Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and 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.

## Riddler Express

From Chris Horgan, a medieval matrimony puzzle:

You’re the most eligible bachelorette in the kingdom, and you’ve decided to marry a prince. The king has invited you to his castle so that you may choose from among his three sons. The eldest prince is honest and always tells the truth. The youngest prince is dishonest and always lies. The middle prince is mischievous and tells the truth sometimes and lies the rest of the time. Because you will be forever married to one of the princes, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with each. But there’s a problem: You can’t tell the princes apart just by looking, and the king will grant you only *one* yes-or-no question that you may address to only *one* of the brothers.

What yes-or-no question can you ask that will ensure that you do not marry the middle prince?

## Riddler Classic

From Stephen Carrier, a puzzle about domestic boundaries:

Consider four square-shaped ranches, arranged in a two-by-two pattern, as if part of a larger checkerboard. One family lives on each ranch, and each family builds a small house independently at a random place within the property. Later, as the families in adjacent quadrants become acquainted, they construct straight-line paths between the houses that go across the boundaries between the ranches, four in total. These paths form a quadrilateral circuit path connecting all four houses. This circuit path is also the boundary of the area where the families’ children are allowed to roam.

What is the probability that the children are able to travel in a straight line from any allowed place to any other allowed place without leaving the boundaries? (In other words, what is the probability that the quadrilateral is convex?)

## Solution to last week’s Riddler Express

Congratulations to 👏 Fritz Burkhardt 👏 of South Burlington, Vermont, winner of last week’s Express puzzle!

A rope is tied tightly around the circumference of the Earth. It’s then lengthened by 1 meter and raised up at every point from the ground until it is again a taut circle and every point on the rope is the same height above the ground. How high is the rope off the ground? About **16 centimeters**.

One lousy meter is nothing compared with *the circumference of the Earth*, for goodness sakes, so the intuitive answer would seem to be “not very high at all” or even “so small as to be barely noticeable.” But intuition fails us in this case.

What is the circumference of the Earth in meters? It doesn’t matter! You needn’t know to solve this problem. For now, let’s just call the circumference \(C\) and the Earth’s radius \(R\). (These will drop out of our equations after a bit of algebra.) Let little \(r\) be the added radius from the lengthened string — precisely the value we’re looking for to solve the problem. One last common geometry fact we’ll need: A circle’s circumference is two times \(\pi\) times its radius. (Happy belated Pi Day, by the way.)

After some simple rearranging, we can solve for r, which in this case represents the height of the rope off the Earth’s surface:

\begin{align}

C + 1 &= 2π(R + r)\tag{1}\\

2πR + 1 &= 2πR + 2πr\tag{2}\\

1 &= 2πr\tag{3}\\

r &= \frac{1}{2π}\tag{4}\\

\end{align}

To run you through all that math, Step 1’s left-hand side represents the new rope’s length, in terms of circumference, and equates that to the new rope’s length in terms of radii. In Step 2, we substitute in the formula for circumference on the left and multiply through the parentheses on the right. In Step 3, we subtract \(2\pi R\) from both sides. And in Step 4, we divide both sides by \(2\pi\) and put our variable of interest on the left-hand side.

So the added height of the rope above the Earth is \(1 / (2\pi)\) meters, or about 16 centimeters, or just over 6 inches. That measly meter added to the entire circumference of the Earth bought us enough height for a cat to slip under.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Mikolaj Franaszczuk 👏 of New York City, winner of last week’s Classic puzzle!

Two people are playing a game at a square table. Each has a pile of identically sized coins, and they take turns placing the coins on the table, one after another. The only catch is that coins are not allowed to touch. The winner of the game is the person who places the final coin on the table. One player is guaranteed to win. Who is it, and what is the winning strategy?

The player who goes first is guaranteed to win. Here’s how: She should begin by placing a coin in the exact center of the table. After that, wherever her opponent places his coin, she should place hers in the “opposite” position — specifically, in a radially symmetric position, as if the coin had hopped over the center coin to an equal distance away.

Laurent Lessard animated how a game played in this manner might go:

The proof of the infallibility of this strategy is straightforward: Whenever it is her opponent’s turn, the board is a symmetric sea of coins. So if the opponent has a place for his coin, she is guaranteed to have a place opposite it for hers. Eventually, he will run out of space.

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.