Competition Problem 168a
West leads to South’s three hearts. East-West to defeat the contract.
Successful solvers: None! It seems that the broad hint given by splitting this composition into two setting misfired slightly. The two solvers who successfully tackled 168b both failed to consider the possibility of declarer winning the first trick and playing as shown in line A. below. Most of our regular solvers didn’t submit solutions to either of these problems. It would be good to hear their opinions.
Declarer has eight top tricks via red suit finesses and on a passive defence will come to a ninth by drawing trumps and applying endgame pressure on West. West must therefore attack clubs, East then threatening to score club tricks should declarer draw trumps. In fact West must lead the ♣K, as East might be required to overtake both the ♣9 and the ♣J if declarer opts to win the first trick.
A. Suppose North wins the first trick with the ♣A and leads the ♥5. Then East must play the ♥6! South wins with the ♥7 but note that it is now impossible to play two more rounds of trumps such that the remaining hearts in both North and South are higher than East’s. Declarer might as well exit in clubs now, but East overtakes West’s card and returns a diamond to North’s ♦Q. East then covers North’s ♥9 and West discards a diamond. We now have this position, with South on lead and defenders requiring four more tricks:
1. If declarer now tries one more round of trumps, then West must discard the ♠5, regardless of whether the trick is taken in North or South. Declarer can set up the long spade now but West gets in twice to attack diamonds, East refusing to ruff North’s ♦A. West gets in a second time in spades, cashes the ♦J, and leads the good ♦9. If South’s last heart is the ♥4, East ruffs with the ♥J for the setting trick; otherwise East discards, South has to ruff with the top heart, and the ♥J beats North’s ♥8 at trick thirteen.
2. If instead declarer leads a spade to the ♠A in the above position, East must drop the ♠J. North can now lead the ♥8 and run it if East plays low, thus making sure that both North and South can beat East’s last heart, but in fact it matters not which heart East plays on this trick. West discards a diamond and now:
(i) If the next lead is a spade, West wins as cheaply as possible and leads the ♣J. Declarer’s and dummy’s last trumps make separately but East can ruff the ♦A and West still has a spade tenace over South.
(ii) If it is a diamond, East ruffs the ♦A and leads a spade. West makes two spade tricks and a diamond.
B. If North lets the ♣K hold, West continues the suit. Again East plays the ♥6 on North’s ♥5 and thereafter the play is as in line A except that South leads the first diamond instead of East.
1. If West starts with a lower club, declarer wins the first trick and can play two or three rounds of trumps before exiting in clubs. If East wins the second club declarer will be able to draw trumps and throw West in on diamonds to get a favourable spade lead. And if West plays the ♣K the next lead is to declarer’s advantage one way or another.
2. If East plays the ♥9 on the first heart, South wins, crosses to North on a diamond finesse, and leads the ♥8, followed by the ♥10 if the ♥8 holds. Whatever East does, South’s and North’s last hearts will now both be higher than East’s and West is squeezed on the third heart: the club discard lets declarer draw trumps and throw West in, whereas a spade or diamond discard lets declarer establish a long card in the discarded suit.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2019