Puzzle 8N73 by Richard Pavlicek
Bridge players have long considered an ace to be a worthy asset, and rightly so until Friday Feb 23, 2018. This was the day Professor Freebid accidentally dropped a deck of playing cards into his large hadron collider. What happened next was unprecedented. Accelerated card particles defied the laws of quantum physics, thus giving credence to the Professors boson deflection theory. Bridge had begun a new era!
Ill spare you the math. Suffice it to say that after billions of collisions, ace particles would expectedly surge to the fore, but it didnt happen. Lower-rank particles continually kept pace! Could it be that our perception of card ranks is just an illusion? Could all card ranks be the same? The Professor wouldnt go so far as to make that contention, attributing the phenomenon to obtusely deflected bosons. Well, sure; he took the words right out of my mouth. If youre still a bit fuzzy, this puzzle may shed some light.
Click for a list of successful solvers
As declarer in notrump, suppose you decide to play the spade suit below. How many tricks can you win?
With the king offside, any finesse is doomed, so you must lose two tricks. Hence you will win two tricks, the ace plus the long card in dummy. Now just say the magic words, Yabba dabba doo no, wait, thats too prehistoric how about Abracadabra, let the ace be nine (and vice versa) to morph into this layout:
Leading from South and finessing the nine, then subsequently toward the queen, will lose two tricks, and win two tricks, same as before. Evidently the nine is equivalent to the ace. Quick! Lower your notrump ranges 4 points while the Professors bosons are flying amok.
How far can this paradox go? Can a card below a nine render an ace benign? The Professors theory indicates yes, but he doesnt have the time to prepare documentation. Your help is requested to prove it. Dont let him down!
Create two suit layouts where an eight or lower is equivalent to the ace when the cards are swapped.
Layouts must fit the distributions below, with the ace fixed in North as shown. Fill in the remaining 12 cards (one per box) for each layout. Tab key will advance to the next box. Now for the easy part: just click Verify for an instant analysis. If unsuccessful, try again!
Many solutions exist. Tiebreaking goals are (1) the lowest swappable card rank, and (2) for the North-South hands to be as weak as possible, in that order of priority. Strength is judged by the sum of all card ranks: Ace = 14, king = 13, queen = 12, jack = 11, etc.
If desired, you may submit your solutions using the form below. Solutions may be submitted only once, and doing so will add you to the list of successful solvers, ranked according to the indicated tiebreakers. You will also receive an automatic reply with a copy of your solutions and what Richard believes are the optimal solutions.
© 2018 Richard Pavlicek