My dear Watson, I was skimming through your book this morning and spotted an oversight. On page 476 you state that the best contract for North-South is seven clubs. I must disagree. [Holmes puts down his pipe and pulls out a notepad from his vest pocket].
I wrote down my auction on this pad to reach six notrump, which is ironclad, while seven clubs depends on a normal trump break and even then it could fail against some foul distributions.
Good heavens, Holmes! That was 30 years ago, and Ive had no time for bridge with my medical career. Most of it is long forgotten, but I remember that a 3-2 break is a favorite nearly 68 percent if I recall so seven clubs should make more than twice as often as not.
Yes, but numbers dont lie. Suppose the deal is played 100 times at total points. Ill even give you the benefit of doubt that seven clubs, 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 six notrump, 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 six notrump truly make every time? Surely it might fail against some bad 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.
How would Holmes play to guarantee 6 NT?
Assume a club lead with East following, and subsequently East and West will follow low or discard safely. Further, assume no suit will break if you need it, and no finesse will work if you take it.
Enter the plays you would make and click Verify. Use the help provided to make corrections and repeat. See how many tries it takes you to discover Holmess perfect play.
Tiebreaker: Before the lead, what is the percent chance to make 7 NT with double-dummy play all-around?94.44 95.53 96.67 97.29 98.34 99.29
In July 2016 this puzzle was presented as a contest, inviting anyone who wished to submit a solution. Participation was down slightly from June, with 51 persons giving it a try, of which only the 15 listed found a way to guarantee 6 NT.
Congratulations to Tim Broeken for his umpteenth win (well, its only 10, but far and away the leader). He and Dan Gheorghiu were the only solvers to submit the optimal line of play, not only ensuring 6 NT but allowing 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 asked How would Holmes play
The master sleuth would be expected to make optimal plays, so the superior solution will prevail over the tiebreaker.
The instructive merit of my puzzles is typically zero, but every now and then some useful bridge content slips in. I think my last puzzle to accomplish this was Two-Way Finesses back in 2010, which kind of fits the pattern. (My mother reassured everyone I would do something useful every six years.)
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 youre 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 thats not what were 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:
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 Norths 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.
Tinas 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.
Leighs continuation also differs, but its just as sound. When East guards clubs, all the diamonds can be cashed early.
While the above play guarantees 12 tricks, the best play is to win the club lead in hand, cash two hearts ending in dummy, and then finesse the J. This also guarantees 12 tricks but avoids losing an unnecessary diamond trick when the J drops early and clubs split 3-2. Assume the diamond finesse loses (else establish hearts and claim) to put West on lead in this ending:
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 dummys 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 Wests 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.
The tiebreaker was difficult, as offering anything but a ballpark guess required considerable effort, arguably a waste of time. Fortunately or unfortunately [pick one] several solvers have priorities like I do: If its necessary, it can wait (like taking out the trash); if its useless, get on with it! Four people found the correct answer of 98.34 percent, but only two (Tina Denlee and Charles Blair) were also correct on the main puzzle, a requirement to make the leaderboard.
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 declarers threats. An exception occurs if West has five or more spades, as this means dummys 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 theyre all equivalent. (Specific spots might matter in lower contracts where throw-in plays are relevant, but not in 7 NT.)
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 dummys third spade as the common threat. Close! It does with a heart lead, but double-dummy bridge isnt 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 doesnt 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.
Foster Tom: You asked for my best guess (95.42) so I wasnt precise. I imagined that favorable outcomes in each suit were independent, tacked on a few points for double-dummy play, and chose a Catalan number for the hundredths.
you needed to go back to Windows 98 and append a Fibonacci number (34).
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 just like the logic in your book, or the ground clutter at Colonel Rosss stable.
Acknowledgments to Sir Arthur Conan Doyle (1859-1930)Apologies to bridge author Louis Watson (1907-36)© 2016 Richard Pavlicek