Competition Problem 161a
is declarer in four spades.
Successful solvers: Steve Bloom, Ed Lawhon, Sebastian Nowacki, Zoran Sibinović, Wim van der Zijden.
First we show how the contract is defeated in this layout:
West starts with a low spade to South’s ♠9. Declarer’s best try is to play the top clubs and the ♦A, on which West drops the ♦K. Aiming for a heart ruff in North, South now opens that suit. Leading the ♥4 would allow West to win with the ♥7 and play another trump, so declarer tries the ♥8, but West rises with the ♥K and leads the ♠7 in this position:
North can try winning with the ♠K and ruffing a club, but East discards a heart on the spade so as to have a safe exit on the ♣5 when South exits on a low heart.
If the ♠7 and ♠6 are switched in the above ending, then North can win the spade with the ♠7. Now South can ruff a club, cash the ♦Q and cross to the ♠K to lead the good ♣9. If this holds, South’s ♠A is the tenth trick; otherwise West ruffs and has to give South a heart trick.
If the ♥7 and ♥6 are switched in the original layout, then South leads the ♥4 instead of the ♥8 at trick five. Now when West rises with the ♥K and leads the ♠7 to North’s ♠K we have this:
South ruffs a club and leads the ♥Q. West wins and leads a trump, simultaneously denying the heart ruff and access to the good club. However, South’s two trump winners force East down to the ♥J and ♦J8 such that a heart exit obtains a diamond lead into the split tenace.
Trap: If West leads a club, declarer can always setup a club winner in North and use the third round heart ruff to reach it.
See the solution to Competition Problem #4 for the recommended tabular format if you prefer not to write in English prose.
Hugh Darwen, 2018