Can You Bet Solely on Your Team’s Record?

In Back to the Future Part II, Biff Tannen made his fortune betting on outcomes of sporting events. But what if he only knew a team’s final record without knowing the scores of individual matches?

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. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

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 🎬.

Note: Due to the upcoming holiday, the next edition of Fiddler on the Proof will be coming out on September 8, 2023.

This Week’s Fiddler

From Dean Ballard comes a gambit that’s just in time for American football’s new season:

You’re betting on the final two football games of the season for your home team, the Fiddle-delphia Eagles. Thanks to the wonders of time travel, you happen to know that the Eagles will win one game and lose the other. Unfortunately, you can’t remember in which order they do so. Maybe the Eagles win the first game and lose the second, or maybe they lose the first and win the second.

You have $100, of which you can bet any amount (including fractions of pennies) that the Eagles will win. So if you bet x dollars on the first game and the Eagles win, you’ll have 100+x dollars to bet on the second game. But if the Eagles lose that first game, you’re left with 100−x dollars to bet on the second game.

You want to implement a betting strategy that guarantees you’ll have as much money as possible after both games. If you did so, then after the two games how much money would you be guaranteed to have? (Note: By “guaranteed to have a certain amount of money,” I mean that you will have at least that much no matter which specific order of wins and losses occurs.)

Extra Credit

In the original version of the puzzle that Dean submitted, there were 12 games remaining in the season rather than two. Suppose you know that your hometown Fiddle-delphia Eagles will win exactly eight of the 12 games. As before, you have no idea which eight of the 12 games they will win specifically.

Moreover, instead of only betting on the Eagles to win in each game, you can now bet on either team winning—the Eagles or their opponent. Betting x dollars on either team means you’ll still win x dollars.

Once again, you want to implement a betting strategy that guarantees you’ll have as much money as possible after the 12 games are complete. If you did so, then after the 12 games how much money would you be guaranteed to have?

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 sharing a fascinating observation from science fiction author Greg Egan that I reframe as a puzzle here.

Suppose two people (the red and blue dots, as shown below) are standing between two infinite parallel mirrors (the lines below). There are an infinite number of light paths between them, involving ever more reflections. What is the fewest number of points at which you can place obstructions such that all light paths are blocked? Note that you are not allowed to place blockers at the locations of the red and blue dots themselves.

The answer may surprise you! If you want a spoiler, Greg Egan wrote up the solution and went on to explore polygonal mirror arrangements.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Isaac Wink 🎻, from Lexington, Kentucky. I received 44 timely submissions, of which 31 were correct, good for a 70 percent solve rate. Not too shabby for such a computationally involved puzzle!

Last week, you were the manager for the New York Frets, a baseball team that has woefully underperformed this season. In an effort to right the ship, you were tinkering with the batting order.

Eight of your nine batters were “pure contact” hitters. One-third of the time, each of them got a single, advancing any runners already on base by exactly one base. (Your team was very slow on the base paths. That meant no one was fast enough to score from first or second base on a single—only from third.) The other two-thirds of the time, they recorded an out, and no runners advanced to the next base.

Your ninth batter was the slugger. One-tenth of the time, he hit a home run. But the remaining nine-tenths of the time, he struck out.

Your goal was to score as many runs as possible, on average, in the first inning. Where in your lineup (first, second, third, etc.) should you have placed your home run slugger?

I can still recite the conventional wisdom from before the days of Moneyball: Have your fastest player bat first, have a good contact hitter bat second, have someone who can hit for both contact and power bat third, have your best power hitter bat fourth, and so on. Baseball wisdom has certainly evolved over the last two decades, to the point where Aaron Judge—a slugger who last year set the record for most home runs in a season in the American League—was batting leadoff.

In this puzzle, however, having your slugger at leadoff wouldn’t have made much sense. Because you wanted to maximize the number of runs scored in just one inning, you were better off having your home run hitter come up to bat after a few contact hitters got on base so that he could drive in several runs at once. But was the optimal spot in the lineup second, third, fourth, or even somewhere later?

