Double Dummy Corner

 

Competition Problem 155a

composed by Leigh Matheson
presented for solving in February 2018

DR6

♠ AK5x

 98

 KQ3

♣ AKy2

♠ Q96

 K

 J109876

♣ QJ3

♠ 87

 106543

 A54

♣ 876

♠ J104x

 AQJ72

 2

♣ 109y

South is in six no-trumps. West leads the J.
The cards marked x are the ♠2 and ♠3; the cards marked y are the ♣4 and ♣5.
How must those four cards be placed to enable East-West to defeat the contract?
How is the contract made if those cards are placed differently?

Successful solvers:   Steve Bloom, Ian Budden, Ed Lawhon, Steve McVea, Radu Mihai, Sebastian Nowacki, Zoran Sibinović, Rajeswar Tewari, Wim van der Zijden

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Solution

In all cases the opening lead goes to the Q and A and East returns a diamond.  The defence can prevail only when North has the 2 and the 4.  We will show how the contract is made in the other cases.

A.  North has the 3 and 4.  Then this is the layout:

♠ AK53

 98

 KQ3

♣ AK42

♠ Q96

 K

 J109876

♣ QJ3

♠ 87

 106543

 A54

♣ 876

♠ J1042

 AQJ72

 2

♣ 1095

South discards the 5 on the second diamond, won by North who leads a heart to the A in order for South to lead the J through West, who does best to cover with the Q.  North wins with the K and leads the 5 (not the 3) to South’s 10.  The 10 is covered by the J and K, whereupon the A catches East in a seesaw triple squeeze!

1.      If East throws a heart, South plays the 2 and has an entry on the 4 to score the remaining hearts after North’s 9 is allowed to win.  Those cards and the A take the remaining tricks.

2.       If East throws a diamond, South again plays the 2 and West now has the sole guard against North’s 3.  The 9, which holds, is followed by the 3 to South’s 4, wringing a club from East.  So West now has sole guard of both minor suits and is squeezed when South scores the Q and J.

3.       If East throws a club, South drops the 4 and overtakes the 9 even when East doesn’t cover it!  South cashes the other heart winner, followed by the 9.  When this holds, North has an entry on the 3 to take the last two tricks in clubs.

A.  North has the 3 and 4.  Then this is the layout:

♠ AK52

 98

 KQ3

♣ AK52

♠ Q96

 K

 J109876

♣ QJ3

♠ 87

 106543

 A54

♣ 876

♠ J1043

 AQJ72

 2

♣ 1094

This time South discards a high club on the second diamond.  Play can now follow the same line as before (though other orders of play are possible to lead to the same ending).  East is triple-squeezed as before on the third round of spades and the play is the same as before on a red suit discard—but not when East discards a club.  In that case, when the 2 is led to South’s 4 East is squeezed again, considering that another club discard would allow North to finesse the 5, thanks to South’s careful discard at trick 2.

Clearly, if North has both the 3 and 5, South can discard any club at trick 2 and play as in A or B.  But when South has both of those cards East can safely discard clubs on the spades.  In A.3, North has no entry to the club winners and in B the threatened finesse in clubs is blocked when West refuses to cover the 5.

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: 14 March, 2018