Main Puzzle 8M91 by Richard Pavlicek

6 NT South | K 4 3 A Q 10 9 8 2 A 6 5 4 | E-W Vul West Pass All Pass | North 1 3 NT | East Pass Pass | South 2 NT 6 NT | ||

A J 2 K 2 A K J 10 K Q 3 2 |

“Good heavens, Holmes! That was *30 years ago*, and I’ve had almost no time for bridge with my medical career. Most of it is long forgotten, but I do remember that a 3-2 break is a favorite — nearly 68 percent if I recall — so 7 should make more than twice as often as not.”

“Yes, but numbers don’t lie. Suppose the deal is played 100 times at total points. I’ll even give you the benefit of doubt that 7 , on average, will make 68 times (68 × 1440 = 97,920) and fail 32 times (32 × -50 = -1600) for a net gain of 96,320 points. Now consider 6 NT, which will make every time for a gain of 100 × 990 = 99,000 points, and the occasional overtrick will bring this to over 100,000. So which do you prefer: the club grand, or a hundred grand?”

“Astounding, Holmes. I must say, your deductions never cease to amaze me. But will 6 NT truly make *every* time? Surely it might fail against some foul distributions or by misguessing the layout.”

“Watson, I would stake my reputation on it, with Her Majesty the Queen as my witness. Even after a club lead, which gives away nothing, I could throw it against the Tower of London for 12 tricks.”

Your mission is to determine how Holmes would play to justify his assertion.

After a club lead, how do you play 6 NT to ensure 12 tricks? |

Assume East follows to the club lead, and subsequently East and West will follow low or discard safely if they can’t win the trick. Further, no suit will break if you need it, and no finesse will work if you take it.

As a tiebreaker, include your best guess to this difficult question: Before the lead, what is the percent chance that South can make 7 NT with double-dummy play all-around? Assume the East-West cards are randomly distributed, so any layout is possible.

Congratulations to Tim Broeken for his umpteenth win (well, it’s only 10, but far and away the leader). He and Dan Gheorghiu were the only solvers to submit the optimal line of play that not only ensures 6 NT but also allows for an *overtrick* if the J drops early.

I was torn this month in judging the ranking, because Tim and Dan both were amiss on the percent chance to make 7 NT at double-dummy, while Tina Denlee and Charles Blair were spot on. I reviewed my wording of the puzzle, which said in part “Your mission is to determine how Holmes would play…” The master sleuth would be expected to make optimal plays, so the superior solution will prevail over the tiebreaker.

Rank | Name | Location | Plays To Ensure 6 NT | Percent |
---|---|---|---|---|

1 | Tim Broeken | Netherlands | 4 Q; K 8; 2 Q; 2 J | 99.29 |

2 | Dan Gheorghiu | British Columbia | 4 K; K 8; 2 A; 2 J | 99.40 |

3 | Tina Denlee | Quebec | A 2; 2 10 | 98.34 |

4 | Charles Blair | Illinois | A 3; 2 10 | 98.34 |

5 | Dominique Canneva | France | A 2; 2 10 | 98.52 |

6 | Grant Peacock | Maryland | A 2; 2 J | 98.59 |

7 | Nicholas Greer | England | A 2; 2 10 | 99.23 |

8 | Sherman Yuen | Singapore | A 2; 2 10 | 97.29 |

9 | Leigh Matheson | Australia | A 3; 2 10 | 99.76 |

10 | Gareth Birdsall | England | A 2; 2 J | 99.80 |

11 | Jonathan Mestel | England | A 2; 2 J | 99.80 |

12 | Jacco Hop | Netherlands | A 2; 2 J | 96.50 |

13 | Foster Tom | British Columbia | A 2; 2 J | 95.42 |

14 | Jim Munday | Mississippi | A 2; 2 J | 75.00 |

15 | Jon Greiman | Illinois | A 2; 2 10 | 13.50 |

After the safe club lead, declarer has 10 top tricks. Suppose you attack hearts and concede a trick to the J (unless it drops of course then you’re home). This establishes an 11th trick, and the 12th might come from a 3-2 club break, a finesse in spades or diamonds, or a squeeze. Yes, it *might*, but that’s not what we’re looking for. To obtain a lock, the heart suit must be preserved as a squeeze threat.