To figure this out, solver Angelos Tzelepis ran 20 million simulated innings with the slugger at each spot in the lineup, the results of which are shown below. It turned out that having the slugger bat third or fourth generated nearly the same number of runs on average—about 0.24 runs per inning.

To confirm this empirical result, several solvers, including Greg Tanner, found the precise result using what are called Markov chains. In short, Greg encoded every possible state of the inning (i.e., the current number of outs and the number of runners on base) in the rows of a matrix—plus an additional row for recording runs scored—with the probabilities of transitioning from one state to another in the corresponding columns. Eight of these matrices were for contact hitters, while the ninth was for the slugger. The lineup variations corresponded to a different multiplication order for these nine matrices. Finally, the expected number of runs could be read out in the last row of the limit of the product (i.e., when these nine matrices were multiplied by themselves in sequence, over and over again). Sure enough, the greatest expected number of runs came when the slugger batted fourth.

It was a nice result that the slugger batted fourth, or “cleanup,” as it’s called in baseball parlance. But by no means was this an obvious result. Slightly tweaking any of the probabilities in the puzzle yielded different results. For example, if the slugger hit a home run 5 percent of the time instead of 10 percent of the time, then you would have positioned him ninth (i.e., last) in the lineup, as contact hitters had a better chance of driving in runs.

Alternatively, if your slugger still had a 10 percent chance of hitting a home run but the contact hitters got on base one-fourth of the time instead of one-third of the time, then having your slugger bat third wound up being the best option by far.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Derek Nagel 🎻, from Savage, Minnesota. For the Extra Credit, I received 21 timely submissions, 19 of which were correct. While that’s an impressive 90 percent solve rate, these stats suggest that anyone brave enough to attempt the Extra Credit knew exactly what they were getting themselves into.

For Extra Credit, instead of scoring as many runs as possible in the first inning, you now wanted to score as many runs as possible over the course of nine innings. And instead of just having one home run slugger, you now had two sluggers in your lineup. The other seven batters remained pure contact hitters.

Where in the lineup should you have placed your two sluggers to maximize the average number of runs scored over nine innings?

Thinking back to the conventional wisdom, it might have made sense to have both your sluggers in the “heart” of the lineup, such as batting third, fourth or fifth. At the same time, it also made intuitive sense to separate your sluggers, so that if the first one cleared the bases with a home run, several contact hitters would have an opportunity to load up the bases before the second slugger came up to bat.

As it turned out, both approaches—”stacking” vs. “staggering” your sluggers—offered decent results. But as with last week’s Fiddler, these results were quite sensitive to the specific probabilities at play. To determine the best lineup, you needed to run many simulations or work out the solutions analytically.

The latter is precisely what solver Michael Schubmehl did. Michael found that the optimal lineup had sluggers batting third and fourth, a clustering strategy that garnered approximately 1.9977 runs on average. Solver 🎬 Laurent Lessard 🎬 computed the same result and plotted the 9 choose 2, or 36, lineup variations:

Laurent’s diagram nicely illustrates the stacking vs. staggering debate. While having your sluggers bat third and fourth was indeed the optimum for the given probabilities—with second and third not too far behind (1.9876 runs on average)—staggering the sluggers also did fairly well. In particular, third and seventh earned 1.9843 expected runs, fourth and eighth earned 1.9806 expected runs, and third and eighth earned 1.9799 expected runs. Even over the course of an entire 162-game season, these differences were rather modest.

Solver 🎬 Tom Keith 🎬 further generalized the Extra Credit so that there were anywhere from zero to nine sluggers to be placed in the lineup, as shown below. Interestingly, when adding a third slugger, the best spot was eighth in the order, an apparent win for the staggering strategy. Tom further found that a single slugger was optimally placed third in the lineup (rather than fourth) when optimizing over the course of a nine-inning game (rather than a one-inning game).

Hopefully all this analysis helps the New York Frets, or any other New York baseball teams whose analytical methods have recently been questioned in public. 

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.