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Can You Solve The Puzzle Of The Pirate Booty?

Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here: math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world — including you! You’ll find this week’s puzzle below.

Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏 Michael Loginov 👏 of Menlo Park, California, our big winner. You can find a solution to the previous Riddler at the bottom of this post.

Now here’s this week’s Riddler, a high-seas piracy puzzle that comes to us from Kyle Joecken, a data scientist from Portland, Oregon.

Ten Perfectly Rational Pirate Logicians (PRPLs) find 10 (indivisible) gold pieces and wish to distribute the booty among themselves.

The pirates each have a unique rank, from the captain on down. The captain puts forth the first plan to divide up the gold, whereupon the pirates (including the captain) vote. If at least half the pirates vote for the plan, it is enacted, and the gold is distributed accordingly. If the plan gets fewer than half the votes, however, the captain is killed, the second-in-command is promoted, and the process starts over. (They’re mutinous, these PRPLs.)

Pirates always vote by the following rules, with the earliest rule taking precedence in a conflict:

1. Self-preservation: A pirate values his life above all else.
2. Greed: A pirate seeks as much gold as possible.
3. Bloodthirst: Failing a threat to his life or bounty, a pirate always votes to kill.

Under this system, how do the PRPLs divide up their gold?

Extra credit: Solve the generalized problem where there are P pirates and G gold pieces.

Need a hint? You can try asking me nicely. Want to submit a new puzzle or problem? Email me.

And here’s the answer to last week’s Riddler, concerning a baseball division with five evenly matched teams, which came to us from Nick Keenan.

As many of you quickly noticed, how many wins the division-winner would have rests on assumptions about the teams’ schedules: the number of games they play against teams in their own division, the number of games they play against teams outside of it, and so on. The problem was a mini-meta-lesson in modeling.

A good place to start is with the (over-)simple model in which the schedule doesn’t matter, one in which each game is decided by a coin flip — heads, the team “wins”; tails, it “loses.” Laurent Lessard takes us through that math (the analytic solutions get hairy pretty quickly), as do Christoph Walsh and Andrew Mascioli. Under the “coin-flip” assumption, this is the distribution of the number of wins of the division champ (the teams play 162 games in the season):

The average number of wins by the division champ (i.e., the “solution”) in this case is about 88.4. These are the distributions of wins for all the teams:

You can simulate these simple seasons pretty easily with a few lines of code in the stats software R. Here’s Austin Williams’s version:

```store<-matrix(0,100000,1)

for (i in 1:100000) {
A <- rbinom(5, 162, .5)
store[i] <- max(A)
}

mean(store)```

But what about a more realistic schedule? Nothing above, for example, precludes the possibility, though slim, that every team could win all its games. In real life, this is not only unlikely but also impossible — a win for one team is a loss for another.

Geoff Buchan takes us through two other types of schedules, via computer simulation. First, using the actual 2016 Major League Baseball schedule, he found that the division champ averaged 88.8 wins. Second, he simulated an even more restrictive case, one in which the division’s teams only play one another, which means that every win for a team in the division is a loss for another team in the division. In that case, the champ averaged 89.3 wins. (Buchan’s Perl program for these simulations can be downloaded here. Christopher Long has some baseball-season-simulating Python code.)

However, intricate assumptions may not benefit us much in this case.

Lessard explains that in a typical 162-game season, the five AL East teams play 19 games against each of the other four teams for a total of 76 games. The other 86 games are against non-division opponents. The large share of out-of-division games drowns out the difference between the distributions under the coin-flip assumption and the real-world-schedule assumption. “In other words, the assumption that each team simply flips 162 coins turns out to be a good one,” he wrote.

Zach Wissner-Gross shows how the spread between the first-place team (green) and last-place team (blue) widens as the number of divisional games grows.

The 🏆 Coolest Riddler Extension Award 🏆 goes to Roger Strong. Roger observed that the most wins in a real, 162-game baseball season was 116, by the 2001 Seattle Mariners. He then asked how many teams would have to be in the division for the expected number of wins by the division champ to be at least 117. Very clever! He argued that it would take a division of 261 million evenly matched teams.

Finally, because #RiddlerNation is merely one sovereignty within the weird, wide puzzling world, I’d like to start highlighting some other puzzling goings-on each week, in the interest of strengthened foreign relations. The inaugural bunch:

• There is an internet Stonehenge on Reddit — a mysterious subreddit filled with “seemingly random strings of numbers and text.” [The Kernel]
• “Have you ever encountered a problem that seemed to require a lengthy calculation only to realize that, with a simple shortcut, you can solve it in your head?” [Quanta Magazine]
• Edward Frenkel, Georg Cantor and some puzzles on sets and infinity. [The New York Times]
• Puzzles about dots and lines. [The Guardian]
• Cutely designed problems about averages. [Expii]

Have a wonderful weekend!

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.