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Puzzle 8S83 by Richard Pavlicek

Oh well; as South pretend you experience a temporary return of sanity:

3 NT South | 4 3 2 3 2 Q 5 4 3 Q J 10 9 | Both Vul | West Pass Pass | North 3 NT | East Pass | South 2 NT Pass | |

? K Q J 9 ? ? | ? A 10 8 7 ? ? | ||||||

Lead: K | A Q 5 6 5 4 A K 2 A K 8 7 |

West leads the K, and luckily you lose only four tricks with hearts as shown. Assume all other suit distributions are completely random, except no one can have seven spades, as that player would have bid with 7-4 shape. Assuming you choose the line of play with the best chance:

What is your percent chance to make 3 NT (1) if East wins the fourth heart? (2) if West wins the fourth heart?

Further, assume the defender who wins the fourth heart is just happenstance with no underlying motive, after which East leads a spade; West of course does not (else you could claim).

This contest ran from January 6 to February 28, 2021. There were 36 entries from 25 persons (multiple entries were allowed but only the latest one counted) of which only four were exactly right on both parts, ranked below by date and time of entry. Bridge is hardly rocket science, so I included six others who were close enough for practical purposes, ranked by least total divergence. Neat! Another Top Ten List to boost my Letterman points.

Congratulations to Sam Pahk, who was first of the four perfectas. Sam is only a recent participant and has two second-place finishes prior to this first win.

Rank | Name | Location | Part 1 | Part 2 |
---|---|---|---|---|

1 | Sam Pahk | Massachusetts | 62.81 | 70.16 |

2 | Tim Broeken | Netherlands | 62.81 | 70.16 |

3 | Tom Slater | England | 62.81 | 70.16 |

4 | Charles Blair | Illinois | 62.81 | 70.16 |

5 | Nicholas Greer | England | 62.79 | 70.14 |

6 | Aldert Westra Hoekzema | Netherlands | 62.68 | 70.16 |

7 | Grant Peacock | Maryland | 62.67 | 70.16 |

8 | Brad Johnston | New Zealand | 62.67 | 70.12 |

9 | Phillip Martin | New York | 62.76 | 70.00 |

10 | Sherman Yuen | Singapore | 62.92 | 70.00 |

After winning the fourth heart, East leads a spade. Should you finesse? Some participants thought so, because the finesse (50 percent) is more likely than a 3-3 diamond break (~38 percent); but the latter also avails the chance of a squeeze, which *combined* is much better. On the mark:

**Nicholas Greer**: The best line is to take the A and play for 3-3 diamonds or a squeeze.

**Brad Johnston**: Rising with the A will win any time diamonds are 3-3 or the K is with the long diamonds.

To calculate the exact chance, the first step is to determine the number of *possible* West hands (or layouts since East must have the rest). There are 18 cards to distribute (7 spades, 6 diamonds, 5 clubs) and West must get 9 of them, so there are 18c9 (18 choose 9) = 48620 combinations. *But*, we must subtract from this all hands considered to be impossible; i.e., West cannot have *seven* spades (11c2 = 55 hands) or a spade *void* (11c9 = 55 hands). Hence the number of West hands deemed possible is 48620 - 55 - 55 = 48510.

The next step is to determine the number of *successful* West hands, which I have ordered by diamond length in the following table:

Case | West hand | Combinations | Hands |
---|---|---|---|

1 | - (no K) | 6c0 × 11c9 | 55 |

2 | x (no K) | 6c1 × 11c8 | 990 |

3 | xx (no K) | 6c2 × 11c7 | 4950 |

4 | xxx | 6c3 × 12c6 | 18480 |

5 | xxxx K | 6c4 × 11c4 | 4950 |

6 | xxxxx K | 6c5 × 11c3 | 990 |

7 | xxxxxx K | 6c6 × 11c2 | 55 |

Successful hands | Sum of Cases 1-7 | 30470 | |

Successful percent | 100 × 30470/48510 | 62.81 |

If West is short in diamonds without the K (Cases 1-3) East will be squeezed. If West has exactly three diamonds (Case 4) nothing else matters. If West is long in diamonds with the K (Cases 5-7) West will be squeezed. Summing the cases gives 30470 successful hands (out of 48510).

**Sam Pahk**: I took all the possible distributions and found successful outcomes.

**Tom Slater**: I’ve included the instant claim when East follows instructions and tables his singleton K. If you’re asking the question after East leads a *lower* spade, the chance of success drops slightly — I think to 62.68 percent — because a singleton K in East is a friendly layout.

