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 🎬.

You and your opponent are beginning a best-of-seven series, meaning the first team to win four games wins the series. Both teams are evenly matched, meaning each team has a 50 percent chance of winning each game, independent of the outcomes of previous games.

As the team manager, you are trying to motivate your team as to the criticality of the first game in the series (i.e., “Game 1”). You’d specifically like to educate them regarding the “probability swing” coming out of Game 1—that is, the probability of winning the series if they win Game 1 minus the probability of winning the series if they lose Game 1. (For example, the probability swing for a winner-take-all Game 7 is 100 percent.)

What is the probability swing for Game 1?

Submit your answer

Instead of a best-of-seven series, now suppose the series is much, much longer. In particular, the first team to win N games wins the series, so technically this is a best-of-(2N−1) series, where N is some very, very large number.

In the limit of large N, what is the probability swing for Game 1 in terms of N? (For full credit, I’m expecting an answer that is rather concise!)

Submit your answer

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 sharing a recent article from Quanta about the discovery of a convex polyhedron without the “Rupert property,” meaning there’s no way to orient the polyhedron so that it can tunnel through a copy of itself. (This phenomenon is better explained in the article.)

Here is Quanta’s rendering of the “Noperthedron,” which was discovered (or rather, it was proven to not have the Rupert property) by Jakob Steininger and Sergey Yurkevich:

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.

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.

Congratulations to the (randomly selected) winner from last week: 🎻 Steven Goldburg 🎻 from Norfolk, Virginia. I received 61 timely submissions, of which 52 were correct—good for an 85 percent solve rate.

Last week, you were presented with ESPN’s odds for the two clear favorites in the race for baseball’s American League Most Valuable Player (MVP) award:

Aaron Judge of the New York Yankees, whose odds were -150.

Cal Raleigh of the Seattle Mariners, whose odds were +110.

The “-150” odds meant that for every $150 you bet on Judge to win MVP, you’d earn $100 if he actually won. The “+110” odds meant that for every $100 you bet on Raleigh to win MVP, you’d win $110 if he actually won. Note the differing results (winning $100 vs. betting $100) that came with the odds being negative vs. positive.

While these betting odds may have been informed by an assessment of Judge’s and Raleigh’s real chances, they may also have been informed by how much money people were betting on each player.

Suppose all bettors wagered on either Judge or Raleigh with the odds above. Some fraction f of dollars wagered was in favor of Judge, while 1−f was wagered on Raleigh. For what fraction f would the oddsmaker earn the same amount of money, regardless of which player won the MVP award?

Let’s figure out how much money the oddsmaker would earn in either of the two cases (Judge wins, Raleigh wins) and then set them equal to each other.

For simplicity, let’s suppose $1 was the total combined amount of money wagered on the players, although the reasoning here will apply equally well for other total amounts. For now, the total money wagered on Judge was f dollars and the total money wagered on Raleigh was 1−f dollars.

If Judge wins, the oddsmaker would keep all the money bet on Raleigh, which was 1−f dollars. However, the oddsmaker would also have to pay out winnings to those who correctly bet on Judge. Judge had odds of -150, which meant every $150 bet on Judge earned an additional $100, so the payout was 100/150 (or 2/3) the amount wagered. Thus, the oddsmaker would have to pay 2/3·f. The total income would then be 1−f−2/3·f, or 1−5/3·f.

If Raleigh wins, the oddsmaker would keep all the money bet on Judge, which was f dollars. However, this time the oddsmaker would have to pay out winnings to those who correctly bet on Raleigh. Raleigh had odds of +110, which meant every $100 bet on Raleigh earned an additional $110, so the payout was 110/100 (or 11/10) the amount wagered. Thus, the oddsmaker would have to pay 11/10·(1−f). The total income would be f−11/10·(1−f), or -11/10+21/10·f.

The two expressions are plotted below. Where the lines intersect was the value of f for which the oddsmaker made the same profit no matter which line you were on—that is, no matter who won the MVP award.

Setting the two expressions equal gave you the equation 1−5/3·f = -11/10+21/10·f. Isolating f gave you 21/10 = 113/30·f, and then solving gave you f = 63/113, or about 55.75 percent. If you mixed up f and 1−f, instead submitting an answer of 50/113, I still awarded credit.

When 63/113 of the money was wagered on Judge, the oddsmaker was set to earn 8/133 dollars (a little more than 7 cents) for every dollar that was wagered, no matter which player won. Furthermore, as long as between 10/21 and 3/5 (i.e., between 47.6 percent and 60 percent) of the money was wagered on Judge, the oddsmaker was guaranteed to earn some positive profit no matter what.

Congratulations to the (randomly selected) winner from last week: 🎻 Brook Chuang 🎻 from Saratoga California. I received 38 timely submissions, of which 27 were correct—good for a 71 percent solve rate.

Suppose there were two leading candidates, A and B, for MVP in the Fiddler Baseball League.

Part 1:

The odds for A winning the award were 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 could set the odds for B so that they’d make the same amount of money regardless of whether A or B won the award. However, below a certain value of f, it was impossible for the oddsmaker to do this.

In terms of x, what was this critical value of f?

Part 2:

Now, the odds for A winning the award were set to -100y, where y > 1. Again, for many values of f, the oddsmaker could set the odds for B so they’d make the same amount whether A or B won the award.

In terms of y, what was the critical value of f below which this wasn’t possible?