The sun., 1915-05-30
Bridge problem No. 327. by H. c.
Mankowski, had near solution that
trapped a number OF those who tried to
DO the problem. As usual. there was a
defence TO this variation that they
overlooked. Here is the distribution:
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Clubs are trumps and Il Is in the lead.
Y and Y want six tricks against any
defence.
The solution is for Il to lead the ace
OF hearts. A may trump this trick or
pass Il, but the best defence Is to run
with the deuce OF trumps. Y overtrumps
with the queen and leads the two high
est trumps. 20 as to exhaust A.
Upon these two trump leads 2 may
discard two diamonds or two spades, II
does not matter which. The next step
is for to throw Il into the lead with
the losing trump. Y discarding another
spade or diamond, whichever suit he
has left The only trick that b can
mako after this is the king of hearts.
but A will make the ace OF whichever
suit z has kept at the end.
If A refuses to trump the first tricl;
Y allows the ace OF hearts to win. dis
carding small spade. A presumably
dlscards a spade. Il then leads a dia-
snond and Y trumps the trick with n
high trump and leaus three rounds OF
trumps, high cards first. u discards
hearts on the first two trump leads.
+: b has the best trump left for the
third round Y will discard n spade. get
IN on the hearts and make the ten OF
diamonds. losing the last trick 20 AS
ace OF spades. II A has the best trump
for the third round II will discard
a third heart, make his spade king and
the ten OF diamonds. losing heart trick
at the last to B's king.
The false solutions are the small heart
and diamond openings. The defence TO
the small heart opening is for A to dis
card the five of spades. If Y trumps
with an honor and leads three rounds of
trumps. B will give up the jack and
ten. so that A shall win the third round
with the nine.
If z has discarded three hearts A will
lead the ace and another spade, putting
z in and getting back himself with the
ace of diamonds after u has made his
ace OF hearts. II z has kept two hearts
and one diamond A will make his ace
Of diamonds and put Il in with 4 small
spade. so that B shall make two hearts
or \ shall make two spades at the end.
Several thought the diamond opening
would solve and the defence is more in
tricate. Y trumps the trick with an
honor. but b undertrumps with the jack
or ten. When Y leads a high trump Il
st!ll keeps the seven and \ keeps the
deuce.
Now II continues with his high
trump, a throws in the nine and leaves
y In the lead 50 that y must lose four
spade tricks, as \ will lead a small
spade after making his ace and queen,
putting in, but still holding the major
tenace over him 10 win two more tricks.
II y leads the small trump at the
third trick. A will overtake Rs seven
with the nine and throw Y back in the
lead With the deuce This is the varia-
ton overlooked by those who picked the
diamond opening for the solution.
IL b is allowed TO hold the third trick
Y and z can make their contract, z DIS
cardng n heart. b leads heart, Y
covers and \ trumps with the nine Of
clubs, but Y refuses TO overtrump. Now
nothing AN prevent Y from getting IN
with the spade king and giving two
discards on the established diamond and
the ace of hearts, y trumping The last
trick OF all. This IL the line OF play that
solves. but II Is unsound.
This Is very clever problem. its charm
being in Its innocent exterior. THE SUN
takes grea, pleasure in presenting the
composer to the solvsn: