Double Dummy Corner

 

Competition Problem 153a

composed by Paolo Treossi
presented for solving in December 2017

DR7

♠ AK102

 AK842

 3

♣ AJ8

♠ QJ9x

 75

 K109

♣ 7654

♠ x43

 QJ1093

 Q2

♣ Q102

♠ 87

 6

 AJ87654

♣ K93

The x’s are the 5 and 6.
(a) Place them so that South can make three no-trumps against any defence and
(b) show how the contract is defeated if they are wrongly placed.

Successful solvers:  Radu Mihai, Zoran Sibinović, Rajeswar Tewari

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Solution

(a) West has the 6.

West does best to lead the 7.  Declarer wants to lose this trick but without leaving West on lead for a damaging club return.  So North covers with the 8.  East wins the trick and returns the 3, West’s 5 forcing North to win with the A or K.*  North advances the 3.

A.      If East plays the Q, South ducks.  East does best to continue hearts.  North can either duck or win and continue the suit, South discarding diamonds.  Assume the latter.

1.       If East cashes the last heart, South discards another diamond.  Now there is a spade-diamond squeeze on West.  For example, South wins the A, North throwing a low club, and plays a spade to the J and K.  The A and a club to the K bring about that squeeze.  Note that the same situation arises if North ducks the third heart, wins the fourth, and loses the fifth.

2.       If East keeps a heart winner and exits on a diamond, South wins with the A but this time North discards the 2.  South leads a spade to the J and K and comes to hand on the K to finesse the 10, cash the remaining spade winner, and throw East in on a heart for club into North’s AJ.

3.       If East keeps a heart winner and exits on a spade, North will make the A and three spade tricks, South’s K providing the entry for the finesse of the 10.  That brings West down to a winning spade and K10, so North exits on the fourth spade and South’s AJ take the last two tricks.

B.      If East plays the 2, South wins with the A and leads a spade.

1.       If West plays the J, North plays the 2!  South’s K takes the club return.  The spade finesse is taken and North cashes the major suit winners, following with a heart to East.  South keeps J8 and a club.  This is the three-card ending with East on lead:

♠ none

 4

 none

♣ AJ

♠ none

 none

 K10

♣ 7

♠ none

 none

 Q

♣ Q2

♠ none

 none

 J8

♣ 9

Note that if East has the Q instead of the Q here, North will definitely make two club tricks.  So East has the Q and leads it, but it is of no avail for West to overtake, because then North discards the J and either the 4 or the J will score in addition to North’s A.

2.       If West plays low on the first spade, North wins with the 10 and leads the J (!) to East’s Q and South’s K.  South exits on a low diamond.  If West plays low it will be a simple matter to eliminate East’s spades and throw that hand in with hearts to obtain a club lead into the split tenace.  But West can avoid that by rising with the K and leading a club—unless North discards the A!  In that case North wins the spade return and leads a club towards South’s 93. East wins with the 10 and is on lead in this position:

♠ A2

 A42

 none

♣ none

♠ J6

 none

 10

♣ 76

♠ 5

 QJ10

 none

♣ 2

♠ none

 none

 J876

♣ 9

North makes the A and A, then exits on the 2 so that South takes the last two tricks.

(a) East has the 6.

Now West must lead the 7, and play follows as in (a).*  When North leads the diamond at trick 3, East plays low, so South wins and we are in line B.  On the spade from South West must play the 9, so North wins with the 10 and we are in B.2.  When South exits on the low diamond, West rises with the K and returns a spade or club.  Assuming a spade return, followed by a club from North to East’s 10, we have this:

♠ A2

 A42

 none

♣ none

♠ J5

 none

 10

♣ 76

♠ 6

 QJ10

 none

♣ 2

♠ none

 none

 J876

♣ 9

Now East leads a heart and West discards the J!  In (a) that play is ineffective because North merely exits immediately on the 2 to West’s 6 and South takes the last three tricks (so, assuming North has played the K rather than the A, none of North’s three aces takes a trick, as the composer likes to point out).  Here, however, North can do no better than to win two spade tricks and concede the last two to East.

* East can alternatively lead the 2 at tick two and this also defeats the contract in (b).  In (a), South would win with the A, lead a spade to the 9 and 10. If North now cashes a top heart we have the situation already described, with West’s 6 defeating the contract.  However, North can optionally lead the J to Q and K immediately.   When declarer plays that way in (b), West will have a heart left when winning the second diamond with the K and must lead it to be sure of being able to discard that top spade when East later exits on a heart.

Here is the original problem by Neils Y. Wilson, which appeared as Problem 188 in George Coffin’s Sure Tricks (1948):

♠ AQ102

 AK432

 3

♣ A54

♠ KJ98

 76

 K109

♣ 9876

♠ 765

 QJ1095

 Q2

♣ QJ2

♠ 43

 8

 AJ87654

♣ K103

South to lead at no-trumps.  North-South to make nine tricks.

The problem could of course have been stated: “South to lead and defeat East’s one no-trump contract by three tricks”.  Anyway, South has to lead a spade, but after winning this trick as cheaply as possible North has several immediate choices: lead the 3, cash a top heart followed by the 3, or exit on a low heart.  Several orders of play are available after that, but declarer can always contrive to play a club from North at some stage, forcing East to split the honours, and eventually throw East in with a heart to obtain a club lead into the split tenace.  That was the composer’s idea but the intention was for the avoidance play in diamonds that Paolo Treossi has recovered in this clever revision which also enforces the first round duck in hearts.

See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.

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© Hugh Darwen, 2018
Date last modified: 15 January, 2018