Tom is right on both accounts and decided correctly, since the puzzle says “leads a spade” with no indication of rank. If told a *lower* spade, all East hands with a singleton K (11c8 = 165) will be impossible and must be subtracted from both successful and possible totals, leaving 30305/48345, or ~62.68 percent. Confusion over this evidently led several solvers slightly astray.

**Charles Blair**: An omniscient East, with short diamonds and 4+ spades without the king, might want to return a non-spade to preserve South’s losing option.

True, but an omniscient *declarer* would play for the squeeze anyway; there’s something about Greek gifts.

Without the committing spade lead from East your chances improve, because you can *postpone* the finesse-or-squeeze option until you discover the distribution, and you will always know the exact division of every suit. If *East* has longer diamonds, this is academic, because he’ll be forced to blank the K; but if *West* has longer diamonds, you will play for the squeeze if he also has longer spades, or take the finesse if East has longer spades. Some other explanations:

**Phillip Martin**: If West wins the fourth heart, you make if diamonds are 3-3, if East has long diamonds and the K, or if West has long diamonds and the hand with longer spades has the K.

**Nicholas Greer**: Declarer can cash clubs and diamonds ending in dummy, with the option of dropping the K (after a presumed squeeze) or finessing according to who has longer spades.

**Brad Johnston**: Now I have the luxury to test both minors before touching spades. When West has longer diamonds and longer spades, play for the squeeze; but more likely East has longer spades, then you finesse.

To calculate the exact chance, we can skip the first step, since we’ve already determined there are 48510 possible West hands. Counting *successful* hands is more complicated, because hands with 4+ diamonds need to be broken down by club length to define the number of spades, as Cases 5-16 in the table below. Note that Cases 1-4 are identical to Part 1.

Case | West hand | Combinations | Hands |
---|---|---|---|

1 | - (no K) | 6c0 × 11c9 | 55 |

2 | x (no K) | 6c1 × 11c8 | 990 |

3 | xx (no K) | 6c2 × 11c7 | 4950 |

4 | xxx | 6c3 × 12c6 | 18480 |

5 | xxxx - Kxxxx | 6c4 × 5c0 × 6c4 | 225 |

6 | xxxx x Kxxx | 6c4 × 5c1 × 6c3 | 1500 |

7 | xxxx xx xxx | 6c4 × 5c2 × 6c3 | 3000 |

8 | xxxx xxx xx | 6c4 × 5c3 × 6c2 | 2250 |

9 | xxxx xxxx x | 6c4 × 5c4 × 6c1 | 450 |

10 | xxxxx - Kxxx | 6c5 × 5c0 × 6c3 | 120 |

11 | xxxxx x xxx | 6c5 × 5c1 × 6c3 | 600 |

12 | xxxxx xx xx | 6c5 × 5c2 × 6c2 | 900 |

13 | xxxxx xxx x | 6c5 × 5c3 × 6c1 | 360 |

14 | xxxxxx - xxx | 6c6 × 5c0 × 6c3 | 20 |

15 | xxxxxx x xx | 6c6 × 5c1 × 6c2 | 75 |

16 | xxxxxx xx x | 6c6 × 5c2 × 6c1 | 60 |

Successful hands | Sum of Cases 1-16 | 34035 | |

Successful percent | 100 × 34035/48510 | 70.16 |

Cases 5-9 separate West hands with four diamonds. If West has one or fewer clubs (Cases 5-6) he must have 4+ spades, so you will play for the squeeze. If West has two or more clubs (Cases 7-9) East will have 4+ spades, so you will take the finesse. Similar reasoning applies for five diamonds (Cases 10-13) and six diamonds (Cases 14-16).

Basing your play on who has longer spades is clearly better, but the advantage in actual play is reduced, because a failing squeeze means down one, while a failing finesse means down two. Aha! I can fix this: Change to “plus or fishfood” scoring, then down one or two (or nine) doesn’t matter. It’s lunch time! The Piranha Strike Back

**Sam Pahk**: I hope I didn’t mess up my combinatorics.

**Nicholas Greer**: My arithmetic has probably been rubbish, but these are the numbers I get.

**Grant Peacock**: I know *good* players do not calculate exact odds, as they are better at logical inferences; and I am eager to enter this contest… not sure what that says about me.

**Brad Johnston**: This took long to calculate using my method, so I’m interested in seeing what “clean” solutions people find.

Weightlifting comes to mind, so the “clean” will be followed by the jerk…

**The Donald**: How dare you not list me, *Mister RP loser!* A finesse might be 50-50 for some halfwit Democrat, but I have clout. For me it’s closer to *80 percent*, which tops your friggin’ leaderboard! Rudy and I will sue!

© 2021 Richard Pavlicek