.

Four of a minor is obviously a slam try, but it's not clear what it means. Does it suggest concern about the other minor? Might four diamonds, for example, be this hand:

♠ A K 10 9 5  ♥ A K Q J 6 4  ♦ --  ♣ x x  ?

Perhaps the right bid is three notrump. That should be some kind of slam try, and it avoids focusing attention on a specific minor. If partner has a minor-suit ace, he can cue bid it, and I can bid four hearts. Now, if partner has the spade queen, he should bid on. Two sure cover cards must be enough if I have slam interest after two hearts.

Of course making up bids in the middle of an auction is a bad idea even with a non-robot partner. However obvious it may seem to you that an undiscussed bid should mean what you want it to mean, it may not be obvious to partner. Slam is sufficiently remote that I'll just give up on it and bid a practical four hearts.

I bid four hearts, everyone passes, and RHO leads the diamond king.

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

West	North	East	South
Robot	Phillip	Robot	Robot
			Pass
1 ♦	2 ♦	Pass	2 ♥
Pass	4 ♥	(All pass)

We didn't miss a slam. I'm off two minor-suit tricks. I need to avoid two spade losers to make this. If trumps are two-two, that should be easy. If not, I need to find spades three-two or find a singleton honor somewhere.

Can I handle a four-one spade break if an honor doesn't fall? If the hand with four spades has three hearts, I can draw one trump, then play ace-king and a spade. They can't stop me from taking a spade ruff. That would be poor line, however, if spade are four-one but trumps two-two. Now I'm letting the defense score a ruff when I'm cold if I just cash two trumps.

If someone does hold a stiff spade, he is likelier to hold two hearts than one. So unless something strange happens to convince me otherwise, drawing two rounds of trumps looks best.

East follows to trick one with the five of diamonds. I play the deuce. West cashes the club ace and taps dummy with another club.

I draw trump. East follows to three rounds. When I cash the spade ace, West follows with the jack. I'm making this. In fact, if that was queen-jack doubleton of spades, I'm making five.

I cash the king of spades. West drops the queen, and I claim. Making five.

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

Plus 650 is worth 64%.

What would have happened had I overcalled with one heart? Some players tried that. The auction proceeds pass--pass to West, who balances with two clubs. Most players now bid three spades. Partner should correct this to four hearts, but of course, he doesn't. He passes three spades.

This is a robot quirk I've noticed before. They don't like correcting with bad hands. A robot once left me in my second suit with a doubleton, holding four cards in my first suit. Maybe, knowing that's a danger, two diamonds is the right bid after all.

So far as not balancing as opener with a stiff spade goes, my favorite example is this hand, which I held years ago in a Regional Open Pairs in Denver:

♠ x  ♥ A K Q J x x x x  ♦ x x  ♣ x x

Vul against not, I opened with one heart. LHO overcalled with two diamonds--pass--pass back to me. It felt funny to sell out below two hearts with eight tricks in my own hand, but the stiff spade worried me. At this vulnerability, I didn't want to get pushed too high, and I suspected the opponents could do a lot of pushing. So I passed. I still remember the look on partner's face when he led a heart and declarer ruffed it.

Selling out was right, but not in the way I expected. The opponents were not cold for four spades. They were cold for five clubs. 

Free Weekly Instant Tournament - July 5 - Board 4

Board 4
Both sides vulnerable

♠ 2  ♥ 8 6 2  ♦ A K Q J 3 2  ♣ K J 6

Three passes to me. I open with one diamond, and partner bids one spade. This is a good hand, but not quite good enough to rebid three diamonds. Three diamonds shows seven and a half to eight tricks. This hand has about seven.

I bid two diamonds, and partner bids two hearts. The tooltip says this bid is forcing to three notrump. Obviously it can't be, since partner is a passed hand. But I'm just as happy partner thinks it is. That means I can bid two notrump and give him a chance to rebid a five-card heart suit. If two notrump were not forcing, I would have to bid three notrump, possibly leaving partner to guess whether to pass or bid four hearts.

I bid two notrump, and partner bids three diamonds. Partner is 5-4-3-1 or 5-4-4-0 and is presumably concerned about three notrump because of his club shortness. While I don't have a sure double club stopper, the fact that my diamonds are solid may mean that a single stopper is good enough. So I bid three notrump. Everyone passes, and LHO leads the three of clubs.

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

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

I don't care for partner's three diamond bid. With a stiff ace of clubs, he should raise two notrump to three. Three diamonds will worry me unnecessarily about the club situation and will often steer us away from three notrump when we belong there.

I have nine top tricks. I have two ways to try for a tenth in hearts: (A) I can lead a heart to the queen, or (B) I can lead a heart to the nine, in case jack-ten or onside, then lead to the queen later if that fails. 

