Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
This week’s puzzles are both baseball-themed, so get your glove and thinking cap ready!
Riddler Express
Riddler Nation’s Jibriel Taha, an avid baseball fan, saw the following tweet from the Milwaukee Brewers’ beat writer Adam McCalvy:
Inspired by the Brewers’ apparent mediocrity (they’ve since gone on a roll to clinch a playoff spot) Jibriel asks the following:
If a baseball team is truly .500, meaning it has a 50 percent chance of winning each game, what’s the probability that it has won two of its last four games and four of its last eight games?
Riddler Classic
Riddler League Baseball, also known as the RLB, consists of three teams: the Mississippi Moonwalkers, the Delaware Doubloons and the Tennessee Taters.
Each time a batter for the Moonwalkers comes to the plate, they have a 40 percent chance of getting a walk and a 60 percent chance of striking out. Each batter for the Doubloons, meanwhile, hits a double 20 percent percent of the time, driving in any teammates who are on base, and strikes out the remaining 80 percent of the time. Finally, each batter for the Taters has a 10 percent chance of hitting a home run and a 90 percent chance of striking out.
During the RLB season, each team plays an equal number of games against each opponent. Games are nine innings long and can go into extra innings just like in other baseball leagues. Which of the three teams is most likely to have the best record at the end of the season?
Solution to last week’s Riddler Express
Congratulations to 👏Michael DeLyser 👏 of State College, Pennsylvania, winner of last week’s Riddler Express.
Last week, Riddler Nation needed help designing the logo for the new Riddler Express™ credit card, which consisted of two overlapping circles with radius 1 inch, creating three distinct regions: one region that’s shared between the two circles and two regions that are part of one circle, but not the other.
In particular, the areas of all three regions had to be exactly the same. How far apart must the centers of the two circles have been?
If the two circles are far apart, the area of the overlapping region is very small. As you move the circles closer together, at some point the area of the middle region will precisely equal the areas of the two outer regions. By moving the circles in this way, Dogan Kazakli determined that the areas are equal when the circles are approximately 0.81 inches apart.
Solver Brian Corrigan gave this exact problem to his high school students late last year, and he was able to nail down the distance with greater precision using calculus — integrating the areas of the different regions and setting them equal. Alan Greenburg took a trigonometric approach, splitting the middle region into two equal circular segments. This leads to an equation for the distance d between the circles:
\(2\cos^{-1}\left(\frac{d}{2}\right)-\frac{d}{2}\sqrt{1-\frac{d^2}{4}}=\frac{\pi}{2}\)
Unfortunately, there’s no simple expression for d beyond this equation, but a computer will tell you that d is approximately 0.8079455 inches.
For extra credit, you were asked to arrange three circles so that the areas of the largest and smallest of the seven resulting regions were as close to equivalent as possible. Here’s an animation showing how the difference between the largest and smallest areas changed along with the distance between the circles:
If you watch very carefully, you can see that the difference is minimized when the circles are a little more than 0.8 inches apart, a distance that closely resembles the answer from the two-circle problem. Surely the answers aren’t the same for both two circles and three circles. But as solver Grant Larsen found, they are indeed the same! He reasoned that the areas are closest when regions I and III in the image below have the same area. That means the combined area of regions I and II (the middle region from the two-circle problem) must equal the combined area of regions III plus II (the outer region from the two-circle problem). So the answer to the extra credit was, once again, approximately 0.8079455 inches.
By the way, if you’d like to know the answer to even more decimal places, you can find it here. Apparently this sequence of digits is kind of a big deal.
Solution to last week’s Riddler Classic
Congratulations to 👏Gavin Lee 👏 of New York, New York, winner of last week’s Riddler Classic.
In last week’s Riddler Classic, you were the coach for Team Riddler at the Tour de FiveThirtyEight, where there were 20 teams. Your objective was to win the Team Time Trial race, which had the following rules:
- Each team rode as a group at a fixed pace, specified by that team’s coach. Teams that couldn’t maintain their pace had “cracked” and didn’t finish the course.
- The fastest team to finish the course was the winner.
- Teams rode the course one at a time and were aware of the results of all previous teams.
All teams were of equal ability, and the faster the pace, the greater the probability of cracking. Team Riddler was the first team to attempt the course. How likely were they to finish the course while maximizing their chances of winning?
Gavin assumed Team Riddler would finish the course with probability p and that there were N total teams. If Team Riddler cracked, they’d certainly lose, giving that event probability 1-p. If Team Riddler finished the course, then the second team would ride at a pace that was ever so slightly faster than Team Riddler’s, giving it a similar probability p of finishing the course. Therefore, in order to win, Team Riddler needed to finish the course, with probability p, while the remaining N-1 teams each needed to crack, with probability 1-p. Since teams’ chances of cracking were independent, the probability of all this occurring was p(1-p)N-1. To maximize this probability, you can set its derivative equal to zero, finding that p = 1/N. When there are 20 teams, that means the probability that Team Riddler finishes the course was 1/20, or 5 percent. Plugging in this value of p reveals that Team Riddler’s chances of winning stood at a paltry 1.89 percent.
To solve the extra credit, you first needed to determine the optimal strategy for each team. If no team before you successfully finished the course, and there are N teams remaining (including them), then it’s just like being the first to compete among N teams, so you should select a pace such that your chances of finishing the course are 1/N. However, if any prior team finished the course, then you’re forced to attempt an ever so slightly faster pace than the current winning pace.
Now, to find Team Riddler’s chances of victory when it was the last to attempt the course, solver Jason Shaw broke things down into cases in his writeup. If the first team finishes the course, which we already know happens 1/20 of the time, then Team Riddler must ever so slightly beat them, also with probability 1/20. If the first team cracks, with probability 19/20, and the second team finishes the course, now with probability 1/19, then Team Riddler must beat that second team, again with a probability of 1/19. Continuing with this reasoning and working out all the cases, Jason found that when Team Riddler is the last to attempt the course, its chances of winning were 1/20×(1/20+1/19+1/18+…+1/1), a sum that’s just shy of 18 percent. Here are the chances of winning based on the order in which teams attempted the course:
In the Tour de FiveThirtyEight, it definitely pays to go last.
Finally, solver Sawyer Tabony extended the problem, considering what the first and last team’s chances of winning would be as more and more teams compete. For N teams, the solutions are beautifully concise: 1/(eN), and ln(N)/N, respectively. Surprise, surprise, this is yet another Riddler in which Euler’s number magically appears.
Want more riddles?
Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!
Want to submit a riddle?
Email Zach Wissner-Gross at riddlercolumn@gmail.com.