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Puzzle 8S85 by Richard Pavlicek

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

5 East | K Q 10 9 8 7 6 4 K 9 10 9 4 | None Vul | West1 Dbl | North3 4 | East4 5 | South4 All Pass | |

A 5 3 2 Q 9 8 5 A Q 4 2 5 | J — J 7 6 3 A K J 8 7 6 3 2 | ||||||

Lead: K | 4 A K J 10 7 6 3 2 10 8 5 Q |

Board 8 was played 125 times with 26 different results. 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. But as usual when both sides have wild distribution, competition was fierce, and N-S were often set multiple tricks doubled.

*But who really cares!* Instead, I offer you a probability puzzle:

What is the percent chance that a deal contains: | |

1. Exactly two eight-card suits | 0. |

2. At least two eight-card suits | 0. |

3. At least two eight-card or longer suits | 0. |

All percents are 0.xxxx (you guess the xxxx).

This puzzle contest ran August 3-31, 2021. There were 32 entries from 20 persons (multiple entries were allowed but only the latest one counted) of which only three were exactly right on all three parts, ranked below by date and time of entry. Another four came close enough for practical purposes, ranked by least total divergence. The rest were way off, and some appeared to be random guesses.

Congratulations to Peter Randall of England, who now holds the unbeatable record of winning *every contest* he entered at this web site. Yes, this was his only entry, but still remarkable to win on the first try. Talk about having 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 |

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 | 35332 | 15717541316474557292763888 ~ 15.7 septillion | 0.029299316151 |

2. Two+ 8 card suits | 39847 | 15983279046544308018793416 ~ 16.0 septillion | 0.029794682037 |

3. Two+ 8+ card suits | 128398 | 19392355630901103879273984 ~ 19.4 septillion | 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.

And I suspect too many vodka martinis.

© 2021 Richard Pavlicek