Free Weekly Instant Tournament - May 30 - Board 8

Board 8
Neither side vulnerable

♠ 10 7 3  ♥ A K 9 6  ♦ A K Q 6 4  ♣ K

Partner opens with one club in second seat. I bid one diamond, and partner bids one notrump. 19 HCP opposite 12-14 is worth an invitation to slam. It's possible six diamonds will make when six notrump goes down. But it would be hard to diagnose that. So there isn't much point in bidding two hearts to initiate some complicated auction no one will understand anyway. I might as well just make a quantitative invitation in notrump.

I bid four notrump, and partner raises to six. RHO leads the four of spades.

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

West	North	East	South
Robot	Phillip	Robot	Robot
		Pass	1 ♣
Pass	1 ♦	Pass	1 NT
Pass	4 NT	Pass	6 NT
(All pass)

I play low from dummy, East plays the king, and I win with the ace. The lead of the four doesn't tell me much. It could be from four-deuce doubleton, or it could be low from three or four cards. Or fourth best from five. At least I know it's not fourth best from six, since there is only one lower card. So East must have at least two spades.

I have twelve top tricks with lots of possibilities for a thirteenth. Hearts could come home, or I could execute one of a variety of squeezes. 

Let's say I come down to this position:

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

(A) If either East or West holds four hearts and the queen of clubs, he's squeezed down to a stiff club when I cash the spade queen. The hand with long hearts is roughly even money to hold the club queen, so this line offers about a 50% chance (after excluding three-three hearts).

(B) If West holds four hearts and the jack of spades, he is squeezed in the majors. I must be careful to cash the spade queen first to avoid guessing what to pitch from dummy. When I cash the club ace next, I can pitch the spade ten if I haven't seen the jack.

This line offers roughly a 25% chance, since it works only against one opponent. But, since the same sequence of plays works for both A and B, I can combine my chances. The combination A/B offers roughly a 62.5% chance.

(C) If West holds four hearts and East holds the club queen, I can execute a double squeeze with spades as the double threat. I cash the club ace, pitching a spade from dummy, then cash hearts, reaching this position:

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

West must hold two hearts, so he has already been squeezed down to a stiff spade. Now, when I cash the king of hearts, East is squeezed in the black suits. 

West is 50% to hold the heart length, and East is 50% to hold the club queen. So C works about 25% of the time. It doesn't work if the heart length and club queen are reversed, because then I have no threat behind West. When I cash the heart king, South gets squeezed before West does.

Unfortunately, I can't combine C with either A or B. A or B requires my cashing the spade queen. C requires my retaining it as an entry to the double threat.

Are there any other possibilities? I can think of one more:

(D) If East has both the club queen and the spade jack, I can cash the club ace, pitching a heart from dummy, then cash three hearts. On the last heart, I'm down to

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

East must either pitch his club queen or stiff his spade jack.

This line, however, requires me to give up on three-three hearts. Furthermore, it's never necessary. If East has the club queen, then I can execute A if East has heart length and C if West has heart length. So we can forget about D.

Which squeeze should I play for? The A/B combination is clearly the right line a priori. But I don't have to decide just yet. Perhaps I can gather some clues as I run diamonds.

I cash the jack and ten of diamonds. West plays three, eight; East plays five, deuce. I play a diamond to the queen. West discards the deuce of clubs. The robots tend to discard count when not following suit, so that's probably from a five-card suit. Assuming hearts are not three-three (since I don't need a squeeze if they are), that makes West either 2-4-2-5 or 4-2-2-5.

I cash the king of diamonds, pitching a club. East pitches the deuce of spades; West, the deuce of hearts. While the robots typically discard count cards, they sometimes pitch low from doubletons, so it's still possible West is 4-2-2-5. It's also possible West has five hearts and his first discard was from three small clubs. That makes him 3-5-2-3. 

Let's take stock. There are two layouts to consider:

(1) West is 4-2-2-5. In that case, I must hope East has the club queen and play for A. If West has the club queen, no squeeze works.

(2) West is 3-5-2-3. West has pitched a club, so if he has the club queen, it's dropping. If East has it, I must play for C.

I cash dummy's last diamond, discarding a heart. East pitches the five of clubs. Well. That changes things. If East started with queen third of clubs, the queen is now dropping, so I needn't worry about case (1) any more. The only case that matters is case (2), so I must go for the double squeeze.

I cash the club king, reaching the position above. Now I lead a heart to my queen, cash the ace of clubs, pitching a spade, and lead a heart to dummy. Both opponents follow to the second round of hearts, so the nine of hearts is good. Making seven.

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

Hearts were three-three all along, so no squeeze was necessary.

Only three other players were in slam, and only one of those managed to cash his thirteen top tricks to make seven. So plus 1020 is worth 96%.

I'm surprised so many missed slam. I would have thought the field wouldn't even bother to invite but would simply blast six notrump. Weak players tend to overbid hands with lots of high-card points out of irrational exuberance.