ecn-ec:finn@pur-ee.UUCP (08/04/83)
Here is the problem presented again.
North
S:AJ65
H:KT2
D:
C:AKQJ65
West East
S: S:KQT987
H:J987654 H:
D:987654 D:KQ2
C: C:T987
South
S:432
South: H:AQ3
6NT D:AJT3
C:432
If West leads a heart, then South can make his 6NT with no problem (well...
minor problems), the way that the cards are distributed. North takes the
first trick, and then leads out 6 club tricks. Leading hearts back to South
(2) leaves 4 tricks still to go. North has 4 spades, South has 2 diamonds and
2 spades, West is irrellevant, and play now depends on what East has already
thrown. East could have any of the following: 4S, 3S 1D, 2S 2D, or 1S 3D.
Case by case: East has 4S:
You have seen the king and queen and 2 of diamonds fall on the club
and heart tricks, so you play the A & J of D, lead a spade back to
North and lose only the 13th trick.
East has 3S & 1D:
The same as above. You have seen the Q & 2 of D fall. Your Ace
will take the remaining king.
East has 2S & 2D:
The king/queen is out in both suits still, so you lead a spade from
South, with north going low, to force East to take the trick. If
East leads back his remaining spade, North takes it and runs out
spades. Otherwise, South takes the Diamond trick and leads his spade
back to North, so North can do so anyways.
East has 1S & 3D:
The queen of spades has fallen, leaving the king vulnerable. South
leads a spade to North, who promptly takes the last 4 tricks, for a
total of all 13 tricks.
Any way you look at it with the current distribution, if West leads a heart,
South makes his bid. Lots of fun. This one wasn't *too* difficult. I thought
that more people would see this line of play. Oh, well. Have fun all.
The Arch-Druid of the Midwest Bridge...
David Hesselberth, Purdue University Computer Center
{harpo, decvax, ucbvax} !pur-ee!finn (or) !pur-ee:pucc-h:adr