Feeling up to par? by Zia Mahmood
On 26 September, 2014 At 11:58
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Thursday 14 July 2011
A fiendish problem from the late Pietro Bernasconi
Jean Besse and Pietro Bernasconi will be remembered chiefly for their role as torturers, for they composed the fiendishly difficult problems with which the world’s best declarers used to wrestle at the Par contests that accompanied the world championships.
On the tenth anniversary of Bernasconi’s death, it is appropriate to show you one of the reasons he will remain immortal in the bridge world, and to ask a simple question: why would you duck a trick? To exhaust a dangerous opponent of cards in his partner’s long suit, or to rectify the count for a squeeze, or to conserve entries in your own long suits, or to prepare the ground for a cross-ruff, or… East-West vulnerable, dealer South:
North-South play in six no trumps and West leads the 10. What now?
If East has four or more clubs to the jack, there is nothing you can do, so you win the opening spade lead with the K and play a club to the A. East discards a heart, which is good news because you can now guarantee four club tricks.
All you need now is four diamonds, and since you can cater for either opponent having the guarded jack, your troubles are almost over. I am pretty sure that 99.9% of declarers would think:”West has five clubs and East has none, so East is far more likely than West to have long diamonds.” But that kind of reasoning was abhorred by Besse and Bernsaconi – why take a guess when you have a mathematical certainty?
Play a diamond to the K, all following, run the 10 and cash the Q, then duck a trick in whichever major suit East has discarded fewer cards (he has had to make three discards on the clubs).
Win whatever the defence returns and cash the rest of your major-suit winners. If West follows to five rounds of majors, he cannot have four diamonds (since he has five clubs). If he discards on a major-suit winner, you will know the complete distribution of that suit – East will be marked with at least six cards in it and four in the other major (from which he will have discarded twice and followed suit twice), so cannot have four diamonds. Beautifully simple once you see it, horribly complex when you don’t – the essence of all truly great problems.
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