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. 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.

## Riddler Express

As you may have seen in FiveThirtyEight’s reporting, there’s an election coming up. Inspired, Vikrant Kulkarni has an electoral enigma for you:

On Nov. 3, the residents of Riddler City will elect a mayor from among three candidates. The winner will be the candidate who receives an outright majority (i.e., more than 50 percent of the vote). But if no one achieves this outright majority, there will be a runoff election among the top two candidates.

If the voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what is the probability of a runoff?

*Extra credit:* Suppose there are *N* candidates instead of three. What is the probability of a runoff?

The solution to this Riddler Express can be found in the following week’s column.

## Riddler Classic

This week, we return to the brilliant and ageless game show, “The Price is Right.” In a modified version of the bidding round, you and two (not three) other contestants must guess the price of an item, one at a time.

Assume the true price of this item is a randomly selected value between 0 and 100. (Note: The value is a real number and does not have to be an integer.) Among the three contestants, the winner is whoever guesses the closest price *without going over*. For example, if the true price is 29 and I guess 30, while another contestant guesses 20, then they would be the winner even though my guess was technically closer.

In the event all three guesses exceed the actual price, the contestant who made the lowest (and therefore closest) guess is declared the winner. I mean, *someone* has to win, right?

If you are the first to guess, and all contestants play optimally (taking full advantage of the guesses of those who went before them), what are your chances of winning?

The solution to this Riddler Classic can be found in the following week’s column.

## Solution to last week’s Riddler Express

Congratulations to 👏 Nick Russell 👏 of Vancouver, Canada, winner of last week’s Riddler Express.

Last week, you were helping me park across the street from a restaurant for some contactless curbside pickup. There were six parking spots, all lined up in a row.

While I *could* parallel park, it definitely wasn’t my preference. No parallel parking was required when the rearmost of the six spots was available or when there were two consecutive open spots. If there was a random arrangement of cars occupying four of the six spots, what was the probability that I had to parallel park?

With six spots and four cars, there were 6 choose 4, or 15 cases to consider. Some solvers, like Lisa Fondren of Montrose, Michigan, listed them all out and counted how many required parallel parking.

But there were other ways to find the solution that didn’t require working through every case. Rather than considering where the four cars were, solver Libby Aiello equivalently looked at where the two empty spots were. Among the total 15 combinations, there were five in which the two empty spots were adjacent: the first and second spots, the second and third, the third and fourth, the fourth and fifth, and the fifth and sixth.

There were also five combinations in which the last spot was open, since there were five spots from which to choose the *other* open spot. Combining these two cases (having two consecutive open spots and having the sixth spot open), there appeared to be 10 combinations that didn’t require parallel parking.

But that wasn’t quite right. As Libby noted, one combination — when the fifth and sixth spots were open — was counted twice, since two consecutive spots were open *and* the sixth spot was open. Subtracting one to account for this double counting meant there were nine combinations that didn’t require parallel parking, and six combinations that did. Therefore, the probability I had to parallel park was 6/15, or **40 percent**.

I was pleased to see how many readers solved this combinatorics challenge. Now if only that many drivers could successfully parallel park…

## Solution to last week’s Riddler Classic

Congratulations to 👏 Douglas Thackrey 👏 of Loulé, Portugal, winner of last week’s Riddler Classic.

Parking cars was one thing — parking trucks was another thing entirely. Last week, you looked at a *very long* truck (with length *L*) with two front wheels and two rear wheels. (The truck was so long compared to its width that you could consider the two front wheels as being a single wheel, and the two rear wheels as being a single wheel.)

You were asked to determine the truck’s turning radius, given the angles by which you could turn the front or rear wheels. First, you considered what would happen if the front wheels could be turned up to 30 degrees in either direction (right or left), but the rear wheels did not turn.

There was no doubt among readers that this was a geometry puzzle, but the challenge lay in translating the constraints on the wheels (i.e., only turning a certain amount or not at all) into the resulting motion of the truck.

As suggested by the term “turning radius,” the key was to think about the circular motion of the truck. When the driver rotated the front wheels a full 30 degrees in one direction and drove forward, both the front and the rear of the truck would move in circles. The front of the truck always made a 30 degree angle with the tangent line to the circle it was moving around.

Meanwhile, the rear wheels of the truck couldn’t turn. That meant the truck’s rear was always tangent to the circle it was moving around.

If that wasn’t clear, here’s an animation to illustrate how the truck was moving:

The green line segment represents the truck, and the circles represent the paths of the truck’s front and rear. Sure enough, the angle between the truck and the tangent line (represented by the white segment) is always 30 degrees. This means that the front of the truck is moving around a *wider* circle than the rear — and, consequently, that the front of the truck moves *faster* than the rear!

At this point, calculating the turning radius was a matter of geometry and trigonometry. If the green segment was doubled in length so that it formed a chord within the larger circle, the 30 degree angles meant that this chord was one side of an inscribed regular hexagon, whose sides all equal the circle’s radius. And so if the truck had length *L*, the turning radius — that is, the radius of the circle around which the front of the truck moved — was **2**** L**. (Solvers who gave the turning radius for the truck’s midpoint or rear and explained their reasoning were also given full credit.)

That was the case when you could only turn the front wheels. You were also asked for the turning radius when both the front and rear wheels could be independently turned up 30 degrees in either direction. To make the tightest possible circle, the front and rear wheels were both rotated the full 30 degrees, but in opposite directions, allowing the front and rear to move along the same circle. Again, here’s an animation:

This time, the truck made up a complete chord that was a side of an inscribed regular hexagon. That meant the turning radius was equal to the truck’s length, ** L**. (Again, solutions for different locations on the truck were also accepted.)

A few solvers, including Laurent Lessard and Josh Silverman, tackled the general version of this problem, in which the front wheels could be turned an angle *θ*_{1} and the rear wheels could be turned an angle *θ*_{2}. The turning radius at the front of the truck was *L*·cos(*θ*_{2})/sin(*θ*_{1}+*θ*_{2}), while the turning radius at the rear was *L*·cos(*θ*_{1})/sin(*θ*_{1}+*θ*_{2}).

These formulas checked out for both questions in the riddle. And when neither wheel could turn (i.e., *θ*_{1} and *θ*_{2} were both zero), the turning radius went to infinity, which also made sense. In that case, you’d just have to keep on truckin’.

## 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