Zenith Daylong Reward - April 1 - Board 12

Board 12
Our side vulnerable

I don't usually pick and choose the boards I discuss in Gargoyle Chronicles. I follow a pre-determined procedure to select a board before I play it, effectively randomizing my selection. I have two reasons for this approach:

(1) Michael Rosenberg once told me that all bridge deals have something interesting about them. If you can't see it on a particular deal, you just have to dig deeper. Selecting a deal at random to write about forces me to do that, and I sometimes discover things I wouldn't otherwise see.

(2) If I choose the deals I discuss, it is hard to avoid focusing on deals where I do something good. There is much to be learned from examining errors, and choosing boards at random ensures that a fair portion of the deals I discuss will include errors. 

Nonetheless, after 382 posts following this approach, I can't resist making an exception this week to discuss a deal I particularly enjoyed:

♠ K  ♥ K 8 6 3  ♦ A K 8 3  ♣ A 7 6 5

LHO passes, partner opens with one club, and RHO bids two spades, weak. I make a negative double, and partner bids three clubs. Since spades weren't raised, there is good chance partner has enough in spades to provide a stopper opposite my stiff king. I could bid three spades to make sure. But I doubt partner would bid three notrump with, say, jack third, which may suffice. Even three small may suffice, since RHO might duck the opening spade lead to preserve communication. So if I'm content to play three notrump, I should probably just bid it.

Am I content to play three notrump? What do I need to make a club slam?

♠ x x x  ♥ A x  ♦ Q x x  ♣ K Q x x x

is enough to make slam cold on normal breaks. Since that is a perfect minimum--actually a jack less than a minimum--I have enough to invite a slam by Culbertson's Rule.

But it's hard to construct an intelligent auction to investigate slam opposite a robot. Part of the problem is we have no way of stopping in four notrump. If I bid past three notrump, I'm pretty much committed to bidding a slam, since I have no desire to play five clubs at matchpoints. I don't have a slam drive, so, as a practical matter, I think I should settle for three notrump, though I'm not happy about it.

I bid three notrump, everyone passes, and LHO leads the spade five.

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

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

Six clubs looks pretty good, and some will bid it. But there is no sense in worrying about whether it makes or not. I'm not competing against those pairs any more. My only concern is outplaying the other declarers in three notrump.

I have eleven tricks off the top, assuming clubs come home. Squeeze chances don't look good. RHO stops spades. If West guards one of the red suits, I would have a double squeeze except for the fact that both red-suit threats are in my hand. One of the threats must be behind your opponent for a squeeze to operate.

Maybe I can execute a simple squeeze against East. If he has four cards in one of the red suits, can I squeeze him somehow? I don't see how. I need to duck a trick both to correct the count and to isolate the threat. And if I do that, they can attack my entries. Say, for example, I duck a diamond, playing East for four diamonds. Now the defense can simply play another diamond, and I don't have an entry in either threat suit. Another necessarily condition for a squeeze is that you must have an entry in one of the threat suits.

I might as well get a better picture of the hand by running some clubs. Once I'm dealing with specifics rather than generalities, something might occur to me. I win the spade lead in my hand as East plays the spade six. I play a club to dummy's king. East discards the spade three. So East is probably 6-4-3-0 or 6-3-4-0. I play a club to my ace, and East discards the diamond deuce. He wouldn't be discarding from his four-card red suit, so he must be 6-4-3-0.

Now that I know what he has, what happens if I run the clubs? East must come down to seven cards. He must hold three spades and four hearts, else I can duck out a trick in one of those suits. So he must discard all his diamonds. If he started with two diamond honors, I'm home. I can lead the diamond nine and drive the remaining honor, setting up my eight. In fact, maybe all I need is for him to have started with one diamond honor. If West holds two honors fourth in the four-card end position, I can endplay him.

I run all the clubs. East, perforce, holds three spades and four hearts, pitching both his remaining diamonds, the seven and the jack. I don't need all my hearts anymore, so I pitch the heart three. We are down to this position:

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

Now it is West who is squeezed. I cash the spade ace, pitching another heart. West must hold four diamonds, else I can duck out a diamond trick, so he has to come down to a doubleton heart. I cash the king and ace of hearts and float the nine of diamonds to endplay him. Making six.

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

This is not a particularly difficult deal to play. You just run the clubs and it plays itself. I like it because it combines so many different types of squeezes. It is a delayed ducked squeeze, with the threat of establishing a trick by ducking in all three threat suits. It is also a double squeeze, involving a triple squeeze on one opponent and a simple squeeze on the other. It has elements of a guard squeeze (where an opponent is squeezed out of an honor that protects his partner against a finesse), although in this case the "finesse" must be accomplished by endplaying that partner. Finally, the concluding simple squeeze is non-material. West is squeezed out of an exit card, not a guard. So what should we call it? A delayed-duck, quasi-guard, non-material double squeeze-endplay?

Plus 690 is worth 62%. Four pairs did reach six clubs. I was curious if anyone had found a sensible way to investigate, but everyone who bid six clubs got there by blasting. So I don't blame myself for not reaching slam. If you can score 62% simply by playing three notrump correctly, I don't think the risk-reward ratio is right for blasting to a speculative slam.