**Foster Tom**: Begin with a diamond finesse to establish an 11th trick and rectify the count. Then if clubs or hearts break we are done; if the same player guards both, we have a simple squeeze; if the guards are split, we have a double squeeze to win a spade.

There are two initial lines of play that ensure 12 tricks. The simplest, found by 13 of 15 successful solvers, is to win the A and immediately finesse in diamonds. If it wins, switch to hearts and claim; so assume it loses and West exits safely in diamonds (pitch a heart). Next cash the K to discover who guards clubs, then two top hearts ending in dummy to reach this position:

Notrump Win all | K 4 3 Q 10 — 6 5 | ||

North leads | A J 2 — A K Q 3 |

Lead the Q then: (1) If East follows and West guards clubs, pitch a club, then Q A K will squeeze whoever holds the Q. Otherwise, pitch a spade then (2) if clubs and hearts are guarded by the same opponent, A A K K will squeeze him; or (3) if West guards hearts and East clubs, Q A K will effect a double squeeze with North’s 4 the common threat.

**Gareth Birdsall**: Pitch a heart on the diamond return, and cash the K to find out which black suit to discard on the Q.

**Nicholas Greer**: West must win and return a minor; win it, cash a trick in the other minor and three top hearts. Before discarding on the third heart, it is known who has four clubs and who may have four hearts. If they are in the same hand, discard a spade and there is an automatic squeeze. If East has clubs and West hearts, discard a spade (vice versa, discard a club) then cross to hand with a club and play diamonds for a double squeeze.

**Tina Denlee**: When the finesse loses, West must return a minor, so win A K K A. If four clubs East, Q discarding a spade; and depending on who shows out, squeeze East in hearts-clubs, or double-squeeze West in spades-hearts, East in spades-clubs, to win the 4. If four clubs West, Q discarding a club; then I squeeze West in spades-clubs on the last diamond and end up knowing whether the Q is unguarded or with East and a winning finesse.

Tina’s continuation differs but is equally correct. If West guards both clubs and hearts, he can keep only one spade; so instead of playing to squeeze him directly, Tina opts for the proved finesse against East after cashing the K. Style points for the lady!

**Leigh Matheson**: Choice of squeeze plays follows. If East has club length, cash all the diamonds (and a club) before the top hearts for a simple or double squeeze, depending on if East follows to the third heart or not. If West has club length, cash all but one top diamond (and a club) before the top hearts for a different simple or double-squeeze option, depending on if East follows to the third heart or not.

Leigh’s continuation also differs, but it’s just as sound. When *East* guards clubs, all the diamonds can be cashed early.

Notrump Win all | K 4 3 Q 10 9 — A 6 5 | ||

West leads | A J 2 — A K 10 Q 3 2 |

If West leads either minor (pitch a heart on a diamond), win the A and lead the Q, essentially reaching the same position as before, resulting in a simple or double squeeze.

If West leads a heart, win the queen (East shows out else claim) and pitch a *spade*. With West known to guard hearts, cash the A to see who guards clubs. If West, you have a simple squeeze; if East, a double squeeze with dummy’s 4 the common threat.

**Dan Gheorghiu**: There are 10 top winners, plus one more by promoting a diamond, which rectifies the count for a squeeze — a branching function of West’s return, with the crux being to win the A and Q to discover which defender protects each suit. [Dan describes how to continue in all variations].

No other initial sequence of plays will ensure success. The most common flawed solution, submitted by eight people, was to cash a second club early; but West (with four clubs) can now lead a *third* club to kill the squeeze entry if West also guards hearts.

