Happily, the Puzzle Corner found me in my little village of Mirabel aux
Baronnies in southern France where I am spending my sabbatical. Maybe I am
too late, but I decided to send you my proposed solution to Larry Kells'
bridge problem, J/F 1. If it turns out that I am right, I may dust off my
rusty card play and my antiquated bidding techniques from forty years ago
and play a little duplicate in some small French villages. If not, I will
assuage my disappointment with some more good wine and cheese, and see if I
can find a truffle or two. Here goes:
1. To get the greatest excess, consider that the shortest trump suit for
N-S and E-W is obviously 7, so the greatest theoretical excess for the two
sides is if each takes all 13 tricks as declarer, in which case the maximum
joint excess is 6 + 6 = 12. The following deal achieves this maximum and
proves that 12 is the answer.
S x,x,x,x,x,x
H x,x,x,x,x,x,x
D -----
C -----
S ------- S A,K,Q,J,10,9,8
H ------- H A,K,Q,J,10,9
D x,x,x,x,x,x,x D --------
C x,x,x,x,x,x C --------
S --------
H --------
D A,K,Q,J,10,9
C A,K,Q,J,10,9,8
South declares 7 Clubs, East declares 7 Spades. South declaring takes
opening D or C lead, draws trumps and is high for 13 tricks. East
declaring, trumps opening D or C lead, draws trump and is again high for 13
tricks. QED.
2. The greatest deficit is a bit tougher. Trump suits of 13, 12, 11, 10,
9, 8, and 7 cards must be considered. It is easy to see that declarer will
lose the most tricks possible (the goal) with his partnership's trumps
split as evenly as possible so they can be reduced by defender's trumps as
efficiently as possible. With 13 trumps, declarer must win at least 7
trump tricks for a maximum deficit of 6; with 12 trumps split 6-6, declarer
must win at least 5 after the defense takes the first trump trick, for a
maximum deficit of 7; similarly with 11, 10, 9, 8, or 7 trumps in the
partnership, declarer must win at least 4, 2, 1, 0, and 0 trump tricks
respectively after the defense strips as many trumps as possible at the
beginning of the play. That is, the maximum possible deficit is 8 for
holdings of 10, 9 or 8 divided trumps in the partnership, 7 for holdings of
12, 11, or 7 trumps, and 6 for the holding of 13 trumps. But there is a
bit of a zero sum game when we look at the potential tricks outside of the
trump suit for the defense for different allocations of the non-trump
cards. That is, non-trump winners for, say, E-W as defenders constrain the
number of non-trump winners N-S will have defending the same layout. The
best I can do on French time is declaring each way with 8 trumps split 4-4,
and managing to lose 9 tricks with best play, for a negative deficit of 4 +
4 = 8. If anyone does better, I say let them eat cake. Here is a sample
deal that works:
S x,x,x,x
H A,K,Q,J
D A,K,Q,J,10
C ---------
S A,K,Q,J S x
H x,x,x,x H x,x,x,x
D -------- D x,x,x,x,x,x,x
C A,K,Q,J,10 C x
S x,x,x,x
H x
D x
C x,x,x,x,x,x,x
N-S must declare spades, their longest suit with 8. Whoever leads for E-W,
leads a spade, the rest of the spades are drawn and West also takes 5 clubs
for 9 tricks, leaving N-S with just the final 4 heart tricks, or 4 below
their trump total of 8.
E-W similarly must declare hearts, with N-S taking the first 4 heart tricks
and N then taking 5 diamonds, leaving E-W with only the last 4 spade
tricks, again 4 below par. QED
Steve Kanter '68