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 the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.
The American League Championship Series of Riddler League Baseball determines one of the teams that will compete in the Riddler World Series. This year’s teams — the Tampa Bay Lines and the Minnesota Twin Primes — are evenly matched. In other words, both teams are equally likely to win each game of the best-of-seven series.
On average, how many games will the series last? (Note that the series ends as soon as one team has won four games.)
The solution to this Riddler Express can be found in the following column.
From Eric Thompson-Martin comes a puzzle involving two more teams from Riddler League Baseball:
Over in the National League Championship Series, the Washington Rationals and the St. Louis Ordinals (known as the “Ords” for short) are also evenly matched. Again, both teams are equally likely to win each game of the best-of-seven series.
You enter a competition in which you must predict the winner of each of the seven games before the series begins. If any or all of the fifth, sixth or seventh game are not played, you are not credited with predicting a winner.
You win the competition if you predict at least two games correctly. If you optimize your strategy for picking winners, what is the probability you will win the competition?
Extra credit: You enter a second competition in which you must pick the winner of the first game and then of each next game, knowing who won in all the previous games. Again, if you optimize your strategy, now what is the probability you will predict at least two games correctly?
The solution to this Riddler Classic can be found in the following column.
Solution to last week’s Riddler Express
Congratulations to 👏 Ramsey Keilani 👏 of San Diego, California, winner of last week’s Riddler Express.
Last week, you traveled one additional week back in time in an attempt to win the lottery. It was worth $10 million, and each ticket cost a dollar. (Note that if you won, your ticket purchase was not refunded.) All of this sounded pretty great.
The problem was, you weren’t alone. There were 10 other time travelers who also knew the winning numbers. You knew for a fact that each of them would buy exactly one lottery ticket. Now, according to the lottery’s rules, the prize was evenly split among all the winning tickets (i.e., not evenly among winning people). How many tickets should you have bought to maximize your profits?
Since you weren’t told about any non-time travelers who had produced winning numbers, it was fair to assume that the only winning tickets belonged to you and the other 10 time travelers.
From there, solver Lucy Stine assumed you bought x tickets and worked out what your expected profit would be. Those x tickets cost x dollars, which you’d never see again. Meanwhile, your chances of winning the $10 million prize was equal to the number of tickets you purchased (x) divided by the total number of winning tickets (x+10). Putting this all together, your expected profit was 107·x/(x+10)−x.
Next, you had to find which value of x maximized these profits. Approaches for figuring this out included spreadsheets and graphs. But a surefire way to find the maximum was to use calculus, finding where the derivative was equal to zero.
The derivative of the function f(x) = 107·x/(x+10)−x was f’(x) = 107·10/(x+10)2−1. Setting this equal to zero and solving for x gave you the equation (x+10)2 = 108, which meant x+10 = 104, or x = 9,990. In other words, you should have bought 9,990 winning lottery tickets.
Traveling back in time and buying that many tickets was a lot of work — especially the traveling back in time part. Was it worth it? Well, when x was 9,990, f(x) was 9,980,010, which meant your expected profit was $9,980,010. Not too shabby a return on investment!
Solution to last week’s Riddler Classic
Congratulations to 👏 Matt Leerberg 👏 of Cary, North Carolina, winner of last week’s Riddler Classic.
Last week, you were walking along a perfectly straight road. One hundred feet in front of you, and then another 100 feet to the left of the road, there was a lamppost (see the diagram below). On the other side of the lamppost, the same distance away, there was a doppelgänger, who was obscured by the lamppost and who moved precisely twice as fast as you at all times.
You started walking along the road, getting closer to the lamppost, but your doppelgänger remained hidden. Unlike you, they were not constrained to a straight road and could move more freely in two dimensions.
You walked a total of 200 feet (always forward, never backward), so that the lamppost was now 100 feet back and 100 feet left of the road. The entire time, your doppelgänger remained obscured by the lamppost.
At this point, what was the farthest the doppelgänger could have been from the lamppost?
First of all, it wasn’t even clear which way the doppelgänger would have moved at the outset. Suppose you walk an infinitesimal distance forward, as illustrated in the diagram below. Since your doppelgänger was twice as fast as you, that meant they must have moved twice as far in some direction. Because they had to remain obscured by the lamppost, their position had to be collinear with you and the lamppost. But there were still two such intersections, indicated by the blue arrowheads.
So should the doppelgänger have initially headed down and to the left, or up and to the right? Or did it not matter? We’ll come back to this question in a moment.
The doppelgänger often had to choose between two directions that satisfied the conditions of moving twice as fast and remaining obscured. However, if they were far enough away from the lamppost, there was no way they could remain hidden — they simply couldn’t keep pace with the angular velocity with which the lamppost appeared to move in your reference frame. In the illustration above, this corresponded to the line and the circle having zero intersections. Between where there were two possible directions and zero possible directions, there was one — where the line was tangent to the circle. This upper bound on how far away the doppelgänger could have been was a curve in the coordinate plane, which solver Laurent Lessard plotted below:
With that upper bound in place, which path should the doppelgänger have chosen? Surprisingly, solver Allen Gu found that either path was fine — at least, at first. But sticking with the lower path meant the doppelgänger eventually hit (and passed through!) the lamppost. Oops! Meanwhile, sticking with the higher path meant the doppelgänger would eventually hit the upper bound and then have nowhere left to hide.
The key insight to this riddle — and why I hinted that it was so challenging — was that the doppelgänger could switch between these paths at will. Avoiding both the lamppost and the boundary curve was a balancing act, and finding the maximum distance was one more challenge on top of that.
In the end, the doppelgänger’s goal was to reach the horizontal line that was 200 feet north of the lamppost. This way, they could safely coast along at twice your speed, always remaining hidden. Allen animated a few of these paths, each of which had the same end result:
The sharp change of direction in the different paths corresponded to when the doppelgänger switched between the two possible paths at a given point.
Other solvers, like Jim Crimmins, found paths that were slightly different but always ended in the same exact spot. And that spot was twice your own distance from the lamppost, meaning the answer was 200√2, or about 282.84 feet. (Note: If you instead gave the distance between the doppelgänger and you, which was 300√2 feet, I still counted that as correct.)
Now, proving that this was the farthest the doppelgänger could be required digging into some differential equations. Laurent and Eli Wolfhagen both found such an equation. Along the way, both solvers realized that there were two solutions for many points in the coordinate plane, which correlated with the two directions the doppelgänger could choose from. Sure enough, the doppelgänger’s best move was to pass through the point 200 feet directly north of the lamppost, after which they were mathematically confined to a straight path.
Finally, Laurent explored what would happen if the doppelgänger moved at different speeds relative to you:
Interestingly, beyond a critical relative speed of 4.4, the doppelgänger no longer had to switch between the two possible directions, always opting for the northern one.
So was the doppelgänger real, or just a figment of my imagination? Hopefully, you’ll learn the truth in your next pursuit.
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 firstname.lastname@example.org.