There are two problems with (B). For one, it can't work double dummy (unless spades are blocked), since the opponents will be able to cash three spades when East wins the heart trick. I'm not too worried about that, though, because the opponents are unlikely to find the winning defense. East is apt to return a club after winning the heart. A more serious problem is that, if East wins with the ten or jack and returns a club, a heart to queen now risks my contract. The only way to finesse against the heart king safely is to do so right away.

I'm inclined to try (B) anyway. The robots are bad at deceptive discarding. After I run six diamond tricks, I'll probably know whether the heart king is onside or not. So I might as well give myself the extra vig of finding jack-ten onside.

I win the club ace in dummy, as East plays the four. The deuce is still out. If East wanted to play a low a club, he might have played the deuce rather than the four, so West probably has the deuce. That means clubs are probably five-four. Although the robots do card strangely, so that's hardly a sure thing.

I lead the five of diamonds from dummy. East plays the eight. Which card should I win with? The ace is clearly wrong. It telegraphs that I have no finesses to take. The king is marginally better, since I could be missing the ace. But the queen is better yet. It leaves open the possibility that I'm missing the ace or the king.

I win with the queen, and West follows with the six. I lead the six of hearts, and West inserts the ten. I guess I didn't have to worry about whether to finesse the nine or not. I cover with the queen, and East follows with the five. We've reached this position:

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

I have ten tricks. Can I find an eleventh? I could play ace and a heart, hoping the ace drops West's king or that the suit is three-three. But that again risks the defense's finding a spade shift and holding me to nine tricks. Will they find it?

If hearts are three-three and West wins the trick, he knows there is no future in the club suit. So he may find a spade shift. What if East wins the heart with jack fourth? Then West gets a chance to pitch a club. That should suffice to get a spade switch. Setting up the heart isn't likely to work. Is there a safer way to try for another trick?

The count isn't right for a squeeze. I just have to hope the opponents mis-discard. I'll run all my winners except for the club king. Then, if I judge it's safe to do so, I'll exit with a heart or a spade and hope to score two club tricks. If that's my plan, I must cash the heart ace before running diamonds. I want to be in my hand when I make the critical decision in the end game.

It doesn't hurt to cash one more diamond, though. I lead the nine of diamonds from dummy. East discards the seven of spades, and I play low from my hand. The six of spades is still out. The robots tend to discard count. So if East has the six, he probably has four spades. If not, he probably has five.

I cash the ace of hearts. East plays the four; West, the king, the card he's known to hold. I can't be sure whether West began with king-ten or king-jack-ten of hearts.

I play a diamond to my hand, as East discards the seven of clubs.

On the fourth diamond, West discards the five of clubs. I still haven't seen the deuce. But I'm still assuming West has it for the time being. I pitch a spade from dummy, and East discards the spade nine. Still no six. If East had four spades, he might have completed his echo or, more likely, pitched another club or a heart. I'm inclined to think he has five spades and is 5-3-1-4, making West 2-3-3-5.

On the penultimate diamond, West discards the spade eight; East, the club eight. If West started with ace doubleton of spades, I have him now. He's down to

♠ A  ♥ J  ♦ --  ♣ ? x x

I can cash the last diamond and toss him in to get a club lead into my king-jack. No, wait. He can't have ace doubleton of spades. I already concluded that if spades are 2-5, West must have the six. So the only way West can have a doubleton spade is if he has eight-six.

I cash the last diamond. West finally pitches the elusive club deuce, confirming he started with five clubs. East pitches the heart three.

I'm still not sure of the spade count. But I'm pretty sure West has the club queen. East wouldn't have stiffed it if he had it. So there are two possible layouts:

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

or

	NORTH Robot ♠ J 10 ♥ 9 7 ♦ -- ♣ --
WEST Robot ♠ ? ? ♥ -- ♦ -- ♣ Q x		EAST Robot ♠ ? 6 ♥ J  ♦ -- ♣ x
	SOUTH Phillip ♠ 2 ♥ 8 ♦ -- ♣ K J

Either way it can't hurt to exit with a spade. I can't lose the club king, and if East can't or doesn't gain the lead in spades, West will have to lead into my club tenace at the end. A heart exit would be a mistake, however. In the first layout, East could pitch his last club on the heart jack and take the last three tricks.

I exit with the spade deuce. West plays the queen. East overtakes with the king, taking his partner off the endplay. Making four.

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

Plus 630 is worth 71%. I'm not sure what East would have done if I had played ace and a heart after the heart finesse won. Would he have returned his partner's suit or would he have found the spade shift?

Since one measly overtrick was worth 71%, I was right not to find out. I wish I could give you a reason one overtrick was such a good result, but I can't. Everyone was in three notrump. And, for reasons I can't fathom, six declarers simply cashed out and never bothered with a heart finesse.