Puzzle 8S83 Main |
| by Richard Pavlicek |

(and if you believe that, chances are you’re crazy)

If by chance you dropped in for a bridge puzzle, chances are you’ll get one.

If by chance you think clearly, chances are you’ll solve it. *So take a chance!*

As South, you are declarer in 3 NT with a heart lead:

IMPs | 4 3 2 | West | North | East | South | |

N-S vul | 3 2 | 2 NT^{1}
| ||||

Q 5 4 3 | Pass | 3 NT | Pass | Pass | ||

Q J 10 9 | Pass | |||||

? | ? | |||||

K Q J 9 | A 10 8 7 | 1. 20-21 | ||||

? | ? | |||||

? | ? | |||||

A Q 5 | ||||||

6 5 4 | ||||||

Lead: K | A K 2 | |||||

3 NT South | A K 8 7 |

Fortunately, East-West cash only four tricks with hearts splitting as shown. How do you like your chances?

Assume all other suit distributions are random, except no one can have seven spades (that player surely would have bid with 7-4 shape). Further, whoever wins the fourth heart is merely happenstance with no underlying motive, after which East will lead a spade, but West of course will not (else you could claim). Assuming you choose the best line of play:

What is your percent chance to make 3 NT…

1. If East wins the fourth heart?

A. 62.81 B. 64.55 C. 70.16 D. 72.33 |

2. If West wins the fourth heart?

A. 62.81 B. 64.55 C. 70.16 D. 72.33 |

Top Chances Are… |

Congratulations to Samuel 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 | Samuel 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 |

Puzzle 8S83 Main | Top Chances Are… |

What is the chance to make 3 NT after East-West win the first four heart tricks?

IMPs | 4 3 2 | West | North | East | South | |

N-S vul | 3 2 | 2 NT^{1}
| ||||

Q 5 4 3 | Pass | 3 NT | Pass | Pass | ||

Q J 10 9 | Pass | |||||

? | ? | |||||

K Q J 9 | A 10 8 7 | 1. 20-21 | ||||

? | ? | |||||

? | ? | |||||

A Q 5 | ||||||

6 5 4 | ||||||

Lead: K | A K 2 | |||||

3 NT South | A K 8 7 |

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 = 48620 combinations. But we must subtract from this all hands deemed to be impossible: West cannot have *seven* spades (11c2 = 55 hands) or a spade *void* (11c9 = 55 hands). Hence the number of possible West hands becomes 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 | 18,480 |

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 | 30,470 | |

Successful percent | 100 × 30,470/48,510 | 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 30,470 successful hands out of 48,510, or 62.81 percent (to the nearest 100th).

**Samuel 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 the question is after East leads a *lower* spade, the chance 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. What’s that story 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 | 18,480 |

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 | 34,035 | |

Successful percent | 100 × 34035/48,510 | 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 the play on who has longer spades is clearly better, but the advantage is reduced in actual play. A failing squeeze means down only one, while a failing finesse means down two. Aha! I can fix that: Change the scoring to *Plus or Fishfood*, and down one or two or *nine* won’t matter. It’s lunch time! The Piranha Strike Back

**Samuel 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,

yet I am eager to enter this contest. Not sure what that says about me!

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

Weightlifting comes to mind, so what else but to follow “clean” with 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!

Puzzle 8S83 Main | Top Chances Are… |

© 2021 Richard Pavlicek