Can You Throw the Hammer?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “Extra Credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

This Week’s Fiddler

Chris Payne recently introduced me to “TGL” golf, which looks bonkers. Here’s a puzzle that’s inspired by one of the rules from TGL:

You and your opponent are competing in a golf match. On any given hole you play, each of you has a 50 percent chance of winning the hole (and a zero percent chance of tying). That said, scorekeeping in this match is a little different from the norm.

Each hole is worth 1 point. Before starting each hole, either you or your opponent can “throw the hammer.” When the hammer has been thrown, whoever did not throw the hammer must either accept the hammer or reject it. If they accept the hammer, the hole is worth 2 points. If they reject the hammer, they concede the hole and its 1 point to their opponent. Both players can throw the hammer on as many holes as they wish. (Should both players decide to throw the hammer at the exact same time—something that can’t be planned in advance—the hole is worth 2 points.)

The first player to reach 3 points wins the match. Suppose all players always make rational decisions to maximize their own chances of winning.

Good news! You have won the first hole, and now lead 1-0. What is your probability of winning the match?

This Week’s Extra Credit

Instead of playing to 3 points, now the first player to 5 points wins the match.

Good news (again)! You have won the first hole, and now lead 1-0. What is your probability of winning the match?

Making the ⌊Rounds⌉

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. Last week, I shared a question from LearnedLeague’s ongoing mathematically-themed MiniLeague. This week, I’m sharing another one I encountered this past Monday:

Magnus Carlsen is seemingly bored with classical chess but recently played a showcase series of Fischer Random Chess, a variant where players start with their pawns in the normal position, but their pieces on the back row in a random order. The only constraints on the position are that the bishops must be on differently colored squares, and the king must be between the two rooks. This variant is also called ChessN, where N is the number of possible starting positions for one player. What is N?

Feel free to discuss this puzzle and your approach in the comments below!

Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

Standings

I’m tracking submissions from paid subscribers and compiling a leaderboard, which I’ll reset every quarter. All timely correct solutions to Fiddlers and Extra Credits are worth 1 point each. At the end of each quarter, I’ll 👑 crown 👑 the finest of Fiddlers. (If you think you see a mistake in the standings, kindly let me know.)

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Sander & Bas 🎻 from Voorthuizen, Netherlands. I received 82 timely submissions, of which 76 were correct—good for a 93 percent solve rate.

Last week, you had three colors of hibiscus: red, orange, and yellow. You wanted to plant them in a straight hedge of shrubs (each of which was one color) so that the order appeared somewhat random, but not truly random. More specifically, you wanted the following to be true:

• No two adjacent shrubs had the same color.

• No ordering of three consecutive shrubs appeared more than once in the hedge. (But a prior ordering could appear in reverse. For example, ROYOR was an acceptable hedge, but ROYROY was not.)

What was the greatest number of shrubs your hedge could contain?

A good first step was to work out an upper bound, or a reasonable maximum number of shrubs the hedge could contain. To do that, you could count up how many ways you could have three consecutive shrubs, without adjacent shrubs of the same color. There were three possible colors for the first shrub, two possible colors for the second shrub (i.e., you couldn’t repeat the first shrub’s color), and two possible colors for the third shrub (i.e., you couldn’t repeat the second shrub’s color). In total, there were 3·2·2 = 12 distinct trios of shrubs.

Another way to count them all was to note the six trios in which all three colors were different:

• ROY

• RYO

• ORY

• OYR

• YRO

• YOR

And the six trios in which the first and third colors were the same:

• ROR

• RYR

• ORO

• OYO

• YRY

• YOY

The longest possible hedge included all 12 of these trios. A 13th trio would have been a duplicate of one of these. How many shrubs were in a hedge with 12 trios? It took three shrubs to make the first trio, and then each additional shrub after that resulted in another trio. That meant there were 3+11, or 14 total shrubs.

Remember, this was an upper bound, the longest possible hedge that contained every possible trio of shrubs. So the answer was at most 14. But you still had to demonstrate that such a hedge was possible. If it was, the answer was 14; otherwise, it was some value less than 14.

Before revealing the solution, let’s look at an alternative framing of the problem that made use of graph theory. Like the puzzle from two weeks ago, this problem was equivalent to traversing a graph. Graphs are made up of vertices connected by edges. In our graph, each trio of shrubs could be represented by a vertex. And one vertex pointed to another (along a “directed” edge) if the last two colors of the first were the same as the first two colors of the second.

That’s a lot to take in at once, but it made more sense when you studied the following graph, courtesy of solver 🎬 Dennis Okon 🎬 (who used the letters “ABC” instead of “ROY”):

When one vertex pointed to another, that meant the trio being pointed to could immediately follow the trio doing the pointing. For example, ABC (ROY) pointed to BCB (OYO), which meant they could be combined to form ROYO. Meanwhile, ROY also pointed to OYR, which meant they could be combined to form ROYR.

As you can see, there are 12 vertices in the graph, corresponding to the 12 trios we listed above. The question was now to find the longest path in the graph that visited each vertex at most once. After all, visiting a vertex twice meant repeating a trio of colors in the hedge. Ideally, there was a path that visited every vertex in the graph once. Such a path, which was similarly encountered in the puzzle from two weeks ago, is called a Hamiltonian path.

As it turned out, this graph had quite a few Hamiltonian paths, each corresponding to a valid hedge with 14 shrubs. The yellow directed edges in Dennis’ graph above show one such path: RORYROYRYOYORO.

Solver Sameer Gauria went further and computed the total number of valid hedges—a question that would have made a suitable Extra Credit! According to Sameer, this number turned out to be 144.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Ryan Lafitte 🎻 from Tucker, Georgia. I received 51 timely submissions, of which 46 were correct—good for a 90 percent solve rate.

In addition to red, orange, and yellow hibiscus flowers, a fourth color was now in play: pink. Again, you wanted to plant a straight hedge of shrubs that appeared somewhat random. Here were the rules for ordering the shrubs this time:

• No two adjacent shrubs had the same color.

• No ordering of four consecutive shrubs appeared more than once in the hedge. (Again, a prior ordering could appear in reverse.)

• Among any group of four consecutive shrubs, at least three distinct colors were represented.

What was the greatest number of shrubs your hedge could contain?