My phone buzzed at 1:20 am on Monday morning: it was Tom Townsend asking "How do you play 8765 in hand opposite AKJ2 in dummy for four tricks?". Not because
he wanted to be told the answer, which he'd already worked out (and checked with
SuitPlay), but because I was the only conscious person he could find who might be interested in such
esoterica.
First, suppose you lead to the ace, LHO follows small, and RHO plays the nine. Then you come back to hand and lead the six, LHO playing small again.
The queen and ten are still out, and there are three options: play the king, succeeding against RHO's Q9; play the jack, succeeding against 109; or play the three,
succeeding against singleton 9. Q9 or Q10 are two distributions, so are 9 or 10, 109 is one distribution. So apparently one shouldn't finesse the jack, but it's
roughly a toss-up between playing for the drop and playing the deep finesse. (It's relevant that RHO can't afford to play the Q from Q9 even if he knows you need
four tricks from the suit, because he doesn't know you've got the spots for the deep finesse.)
Second, on any other combination of cards (unless the queen has dropped or someone has discarded), the normal and sensible line is to finesse the jack on the second
round (detailed analysis, of the sort I've laid out here before, shows that some unlikely defensive strategies can make a different approach better, but they can't make
this approach worse).
However, if one chooses to play for the drop in the first case, there's a problem: LHO can deflect you from this line into a losing finesse by playing high on
the second round with 1043 (or 943). So the right thing to do is to take the deep finesse.
This does have some practical relevance: almost the exact combination came up on board 118 of the 120-board US teams trials final. At the time
Diamond was leading
Nickell by 5 IMPs, and the last two boards were to be flat, so this hand decided the match.
Weinstein
Greco
Levin
Hampson
- Three or more clubs
- Natural and forcing
- Minimum, three or four clubs
- RKCB for clubs
- One ace
- Queen of clubs?
- No
In the closed room, Levin-Weinstein were playing strong no trump, 5-card majors, and three-card minors. When Levin
opened one club, Weinstein chose to agree clubs: Levin showed a minimum then one ace and no queen of clubs, so Weinstein
gave up in five clubs. The trouble with that was that Levin could easily have had only three clubs (unless the system
is no longer as described
here). But he had four this
time, and five clubs made easily enough, losing the ace of spades and a club.
Moss
Meckstroth
Bathurst
Rodwell
- Precision: 11-15, no 5-card major
- Usually denies a 4-card major
- 11-13 balanced
- Intended as a slam invitation with 4-4 in the minors
- Pick a slam - Bathurst didn't know what 4 meant, but he was a bit better than minimum, so...
- Moss knows what he intended by this
In the open room, Bathurst and Moss had a murky auction to a poor 6NT, and Rodwell led the ten of hearts. Bathurst
won with the ace and led a spade to the four, king and ace.
Meckstroth returned the three of spades to the queen and five. Now declarer led the five of clubs to the two, ace and nine,
came back to the queen of hearts, and led the six of clubs. Rodwell played the four, and Bathurst was
looking at the situation as discussed above. (I'm sparing you my thoughts on the difference between AKJ2 and AKJ3.)
So, giving expert consideration to the arguments, he ran the six? No he did not, he played the jack, and with good
reason. Suppose you run the six and it holds. Now you have eleven tricks, and the best chance for a twelfth seems to
be the diamond finesse. But if you're going to take the diamond finesse anyway, putting the jack in may well not cost
even when clubs are 4-1: If South has three diamonds you've got four tricks in the suit once the finesse succeeds, if he
has four or more he'll get squeezed in the minors, and if he's got two North may get squeezed in the pointed suits.
North discarded a heart on the jack of clubs, so declarer cashed the king of hearts throwing his club, and the
king of clubs, crossed to the king of diamonds, and cashed the jack of spades throwing dummy's club. Evidently there had
been no squeeze, but the diamonds could still be 3-3. Declarer led a diamond towards dummy's AJ5, and North, holding a
spade winner and Q8 of diamonds, claimed two off. That was 13 IMPs and the match to
Nickell.
That's a reasonable line, but Bathurst could have made his contract by running the six of clubs, then squeezing North
in spades and diamonds - worth considering once South shows four clubs and probably some heart length. South, on the other
hand, could have prevented this by covering the six of clubs: it's unimaginable that he would do that since he can't know
that declarer needs to do more than make four clubs tricks (QJ9x of spades would be enough). East, on the third hand, could
have made by cashing a top club at trick two then playing the king of spades from dummy, saving an entry, before getting
everything else right. And South, on the fourth hand, could have beaten
to contract at trick one by leading a diamond to break up the squeeze. "It's poor that they're still playing single dummy
in such a big event", as my correspondent remarked.
