Bridge players have long considered an ace to be a worthy asset, and rightly so until Feb 23, 2018.
This was the date Professor Freebid accidentally dropped a deck of cards into a large hadron collider.What happened next was unprecedented but gave credence to the Professor’s boson deflection theory.Accelerated card particles defied the laws of quantum physics!
“I’ll spare you the math,” lectured the Professor, “but after billions of collisions, I expected ace particles to surge to the fore, but they didn’t. Lower ranked particles continually kept pace. Card play had entered a new era.”
Could our perception of card ranks be an illusion? Are all card ranks the same? The Professor wouldn’t go so far as to make that contention, but instead attributed the phenomenon to “obtusely directed bosons.”
Well, sure… took the words right out of my mouth. If you’re still fuzzy on this, consider the following spade suit.As declarer in notrump, 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, that’s too prehistoric — how about Abracadabra, let the ace be nine! That is, swap the A and 9 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, just like before. So the nine is equivalent to the ace. Quick! Lower your notrump ranges by 4 points while the Professor’s bosons are flying amok.
How far can this phenomenon go? Can a card below a nine render an ace benign? The Professor’s theory indicates so, but he doesn’t have time to prove it. Your help is greatly needed. Don’t let him down!
Create two suit layouts where an eight or lower is equivalent to the ace when the cards are swapped.
In the original contest, solvers had to construct their own layouts (matching specific patterns) but I’ll show the two winning layouts below. See if you can determine the lowest East-West card to swap with the ace, yet win the same number of tricks.
1. What is the lowest spade to swap with the A?
2. What is the lowest heart to swap with the A?
Quit
Congratulations to Stephen Merriman — indeed a merry man — who stands alone at the top with his first win. Stephen is a more recent participant than most of the others listed, having several high finishes before this.
Ranking is by the lowest card ranks swapped and the North-South sums (before the swap). Optimal for Problem 1 is 5:60, and Problem 2 is 6:56, so a perfect total is 11:116. Further ties are broken by date and time of submission (earliest wins).
How low can you go? The five-spot, and this solver’s story is priceless:
Tina Denlee: The card to exchange with the ace must be a trick-winning card, else it would change the result. Moreover, the key position should be reached after two tricks, which means five cards remain, and the key card is the fourth lowest (opposite of usual notrump leads because bosons are upside-down) which is the five. So I’ve just made an ace be-five, and I’ve never seen an ace be-fore.
Very convincing — maybe not for bridge, but at least for Quebec as the B.S. capital of North America.
Tina was clever, but her pip count was one off the optimal (rank sum 60) found only by Stephen Merriman:
The same number of tricks are won when the ace is swapped with the five. In the swapped case, no matter how West defends, leading the nine and eight (each covered) then the two eventually establishes the five.
The pattern for the heart suit required the four-card length behind dummy’s ace, which renders the five-swap impossible by my research. The lowest ace exchange now is the six-spot, as Dan Gheorghiu (British Columbia) showed by these layouts:
Declarer can win only two tricks in the first layout when East covers any card led from dummy. Surprisingly — at least before bridge met the large hadron collider — declarer does just as well by swapping the ace for a six. Who needs high cards any more!
Like Tina in the spade suit, Dan also got pipped by a single point — call it a tough day for Canadians — as shown by this optimal solution (rank sum 56):
Duncan Bell: I debated switching the jack and 10 but eventually decided against it.
Good decision! Either way can swap the ace and six-spot, but the given layout is necessary to achieve the lowest North-South sum of 56.
Optionally, the North-South cards can be A-10-9-5 opp. K-3-2 (swapped to 10-9-6-5) for an equally perfect solution, as submitted by our winner, Stephen Merriman.
Samuel Pahk: Sigh. My only stumbling block was the spade suit… and the heart suit…
Charles Blair: A computerized search would be feasible, but as probably obvious from my answers, I didn’t do this.
Konrad Majewski: Great problems! I am not sure whether my solutions are optimal, but it was a lot of fun solving them.
James Martin: Thanks for the ingenious puzzle!
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