Puzzle 8S85 Main |
| by Richard Pavlicek |

Or a lame ex-president?

The following deal was brought to my attention by Mike Cassel of Roseville, Minnesota. It was Board 8 in the Common Game duplicate on Friday, July 30, 2021. The presence of *two* eight-card suits on *Board 8* caught my fancy.

Matchpoints | K Q 10 9 8 7 6 | West | North | East | South
| |

None vul | 4 | 1 | 3 | 4 | 4 | |

K 9 | Dbl | 4 | 5 | Pass | ||

10 9 4 | Pass | Pass | ||||

A 5 3 2 | J | |||||

Q 9 8 5 | — | |||||

A Q 4 2 | J 7 6 3 | |||||

5 | A K J 8 7 6 3 2 | |||||

4 | ||||||

A K J 10 7 6 3 2
| ||||||

Lead: K | 10 8 5 | |||||

5 East | Q |

Board 8 was played 125 times with 26 different results, which is surely to be expected with wild distribution spawning fierce competition. Most common (30 times) was E-W making 5 or 5 , 20 times with an overtrick. The slam was bid and made 19 times, of which six were doubled. And of course N-S were often set multiple tricks doubled in spades or hearts.

*But who really cares!* Instead, I offer you a three-part puzzle. Challenge yourself, or make your best guesses (A-H).

What is the percent chance that a random deal contains:

1. Exactly two eight-card suits?

A. 0.0198 B. 0.0244 C. 0.0293 D. 0.0298 | E. 0.0307 F. 0.0361 G. 0.0409 H. 0.0436 |

2. At least two eight-card suits?

3. At least two eight-card or longer suits?

Top Two Eight-Baggers |

*In the original contest solvers had to submit their own answers. There was no multiple choice.

Congratulations to Peter Randall of England, who now holds the unbeatable record of winning *every contest* he has entered at this web site. OK, OK, this was his only entry, but winning on the first try is notable. Now he has no where to go but *down!*

Rank | Name | Location | Part 1 | Part 2 | Part 3 |
---|---|---|---|---|---|

1 | Peter Randall | England | 0.0293 | 0.0298 | 0.0361 |

2 | Martin Vodicka | Slovakia | 0.0293 | 0.0298 | 0.0361 |

3 | Paul Gilbert | England | 0.0293 | 0.0298 | 0.0361 |

4 | Grant Peacock | Maryland | 0.0293 | 0.0298 | 0.0362 |

5 | Sherman Yuen | Singapore | 0.0295 | 0.0300 | 0.0364 |

6 | Charles Blair | Illinois | 0.0295 | 0.0299 | 0.0367 |

7 | Charles Brenner | California | 0.0293 | 0.0294 | 0.0356 |

Puzzle 8S85 Main | Top Two Eight-Baggers |

Solving this puzzle by combinatorials and summations is possible but way too messy for my liking, let alone error prone from oversights. I look for shortcuts, and for this task I turned to my compilation of dealprints. Many years ago I discovered there are exactly 37,478,624 specific dealprints (see Dealprints and Matrices) which, if you ignore suit identity and rotations, can be reduced to 393,197 distinct generic dealprints. Therefore, it was a simple matter — for my computer, not for me — to parse the data for qualifying dealprints and calculate the number of deals represented by each. The results:

Requirement | Dealprints | Number of Deals | Percent |
---|---|---|---|

1. Two 8 card suits | 35,332 | 15,717,541,316,474,557,292,763,888 | 0.029299316151 |

2. Two+ 8 card suits | 39,847 | 15,983,279,046,544,308,018,793,416 | 0.029794682037 |

3. Two+ 8+ card suits | 128,398 | 19,392,355,630,901,103,879,273,984 | 0.036149595354 |

Each percent is 100 times the number of deals shown divided by the total number of bridge deals: 52!/(13!)^{4}, which is an incomprehensible number of 29 digits (53+ octillion). To put this into perspective, the number of *seconds* before our Sun dies out, along with all living things, is only an 18-digit number. Percents are shown to 12 places but provide the winning answers when rounded to four places.

**Martin Vodicka**: In Part 1, I also counted deals where there are exactly two 8-card suits *and* a longer suit.

However, [discounting this] only makes a difference in the fourth decimal place, reducing it to 0.0291.

Martin judged correctly. In fact it is possible for “exactly two 8-card suits” to coexist with two *13-card* suits.

**Peter Randall**: [Deal counts identical to my table]. The percentage in Part 3 is rounded down from 0.03614960.

Indeed it is, and the fact that it lies almost midway between 0.0361 and 0.0362 caused consternation for some solvers, or at least ruffled their feathers. Speaking of which:

**Grant Peacock**: I took a Monte Carlo approach… For Part 3, after 1 billion deals, 0.0361 looked like the answer… but after 10 billion it kept dancing around 0.03615…and after 70 billion I think the probability is pretty high it will round up to 0.0362… plus I’m getting tired of hearing my PC fan spin, and I don’t need the extra heat in the summer.

**Charles Brenner**: I probably need to clean up my code to get these right. But maybe I’m lucky.

**Martin Vodicka**: I suspect I forgot to multiply by four somewhere, so trying to be precise will be useless anyway.

You know what I suspect? *Too many vodka martinis!*

Puzzle 8S85 Main | Top Two Eight-Baggers |

© 2021 Richard Pavlicek