Welcome to Fiddler on the Proof, the spiritual successor to FiveThirtyEight’s The Riddler column.
Every Friday morning, I present mathematical puzzles intended to challenge and delight you. Most can be solved with careful thought, pencil and paper, and the aid of a calculator. The “Extra Credit” is where the analysis typically gets 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 puzzles are 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
With the regular season over, there are two clear favorites for baseball’s American League Most Valuable Player (MVP) award according to ESPN:
Aaron Judge of the New York Yankees, whose odds are -150.
Cal Raleigh of the Seattle Mariners, whose odds are +110.
The “-150” odds mean that for every $150 you bet on Judge to win MVP, you’ll earn $100 if he actually wins. The “+110” odds mean that for every $100 you bet on Raleigh to win MVP, you’ll win $110 if he actually wins. Note the differing results (winning $100 vs. betting $100) that comes with the odds being negative vs. positive.
While these betting lines may be informed by an assessment of Judge’s and Raleigh’s real chances, they may also be informed by how much money people are betting on each player.
Suppose all bettors have wagered on either Judge or Raleigh with the odds above. Some fraction f of dollars wagered have been in favor of Judge, while 1−f has been wagered on Raleigh. For what fraction f will the oddsmaker earn the same amount of money, regardless of which player earns the MVP award?
This Week’s Extra Credit
Suppose there are two leading candidates, A and B, for MVP in the Fiddler Baseball League. There are two parts to this Extra Credit, so please read carefully!
Part 1:
The odds for A winning the award have been set to +100x, where x > 1.
Let f represent the fraction of dollars wagered in favor of A. For many values of f, the oddsmaker can set the odds for B so that they’ll make the same amount of money regardless of whether A or B wins the award. However, below a certain value of f, it’s impossible for the oddsmaker to do this.
What is this critical value of f? (Your answer should be in terms of x.)
Part 2:
Now, the odds for A winning the award have been set to -100y, where y >1. Again, for many values of f, the oddsmaker can set the odds for B so they’ll make the same amount whether A or B wins the award.
What is the critical value of f below which this isn’t possible? (Your answer should be in terms of y.)
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. This week, I’m spreading the news about an upcoming event at MoMath (the National Museum of Mathematics in New York City).
The MoMath Annual Puzzle-hunt Series (MAPS) will be taking place on Saturday, November 8 from 2 p.m. to 5 p.m. This appears to be a mini-hunt in the style of the famed MIT Mystery Hunt, but with a mathematical theme.
It looks like a lot of fun, so let me know if you plan on attending! (I’m not sure if I’ll be participating just yet.)
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 correct solutions to Fiddlers and Extra Credits are worth 1 point each. Solutions should be sent prior to 11:59 p.m. the Monday after puzzles are released. 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: 🎻 Dave Polidori 🎻 from Rancho Santa Fe, California. I received 69 timely submissions, of which 61 were correct—good for an 88 percent solve rate.
Last week, you started at the center of the unit square and then picked a random direction to move in, with all directions being equally likely. You moved along this chosen direction until you reached a point on the perimeter of the unit square.
On average, how far could you have expected to travel?
Along the square’s perimeter, the midpoints of the four sides were closest to the center, each a distance 1/2 away. Meanwhile, the four corners were farthest from the center, each a distance √2/2 away. So the average distance had to be somewhere between 0.5 and 0.7071 (or thereabouts). A few readers averaged these two extremes, but it wasn’t quite that simple.
It might also have been tempting to compute the average the distances between the square’s center and all the points along its perimeter, equally weighting along the perimeter. By symmetry, you only had to consider the points along one of the square’s four sides, such as the topmost side. If the square was centered at the origin, then points along the top side had coordinates (x, 0.5), with -0.5 ≤ x ≤ 0.5. By the Pythagorean theorem, the distance between each of these points and the square’s center was √(x2 + 0.25). Averaging this expression over the range -0.5 ≤ x ≤ 0.5 (via integration) gave a numerical result of approximately 0.5739.
However, while this was in the right ballpark (i.e., between 0.5 and 0.7071), it was not the correct answer. Why? Because the puzzle stated that each direction was equally (and uniformly) likely, not each point along the perimeter. And when each direction was equally likely, the middles of the sides were sampled to a greater effect than the corners, since the middles represented greater swings in angle. Thus, the answer had to be less than 0.5739.
By symmetry, let’s just consider the right side of the square. If the positive x-axis represented an angle of 0 radians (0 degrees), then you reached the right side if your angular direction was between -𝜋/4 radians (-45 degrees) and 𝜋/4 radians (45 degrees). For an angle 𝜃 in this region, the distance to the perimeter wound up being 1/(2·cos(𝜃)). Averaging this value over the region -𝜋/4 < 𝜃 < 𝜋/4 gave the following integral expression:
Evaluating this integral gave a numerical result of 0.5611. The exact form wasn’t too unfriendly to look at: 2/𝜋 · ln(1+√2). (If you accidentally doubled this answer because you assumed the square had a side length of 2, I still awarded credit.)
Last Week’s Extra Credit
Congratulations to the (randomly selected) winner from last week: 🎻 Jussi Lehtonen 🎻 from Jyväskylä, Finland. I received 43 timely submissions, of which 24 were correct—good for a 56 percent solve rate.
This time around, you started at the center of a unit cube. Again, you picked a random direction to move in, with all directions being equally likely. You moved along this direction until you reached a point on the surface of the unit cube.
On average, how far could you have expected to travel?
