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In retrospect, the 2
alternative
was to be preferred, tending to deny more than 3 of either major.
North not unexpectedly jumped to the heart game contract.
The
K lead was won in dummy
and South took stock.
North-South assets were a combined 6 spades, 7 hearts, 8 diamonds and 5 clubs,
not to mention 22 HCP in dummy and 4 HCP in hand: 26HCP.
Given West could have no more than the remaining 14 HCP and bid a 2nd time,
introducing a new suit, opposite a passed partner, South expected at least 4
spades and at least 5 diamonds.
That left 4 other cards at most in hearts and clubs.
With 8 outstanding clubs and 6 outstanding hearts, East would have at least
10 cards in those two suits.
How West's "at most" 4 cards outside spades and clubs divide?
Declarer decided that West most likely held no more than 2 hearts - perhaps
even a void.
East then would have at least 4 hearts. On which side would the
J
be? In the hand with at most 2 cards in the suit? In the hand with at least
4 cards in the suit?
Would that 1 HCP of a
J be a
requisite for an opening hand?
Would a
Jx really be counted
as a point in bidding a 2nd time opposite a partner that passed?
Declarer led from dummy towards the
T
at trick two, losing to West's
J.
West cashed the
KQ and exited
a small club.
Declarer drew two more trumps ending in hand and led the
Q,
covered by West and won in dummy.
The last trump was drawn. West's
J
and a spade were conceded. In total the defence won 3 spades, 1 heart and 1
diamond.
West would have done even better by ducking the
Q
lead. A 2nd diamond would be ruffed by East and West would later still win a
diamond and a spade.
How was West to know that East had a fourth trump heart to ruff with?
Trump Suit Preference Count
If playing "Trump Suit
Preference Count", East's first shows preference for clubs, having
nothing in diamonds.
With hearts trump, and spades already discouraged at trick one, clubs and diamonds
remained. Clubs is the lower ranking of those two suits.
By first playing the lowest of
8765,
the
5, East encourages the lower
of clubs rather than a diamond, suit preference.
In next following to trump plays, East "gives count" in the 'lowest
suit of unknown count'. East's clubs number 5, odd. "Odd is signalled low"
and "even is signalled high".
Of the remaining
876, lowest
is the
6, played to show an odd
number.
How does East signal when following to West spade leads?
To partner West's
K lead, East
with neither
Q nor
J
discourages with the lowest spade held, the
7
(from
987).
To the next spade play, with two spades remaining, East plays the
9
(from
98).
High-low shows an even number of remaining cards in the suit (remainder count).
To the 3rd spade lead, East plays the last spade, the
8,
confirming that the
9 was higher.
To West, East signalled odds spades AND 12 spades had been played and West had
the last one still in hand.
When West sees South win the
Q,
West should wonder of the first round
T
finesse from
QT9x opposite
AKxx.
Simple. South didn't have 4 hearts. South wouldn't have finesssed the
T.
South had only 3 hearts. East had 4 hearts.
East then started with 3 spades, 4 hearts and an odd number of clubs.
If East's diamonds and clubs number (13total-3spades-4hearts=) 6, then East
clubs number 5 or 3 or 1.
If East held 1 club, then South would be holding, of the eight clubs that West
does not see combined with dummy the remaining, exactly 7.
South bid 1
on a 3 card suit
holding 7 clubs??? With 5 clubs still bid a 3 card heart suit with only 4 points?
Nah. Must have only 3 clubs then.
With proper and dependable signalling and use of deductive reasoning, West should
find East to have 3 spades, 4 hearts, 1 diamond and 5 clubs.
South then would have, by subtraction, 3 spades, 3 hearts, 4 diamonds and 3
clubs for more than one reason.
Getting to 3N or Avoiding 4![]()
Either South must bid a 3 card suit at the higher 2 level, clubs, or North must
be prepared for a 3 card heart suit in order for the pair to get to the "choice"
3N contract.
There are a number of ways to reach 3N:
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