“Timothy, come here! I have a hand to show you from last night, playing with Marlon.”
[The Professor scribbles a diagram on the back of a pickup slip.]
“I was East and picked up my usual garbage — though not bad for poker, with four ninesagainst Marlon’s four sevens — only to watch our opponents bid and make six notrump.Despite being a complete zero, it fuels my latest theory.”
“A complete zero?” Timothy queried. “Must’ve been a weak field. You have two queens and three jacks between you, so North-South have 33 HCP. They also have no eight-card fit, so bidding and making 6 NT should be a normal result, albeit below average for you.”
“Yes, but Marlon could have beaten it. The remarkable fact is that we could not have beaten a slam in any of their seven-card fits. This follows directly from the Law of Total Bosons, a corollary to my boson deflection theory: The rank sum of all defensive quads modulus 16 (nines plus sevens → zero) equals the sum of suit-slam undertricks, also zero.
“Whatever you say,” agreed Timothy, suppressing his true sentiments that the Professor was a nut case. For amusement he played along, “But what if they bid the slam in spades? Surely there’s an undertrick there.”
“Get serious, Timothy, and don’t make light of my theories! The Law applies to majority trump fits, but I’m working on a general theory to encompass all. In college, I learned never to underestimate the power of trump.”
“I see. Your alma mater must have been the Electoral College.”
Any more nut cases out there? Here’s your chance to practice zero tolerance:
Construct North-South hands to fit the story.
Many solutions exist. As a further goal (tiebreaker for the February 2017 contest) try to give North the best possible poker hand, and South the worst possible poker hand.
1. Before reading further, can you pick the winning South hand?
Quit
Congratulations to Duncan Bell, England, who was the first of only three to find the optimal solution (best poker hand North and worst poker hand South). Duncan also won The Twelve of Spades in December, and despite being a late joiner to this series, might be “for whom the bell tolls” in the future. Placing second was another ringer, Leif-Erik Stabell, Zimbabwe; but the peals draw silent for Grant Peacock, Maryland, in third place. No, wait… [in my best Don Adams imitation] Would you believe a bell-shaped bird?
This puzzle has many solutions, a whopping 137,351 by my analysis, so the crux was to find the best and worst poker hands for North and South, respectively. A few solvers overlooked the possibility of a lowball straight flush (5432A in hearts or clubs) no doubt because the ace is always high in bridge, thus settling on four aces as North’s best poker hand.
I liked this first deal from Dan Gheorghiu, British Columbia, because it was only entry that reduced South’s poker hand to a measly two pair (KKTTQ). Unfortunately, this was only the secondary tiebreaker, so North’s four aces placed him well down the list.
Dan Gheorghiu: In 6 NT declarer has 10 top tricks and can establish a heart by leading twice from dummy (giving up a trick) with potential for a positional squeeze against West’s minors; but this fails because hand entries are only in hearts, so dummy’s A cannot be won before the top spades. Obviously, either major lead will defeat 6 NT, so perhaps Marlon led a minor to let it make.
Six of any suit (except spades) makes against any defense. Most interesting I thought was 6 with a trump lead:
After six tricks South is on lead in this position:
If you cash the Q you fail! Instead you must ruff a spade (the queen if you want to show off) and lead a good diamond. East has no answer: (1) If he ruffs low, overruff and return to dummy in clubs; A, etc. (2) If he ruffs high, you can draw his remaining trumps thanks to the two club entries. (3) If he pitches a club, cash A-K before leading the last diamond.
The story conditions — 6 , 6 and 6 the only makable six-bids — were met by 14 successful solvers, but only one produced an overtrick. While immaterial in the ranking, Charles Blair, Illinois, earns style points for this entry cold for 7 .
Constructions are possible to make both 7 and 7 (obviously not 7 ), e.g., North A-K-Q-2 A A-K-Q-10 A-K-J-2. None of these, however, would score high on the leaderboard, because North cannot be given a straight flush.
After a heart lead, South wins a top card in each suit (proving the lead doesn’t matter) then two top spades, a spade ruff, and two more winners to reach this ending:
A club is ruffed low, and declarer crossruffs the remainder, as West can only underruff and be trump-couped.
Charles Blair: I wonder if we could have six of any suit making but 6 NT going down.
Yes, in fact seven of any suit can be made with 6 NT failing. Examples are shown in Can You Solve a Mystery? (The ‘mystery’ was to create such a deal where even five notrump fails, but it remains unsolved and is now believed to be impossible, though unproved.)
The next entry from Dean Pokorny, Croatia, was the second-best effort, reducing South’s poker hand to a queen-high flush. The required suit slams (6 6 6 ) are relatively easy to make. Six notrump is defeated with a heart lead but requires careful defense by East.
Note East’s dual unblocking plays in clubs to leave the following position:
When declarer gives up a club, West can win and lead a second heart to foil communication. (If East won the third club, he could not lead a heart without losing a trick.)
Also note that declarer would have a double squeeze (treating the Q like a deuce) if the Q were cashed early, but doing so would allow West to cash the setting trick.
Foster Tom: Giving North a straight flush in clubs allows two clubs to be established, which necessitates a blocked position for 6 NT to fail.
The ultimate solution, found by three solvers, reduces South to a jack-high flush. Below is the winning entry from:
Duncan Bell: Leading hearts twice beats 6 NT by disrupting communications, which otherwise makes by establishing clubs. Six hearts makes by ruffing a spade and club in North, finessing a heart (East splitting) and leading a diamond at trick 11. Six diamonds makes, as clubs can be established without losing a trick; and 6 plays similarly to 6 NT, except with trumps there are no communication problems.
The play in 6 is perhaps the most interesting. Suppose West leads a diamond.
After East follows suit to the first 10 tricks, the following three-card ending is reached:
Professor Freebid: Wrong, Richard!I can assure you it’s a four-card ending.Each hand has an invisible boson.
North leads the Q, and East can win only one trick.
Most solvers who found the straight flush for North tried to reduce South’s poker hand further, but soon realized that a flush was inevitable. Any attempt to remove South’s 5-card suit results in 6 NT making or a suit slam failing.
Grant Peacock: If you give North a straight flush, South has a doubleton. I can’t make a 4-4-3-2 hand for South that works, so his poker hand has to be at least a flush, and J-5-4-3-2 is minimal.
Duncan Bell: I couldn’t give South less than a flush, since he can’t have more than two hearts (North needs a straight flush) or more than two diamonds (to threaten a ruff at trick 11).
Dan Gheorghiu: We all are aware of the Professor’s Law of Total Bosons. In the present case, however, Trump bosons might deflect and escape sideways, unless a wall is built completely around them, in which case we Canadians may be coerced to pay for it. Hopefully, this would not block Internet access, so we’d still have the challenges and headaches of RP puzzles.
Tina Denlee: A complete zero, really? Applying your boson theory, I opened 2 (both majors) as East, which West converted to 2 . North doubled for takeout, but South passed with A-K-Q-3-2. Partner took one trick. Now that’s a complete zero — vulnerable, minus a two and three naughts (2000).
A bosonic lesson for Tina, Turner of the dummy, seeing Johnny Cash one trick and Donald Trump the next.
© 2017 Richard Pavlicek
by Richard Pavlicek by Richard Pavlicek