Competition Problem 145b
by Steve Bloom
The contract is six
spades by South. West leads the ♣9.
Successful solvers: Ian Budden, Johnson, Radu Mihai, Sebastian Nowacki, A.V. Ramana Rao, Zoran Sibinović, F.Y. Sing, Rajeswar Tewari, Andries van der Vegt, Dick Yuen, Wim van der Zijden Tables
(a) Assume this is the layout:
After the ♣A and the ♣2 ruffed, declarer finesses and draws trumps, then discards a diamond, as does East, on the good ♣10. North now leads the ♥10, covered by the ♥J and ♥K, then discards the ♥7 on South’s last spade. East is squeezed. Discarding the ♥Q or ♦K is useless, so East throws a lower red card. Declarer plays ace and another of the suited discarded. Again it is useless for East to drop an honour under the ace (thanks to North’s ♦6!), so that player is endplayed at trick 11. Declarer gets two tricks in the suit now led, plus the established winner in the other red suit.
(b) If North has the ♦5 and West the ♦4, then the play in (a) still works. However, the defence prevails in this layout:
East discards the ♦10 on the ♣10, another diamond honour on the ♠6, and drops the remaining honour under the ♦A. West then covers the ♦8 to complete the defence. The same result arises in the two cases North has the ♦4.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2017