At double-dummy, declarer always has four heart tricks, since whoever has the J can be finessed, which means 11 top tricks. If East has both missing queens, two finesses provide 13, so West must have at least one. If East has the Q, that finesse gives 12 tricks, and whoever guards clubs can be squeezed: West in the black suits, or East in the minors. Therefore, West must have the Q.

If East has the Q, declarer also has 12 top tricks. Then if West stops hearts ( J-x-x-x-x) he can be squeezed in the red suits; or if he stops clubs, he can be squeezed in the minors. Hence East must stop *both* hearts and clubs, which means he gets squeezed. Therefore, East cannot have the Q either, so West must have both missing queens, guarded of course.

With West obliged to protect both pointed suits, it is abundantly clear that *East* must protect hearts. If West also had to keep J-x-x-x-x, he would crumble like a chalk pyramid. His only hope would be to unguard hearts on the third club, but this leads to a double squeeze with East guarding clubs, West diamonds, and spades the common suit.

What about clubs? As long as West has both guarded queens and East protects hearts, clubs can split *any way*, because a 3-2 break gives declarer only 12 tricks, and no squeeze will work when the defenders hold stoppers *behind* each of declarer’s threats. An exception occurs if West has *five* or more spades, as this means dummy’s third spade is a threat to squeeze West in the pointed suits (with clubs 3-2) so clubs must then be 4-1 or 5-0 for the defense to prevail.

That about does it, aside from a few anomalies to be noted. The West hand types that will *defeat* 7 NT are listed below, ordered by spade length, then hearts and diamonds, shortest to longest. Cards shown as ‘x’ can be any available card in the suit (except the J) since they’re all equivalent. (Specific spots might matter in lower contracts where throw-in plays are relevant, but not in 7 NT.)

