Paul Cronin

Problem Solving

In quantum mechanics, Heisenberg’s Uncertainty Principle tells us that we can never see “things” as they are, for the very act of our looking at them changes them. Is there a similar principle involved in solving bridge problems? Does the fact that we know that there is a problem change the nature of the problem? For instance, if we ask what is the way to play 10xxx opposite AK8xx to obtain the maximum number of tricks (assuming no other information is available) then the answer is to lead the 10, play the ace, and then play the K. But if we insert this holding as part of a hand in a “problem”, and the success of the contract depends on obtaining the maximum number of tricks in that suit, should we change the way we play the suit simply because it is known that there is a problem? Does knowing there’s a problem tell us that the normal way of playing the suit will not work? I  thought about this while looking at the problem posed by Eddie Kantar, p. 55, in theMay 2010 Bridge Bulletin. The auction has been


West North East South
  1C P 1S
P 2S P 4S
P P P  

 and you hold 



  West leads the H4. Plan your play.

There are three apparent losers off the top, a spade and two diamonds,and the club suit to tackle. Given this, don’t we know for sure that the club finesse will fail? If it worked, there wouldn’t be any problem! Thus any solution involving the club finesse is wrong, and we look for ways to set up red cards for discards. Was Heisenberg a bridge player?


Richard PavlicekJune 11th, 2010 at 11:24 pm

No, this originated in the days when Ira Corn managed the Dallas Aces. Partnerships varied, and “Broadway” Billy was considering moving on to other things. This “Eisenberg uncertainty principle” caused Corn to add Soloway… or something like that. The “H” was added later as a coverup. Trust me.

Craig BiddleJune 13th, 2010 at 5:38 pm

Technically, the way to play 10xxx opposite AK8xx for the maximum number of tricks is to cash the ace and king. That takes 5 tricks over 40% of the time, 4 tricks about 50% of the time, and 3 tricks the remainder, for an average of 4.3 tricks per occurrence.

The play you cite, leading the 10 and passing it, is the best play for four tricks, losing only to a 4-0 break off side. But it takes five tricks only against QJ doubleton on side or singleton 9 off side, so it averages less than 4.1 tricks per occurrence.

JoshuaWexNovember 10th, 2019 at 3:43 pm InemWouckhoaliliolla

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