Can You Even the Odds?

You and a friend are playing a dice game that just isn’t fair (to your friend). With a different order of play, can you make the game a little more fair?

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 🎬.

This Week’s Fiddler

From 11-year-old (!) Kayla Schubmehl comes a question inspired by the book “Math Games with Bad Drawings,” by Ben Orlin:

Suppose you (player A) and a friend (player B) are playing a game in which you alternate rolling a die. So the order of play is AB|AB|AB, and so on. (The vertical bars here are just for organizational purposes, and do not signify anything special that happens.) The first player to roll a five wins the game. As it turns out, whoever goes first has a distinct advantage!

Kayla wondered about other ways you and your friend could take turns, ways that might result in a fairer game. For example, consider the “snake” method, in which the order is reversed after each time you both roll: AB|BA|AB|BA, and so on.

Assuming you are the first to roll, what is the probability you will win the game?

This Week’s Extra Credit

Kayla also offered some Extra Credit:

Another way to take turns is to use the Thue-Morse sequence, where the entire history of the order is flipped to its complement (from A to B and from B to A) after each round. As an illustration, consider the first few rounds:

• Round 1: Player A goes first.

• Round 2: Only A went in the first round. So now player B goes.

• Round 3: Up to this point, the order has been AB. Flipping this to its complement, round 3’s order is BA.

• Round 4: Up to this point, the order has been ABBA. Flipping this to its complement, round 4’s order is BAAB.

• Round 5: Up to this point, the order has been ABBABAAB. Flipping this to its complement, round 5’s order is BAABABBA.

Writing this out as a single sequence of turns, the order is A|B|BA|BAAB|BAABABBA, and so on.

Assuming you are the first to roll, what is the probability you will win the game?

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 the exciting news that the United States has won the 65th International Mathematical Olympiad. Congratulations to the entire US team, and to all the participants in the competition from around the world!

The competition consists of six problems over the course of nine total hours (three problems on the first day, another three on the second day). Here was Problem 1, which was thoroughly solved (meaning the maximum score of 7 was achieved) by 413 of the 609 participants—good for a 68 percent solve rate:

Determine all real numbers α such that, for every positive integer n, the integer ⌊α⌋ + ⌊2α⌋ + … + ⌊nα⌋ is a multiple of n. (Note that ⌊z⌋ denotes the greatest integer less than or equal to z. For example, ⌊−𝜋⌋ = −4 and ⌊2⌋ = ⌊2.9⌋ = 2.)

Messing around a little, it’s clear that α = 1 does not meet the condition, since the sum is n(n+1)/2, which isn’t a multiple of n when n is even. That said, when α is an even positive integer, the sum is n(n+1)·α/2, which is always a multiple of n. Can you find other values of α that work?

By the way, it was also just announced that Google’s AlphaProof and AlphaGeometry 2 together solved four of the six IMO problems, which would have earned a silver medal in the competition. (Note that, unlike Olympic events, which award one gold, silver, and bronze medal, IMO awards multiple medals of each variety.) While it took three days, rather than the human-allotted nine hours, this is still quite impressive.

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.

Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Cass Sackett 🎻 from Charlottesville, Virginia. I received 45 timely submissions, of which 30 were correct—good for a 67 percent solve rate.

Every year or so I try to run a cycling-related puzzle to coincide with the Tour de France. Previous puzzles have involved the team time trial, “King of the Mountain” points, the shape of the peloton, and being paced up a climb.

This time around, a lone rider in the Tour de Fiddler was being pursued by a group of four riders. The four riders had an advantage—they took equal turns being in the lead position, while the other three riders drafted behind. At any given speed, being in the lead position (as well as riding solo) required twice as much power as drafting.

You were asked to assume that every rider had to maintain the exact same average power over time, whether they were the lone rider or in the pack of four. To be clear, their power could change over time, but the time-averaged value had to be the same for every rider. Also, when leading a pursuing group or riding solo, one’s speed was directly proportional to one’s power. When drafting, one's speed matched that of the leader (again, at half the power output).

The pursuers just passed under a banner indicating there were 10 kilometers left in the stage. How far back from the lone rider could they afford to be, such that they still caught them at the finish line?

Suppose the average power that each rider had to maintain was some number P. With this value in mind, let’s look at the solo rider first.

Since they were riding alone, the solo rider’s average power was simply P. And since their speed was proportional to power, let’s label the constant of proportionality k, so that the solo rider’s average speed was kP.

The thorniest part of this puzzle was to work out the average speed of the group of four riders. To do that, you could start from the constraint that their time-averaged power similarly had to be P. Suppose the current rider in front (doing what is known as “pulling”) was operating at some higher power, Q. The other three riders, who were drafting, were exerting half that power, or Q/2.

Because the riders took equal turns pulling, that meant they each spent three-quarters of their time exerting power Q/2 and one-quarter of their time exerting power Q. Therefore, they each had a time-averaged power of 3/4·(Q/2) + 1/4·Q, which was equivalent to 5/8·Q. Since we said this had to equal P, that allowed you to solve for Q in terms of P: Q = 8/5·P. And so whoever was doing the pulling was exerting an average power of 8/5·P, which meant the average speed of the pursing group was 8/5·kP.

Whatever speed the solo rider was going out—that is, whatever numerical value kP turned out to be—the pursuers’ speed was 60 percent greater. For them to all converge on the finish line at the same time, the pursuers had to travel a distance that was 60 percent longer (or 8/5 as long). Since they were currently 10 kilometers front the finish, the solo rider had to be 5/8·10, or 6.25 kilometers from the finish.

But that didn’t answer the question that was originally asked. You had to determine the distance between the solo rider and the pursuers. And that was 10 minus 6.25, or 3.75 kilometers.

By the way, if you’re not into cycling, these sorts of exciting finishes—where one rider or a small group of breakaway riders are pursued by chasers or even the main peloton—happens all the time, occasionally with a very dramatic final few hundred meters. A solo rider burning the last of their energy while a tidal wave of riders is barreling toward them is truly something to behold.

Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 David Ding 🎻 from Natick, Massachusetts. I received 30 timely submissions, of which 27 were correct—good for a 90 percent solve rate.

In a specific stage of the Tour de Fiddler, there were 176 total riders. Some riders were grouped together in a single breakaway, while the remainder were grouped together in the peloton.

The breakaway group was 10 kilometers from the finish, while the peloton was one kilometer behind (i.e., 11 kilometers from the finish).

As before, you were asked to assume that every rider had to maintain the exact same time-averaged power when leading a group, and that drafters could match the leader’s speed using half the power.

What was the smallest number of riders that the breakaway needed to reach the finish line before the pursuing peloton?