Puzzle 8N73 Main |
| 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 cards into his large hadron collider. What happened next was unprecedented. Accelerated card particles defied the laws of quantum physics, giving credence to the Professor’s boson deflection theory. Card play had entered a new era.

I’ll spare you the math. Suffice it to say that after billions of collisions, ace particles were expected to surge to the fore, *but they didn’t*. Lower ranked particles continually kept pace! Could our perception of card ranks be just an illusion? Could all card ranks be the same? The Professor wouldn’t go so far as to expound that contention, but instead attributed the phenomenon to “obtusely deflected bosons.” Well, sure; he took the words right out of my mouth. If you’re still a bit fuzzy on this, an example may shed some light.

As declarer in notrump, suppose you intend to play the spade suit below. How many tricks can you win?

Notrump | A Q 6 4 | |

J 9 7 | K 8 3 | |

N or S leads | 10 5 2 |

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* (and vice versa) to morph into this layout:

Notrump | Q 9 6 4 | |

A J 7 | K 8 3 | |

N or S leads | 10 5 2 |

Leading from South and finessing the nine, then subsequently toward the queen, will lose two tricks, and win two tricks, *just as before*. Evidently 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.

Layouts must fit the distributions below, with the ace originally North as shown. Fill in the other 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.

1. Notrump | A | |

| | |

N or S leads | |

2. Notrump | A | |

| | |

N or S leads | |

To see if your solutions are successful click

This puzzle contest, designated “February 2018” for reference, was open for over a year. Participants were limited to *one attempt*, unlike my usual contests that allowed entries to be revised with only the latest one counting. Participation was good, and there were 17 correct solutions. Tiebreakers were (1) to find the lowest swappable card rank and (2) for the North-South hands to be as weak as possible, in that order of priority. Only one solver was perfect on both problems.

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).

Rank | Name | Location | Card Ranks | N-S Sum |
---|---|---|---|---|

1 | Stephen Merriman | New Zealand | 11 | 116 |

2 | Dan Gheorghiu | British Columbia | 11 | 117 |

3 | Foster Tom | British Columbia | 12 | 122 |

4 | Konrad Majewski | Poland | 13 | 103 |

5 | Duncan Bell | England | 13 | 104 |

6 | Jean-Christophe Clement | France | 13 | 110 |

7 | Tina Denlee | Quebec | 13 | 117 |

8 | Tim Broeken | Netherlands | 14 | 100 |

9 | Richard Jeng | Georgia (US) | 14 | 107 |

10 | Dave Wiltshire | Australia | 14 | 110 |

11 | James Martin | England | 14 | 111 |

12 | Nicholas Greer | England | 14 | 97 |

13 | Martin Vodicka | Slovakia | 14 | 97 |

14 | Charles Blair | Illinois | 15 | 100 |

15 | Alon Amsel | Belgium | 15 | 96 |

16 | Samuel Pahk | Massachusetts | 15 | 96 |

17 | Yiran Liu | Virginia | 16 | 103 |

Puzzle 8N73 Main | Top Abracadabra, Ace Benign |

How *low* can you go? This girl found the answer, a *five-spot*, and her story is priceless:

NT win 3 | A K 10 2 | NT win 3 | K 10 5 2 | ||

Q 9 5 4 | 7 6 | A Q 9 4 | 7 6 | ||

South leads | J 8 3 | South leads | J 8 3 |

**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*.

Tina certainly had the right idea, but her N-S pip count of 61 was one off the optimal (60) found only by Stephen Merriman:

NT win 3 | A K J 3 | NT win 3 | K J 5 3 | ||

Q 10 5 4 | 7 6 | A Q 10 4 | 7 6 | ||

South leads | 9 8 2 | South leads | 9 8 2 |

The same number of tricks are won when the ace is swapped with the five. No matter how West defends, leading the nine and eight (covered each time) eventually establishes the *five*.

The pattern of the heart layout differs, requiring the four-card length *behind* dummy’s ace, which renders the preceding five-swap impossible by my research. This solution is far from optimal, but when the story fits, I can’t resist. O, Canada…

NT win 2 | A J 4 2 | NT win 2 | J 8 4 2 | ||

10 9 | Q 8 6 3 | 10 9 | A Q 6 3 | ||

North leads | K 7 5 | North leads | K 7 5 |

**Tina Denlee**: With the long hand wrong-sided, the layout is ace-bait; *ace be-eight?* The key position is reached after *one* losing trick, which dictates *three* cards higher than the previous five, hence 5 + 3 equals *eight*, the card to swap for the same number of tricks.

Very convincing — not for bridge but to label Quebec as the B.S. capital of North America.

The optimal solution is two notches lower, swapping the *six* without changing the result:

NT win 2 | A 10 9 2 | NT win 2 | 10 9 6 2 | ||

8 7 | Q J 6 4 | 8 7 | A Q J 4 | ||

North leads | K 5 3 | North leads | K 5 3 |

**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!

Puzzle 8N73 Main | Top Abracadabra, Ace Benign |

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