Case | West Hand Type | Combinations | Hands |
---|---|---|---|

1 | Qxx — Qxxxx xxxxx | 6c2 × 5c0 × 7c4 × 5c5 | 525 |

2 | Qxx — Qxxxxx xxxx | 6c2 × 5c0 × 7c5 × 5c4 | 1575 |

3 | Qxx — Qxxxxxx xxx | 6c2 × 5c0 × 7c6 × 5c3 | 1050 |

4 | Qxx — Qxxxxxxx xx | 6c2 × 5c0 × 7c7 × 5c2 | 150 |

5 | Qxx x Qxxx xxxxx | 6c2 × 5c1 × 7c3 × 5c5 | 2625 |

6 | Qxx x Qxxxx xxxx | 6c2 × 5c1 × 7c4 × 5c4 | 13125 |

7 | Qxx x Qxxxxx xxx | 6c2 × 5c1 × 7c5 × 5c3 | 15750 |

8 | Qxx x Qxxxxxx xx | 6c2 × 5c1 × 7c6 × 5c2 | 5250 |

9 | Qxx x Qxxxxxxx x | 6c2 × 5c1 × 7c7 × 5c1 | 375 |

10 | Qxx xx Qxxxxxx x | 6c2 × 5c2 × 7c6 × 5c1 | 5250 |

11 | Qxxx — Qxxx xxxxx | 6c3 × 5c0 × 7c3 × 5c5 | 700 |

12 | Qxxx — Qxxxx xxxx | 6c3 × 5c0 × 7c4 × 5c4 | 3500 |

13 | Qxxx — Qxxxxx xxx | 6c3 × 5c0 × 7c5 × 5c3 | 4200 |

14 | Qxxx — Qxxxxxx xx | 6c3 × 5c0 × 7c6 × 5c2 | 1400 |

15 | Qxxx — Qxxxxxxx x | 6c3 × 5c0 × 7c7 × 5c1 | 100 |

16 | Qxxx x Qxx xxxxx | 6c3 × 5c1 × 7c2 × 5c5 | 2100 |

17 | Qxxx x Qxxx xxxx | 6c3 × 5c1 × 7c3 × 5c4 | 17500 |

18 | Qxxx x Qxxxx xxx | 6c3 × 5c1 × 7c4 × 5c3 | 35000 |

19 | Qxxx x Qxxxxx xx | 6c3 × 5c1 × 7c5 × 5c2 | 21000 |

20 | Qxxx x Qxxxxxx x | 6c3 × 5c1 × 7c6 × 5c1 | 3500 |

21 | Qxxx x Qxxxxxxx — | 6c3 × 5c1 × 7c7 × 5c0 | 100 |

22 | Qxxx xx Qxxxxx x | 6c3 × 5c2 × 7c5 × 5c1 | 21000 |

23 | Qxxxx — Qxxxxxx x | 6c4 × 5c0 × 7c6 × 5c1 | 525 |

24 | Qxxxx x Qxxxxx x | 6c4 × 5c1 × 7c5 × 5c1 | 7875 |

25 | Qxxxx x Qxxxxxx — | 6c4 × 5c1 × 7c6 × 5c0 | 525 |

26 | Qxxxxx — Qxxxxx x | 6c5 × 5c0 × 7c5 × 5c1 | 630 |

27 | Qxxxxx x Qxxxx x | 6c5 × 5c1 × 7c4 × 5c1 | 5250 |

28 | Qxxxxx x Qxxxxx — | 6c5 × 5c1 × 7c5 × 5c0 | 630 |

29 | Qxxxxxx — Qxxxx x | 6c6 × 5c0 × 7c4 × 5c1 | 175 |

30 | Qxxxxxx x Qxxx x | 6c6 × 5c1 × 7c3 × 5c1 | 875 |

31 | Qxxxxxx x Qxxxx — | 6c6 × 5c1 × 7c4 × 5c0 | 175 |

Total number of West hands that can defeat 7 NT | 172435 | ||

Remaining West hands out of 26c13 (10400600) possible | 10228165 | ||

Percent chance of success = 100 × 10228165 / 10400600 | ~98.34 |

Cases 10 and 22 may be surprising, as declarer has 12 top tricks with hearts running. Then with West guarding diamonds and East clubs, a double squeeze might seem to work using dummy’s third spade as the common threat. Close! It does with a heart lead, but double-dummy bridge isn’t horseshoes; the singleton club lead foils communication.

**Charles Blair**: If West is 3=2=7=1 or 4=2=6=1 with both queens and without the J, entries after a club lead prevent a double squeeze. If West is 4=2=7=0, declarer makes 7 NT on a throw-in at trick zero (it doesn’t seem right to use the term “endplay” this early). I hope there are 172,435 West hands which can set 7 NT.

Yes, and West is also “thrown in at trick zero” if he has *two* voids. After accepting a free finesse for 11 tricks, declarer can finesse East in hearts for 12, then squeeze him in clubs and hearts for 13.

**Tina Denlee**: Seven notrump makes when the spade or diamond finesse works (or queen drops), squeezing one opponent in two of three suits; or if West has 3+ hearts (he gets squeezed on the clubs). When East has J-x-x-x, I need West to have 2+ clubs because I miss an entry to handle the hearts then squeeze West — unless West has 5+ spades (then the 4 wins after the squeeze) or West has no club to lead. When East has 5+ hearts, 4+ clubs and 0-2 spades, I lose unless West has two voids and is “begin-played.” When East has 5+ hearts and 3+ spades, it is *game over*.

Too bad… you needed to upgrade to Windows 98 and choose a Fibonacci number.

**Dr. John Watson**: Chances to make 7 NT are *powerful*. Three to the second power is nine, and two to the third power is eight, so we start with 98. For the fractional part, nine to the one-half power is three, and eight to the two-thirds power is four, so we have 98.34 percent.

**Sherlock Holmes**: Yes, my dear Watson, like the logic in your book, or the ground clutter at Colonel Ross’s stable.

Acknowledgments to Sir Arthur Conan Doyle (1859-1930)

Apologies to bridge author *Louis* Watson (1907-36)

© 2016 Richard Pavlicek