Puzzle 8S85 Main

# Two Eight-Baggers

by Richard Pavlicek

### What do you call an eight-card suit?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. 0.0198B. 0.0244C. 0.0293D. 0.0298 E. 0.0307F. 0.0361G. 0.0409H. 0.0436

2. At least two eight-card suits?

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

 Top Two Eight-Baggers

## Peter Randall Wins

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.

*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!

Winner List
RankNameLocationPart 1Part 2Part 3
1Peter RandallEngland0.02930.02980.0361
2Martin VodickaSlovakia0.02930.02980.0361
3Paul GilbertEngland0.02930.02980.0361
4Grant PeacockMaryland0.02930.02980.0362
5Sherman YuenSingapore0.02950.03000.0364
6Charles BlairIllinois0.02950.02990.0367
7Charles BrennerCalifornia0.02930.02940.0356

 Puzzle 8S85 Main Top Two Eight-Baggers

## Solution

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:

RequirementDealprintsNumber of DealsPercent
1. Two 8 card suits35,33215,717,541,316,474,557,292,763,8880.029299316151
2. Two+ 8 card suits39,84715,983,279,046,544,308,018,793,4160.029794682037
3. Two+ 8+ card suits128,39819,392,355,630,901,103,879,273,9840.036149595354

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.

## Behind the 8-ball

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