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 weekâ€™s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Dave Mattingly comes a high-rolling question of craps:

Thereâ€™s a technique for rolling dice called â€śbowling,â€ť in which you place your index finger and thumb on two opposite sides of the die and roll it along the table. If done correctly, the die will never land on the faces on which you were holding the die, leaving you with a 25 percent chance of landing on each of the remaining four faces.

Youâ€™d like to apply this technique to improve your chances of winning a simplified game of craps, in which your goal is simply to roll a 7 or 11 using two dice. With a standard rolling technique, your chances of rolling a 7 or 11 are 2/9, or about 22.2 percent.

Now suppose youâ€™re using your bowling technique, and you roll the dice one at a time (i.e., you know the outcome of the first die before rolling the second). If you play to maximize your chances of rolling a 7 or 11, what will be your chances of winning?

*Extra credit:* Suppose you get one point for rolling a 7 or 11, but now you *lose* a point for rolling a 2, 3 or 12. With a standard rolling technique, youâ€™d average 1/9 of a point. But if you â€śbowlâ€ť to maximize your expected score, what will that average be?

## Riddler Classic

From Adam Wagner comes a curious case of colonies:

You are studying a new strain of bacteria, *Riddlerium classicum* (or *R. classicum*, as the researchers call it). Each *R. classicum* bacterium will do one of two things: split into two copies of itself or die. There is an 80 percent chance of the former and a 20 percent chance of the latter.

If you start with a single *R. classicum* bacterium, what is the probability that it will lead to an everlasting colony (i.e., the colony will theoretically persist for an infinite amount of time)?

*Extra credit:* Suppose that, instead of 80 percent, each bacterium divides with probability *p*. Now whatâ€™s the probability that a single bacterium will lead to an everlasting colony?

## Solution to last weekâ€™s Riddler Express

Congratulations to đź‘Ź Paul Berger đź‘Ź of Forest Hills, New York, winner of last weekâ€™s Riddler Express.

Last week, the astronomers on planet Xiddler, having recently invented the telescope, discovered a new planet in their solar system!

Like Earth, Xiddler orbits its star in a nearly circular path, with an average distance of 150 million kilometers. But *unlike* Earth, there werenâ€™t any other known planets in the solar system â€¦ until now.

Moments after the Xiddlerian sun set below the horizon, three astronomers used their telescopes to find the new planet at the zenith of the evening sky. They then raced to Xiddlerâ€™s Grand Minister to deliver the news.

The first astronomer said that the newly discovered planet orbited their sun with a radius of 50 million kilometers; the second astronomer said the orbital radius was 300 million kilometers; the third astronomer said the orbital radius was 150 million kilometers.

Which astronomer should the Grand Minister have believed?

The keys to this riddle were the relative locations of Xiddlerâ€™s sun and the new planet. While one was near the horizon (the equator of the celestial sphere), the other was at the zenith (the north pole of the celestial sphere). That meant the angle between the sun, Xiddler and the new planet had to be 90 degrees. In other words, as noted by solver (and astrophysicist) Megan Pickett, the newly discovered planet was at quadrature.

If youâ€™re not convinced, then hereâ€™s a helpful visual, courtesy of solver Travis Bishop:

On the left is the sun, while Xiddler is in the bottom right. Because it was sunset, that meant the astronomers were on Xiddlerâ€™s terminator, which was perpendicular to the line between Xiddler and the sun. The new planet was observed at the zenith, and so the sun, Xiddler and the new planet formed a right triangle.

The distance between Xiddler and the sun â€” 150 million kilometers â€” was one leg of this right triangle. Meanwhile, the distance between the new planet and the sun was the hypotenuse, which, by the Pythagorean theorem, had to be greater than 150 million kilometers.

The only astronomer who calculated an orbit greater than 150 million kilometers was **the second astronomer, who said the orbit was 300 million kilometers**. She was the one the Grand Minister should have believed.

By the way, this math works just as well in our own solar system as it does on planet Xiddler. For example, if youâ€™ve ever wondered why Venus is called a â€śmorning starâ€ť or â€śevening star,â€ť itâ€™s because Venusâ€™s orbit lies within the Earthâ€™s orbit, so it always appears relatively close to the sun in the sky. You canâ€™t see any stars or planets when the sun is directly overhead â€” you can only see Venus when the sun is just below the horizon, at dawn or in the evening.

Anyway, the work of the Xiddlerian astronomers has only just begun. Iâ€™m sure weâ€™ll be hearing from them again soonâ€¦

## Solution to last weekâ€™s Riddler Classic

Congratulations to đź‘Ź Harel Dor đź‘Ź of Sunnyvale, California, winner of last weekâ€™s Riddler Classic.

Last week, you were filling in a signâ€™s letters by drawing horizontal lines with a marker. This marker had a flat circular tip with a radius of 1 centimeter, and you were holding the marker so that it was upright, perpendicular to the sign.

Since the diameter of the markerâ€™s tip was 2 centimeters, you decided to fill in the letters by drawing lines every 2 centimeters. However, this was the pattern you got, with apparent gaps between the strokes:

Of course, if you drew many lines all bunched together, youâ€™d have a more uniform shading.

But you didnâ€™t have all day to make the sign. If the lines couldnâ€™t overlap by more than 1 centimeter â€” the radius of the marker tip â€” what should this overlap have been, in order to achieve a shading that was as uniform as possible? And how uniform was this shading?

Before we even discuss the first solution, itâ€™s worth taking a minute to better understand the problem. You might have thought that lines separated by the width of the marker would be nice and uniform. So why were there gaps between the strokes in the first place?

The answer came from the shape of the markerâ€™s tip â€” a circle. Imagine moving a circular tip along a sign at a constant speed. For any given spot on the side, the amount of ink present is proportional to how long the marker was in contact with that point. Points along the midline of a stroke are in contact with the marker the longest, while points that are 1 centimeter away from the midline are only in contact for an instant.

If you looked at a vertical slice of the sign and measured the relative amount of ink along that line, you got a pattern of repeating semicircular â€śhillsâ€ť:

As the strokes began to overlap and move closer together, the â€śvalleysâ€ť between them combined until they overtook the hills, as shown in the animation below:

By the end of the animation, when the overlap was a full centimeter, the distribution of ink was once again quite bumpy. But somewhere along the way, the variance in the distribution of the ink was minimized. And if you thought understanding that there even was a minimum was a challenge, *finding *that minimum was even more challenging.

Solvers Emma Knight and David Zimmerman both used calculus to measure the standard deviation of the ink as a function of the overlap between strokes. Along the way, they both encountered some rather nasty elliptic integrals, turning to their computers for a solution. If the height of each original hill of ink was 1, then the minimum standard deviation of roughly 0.0859 occurred when the overlap between strokes was approximately **0.308 centimeters**.

Josh Silverman also found the same solution via a computational approach, while Jason Ash arrived at a similar result by (reasonably) assuming the sign had a finite number of horizontal strokes.

Instead of minimizing the standard deviation, some solvers instead minimized the *relative* standard deviation (also known as the coefficient of variation), which is the ratio of the standard deviation to the mean. This was minimized when the overlap between the strokes was approximately **0.319 centimeters**: the standard deviation was about 0.08615 and the mean was about 0.9345, which meant the relative standard deviation was roughly 0.0922. Given the ambiguity in the problem, I accepted this answer as equally correct. (Even if you subtracted either of these answers from 2, the diameter of the marker tip, I *still* counted it as correct.)

So what does this optimally uniform shading look like, in the end? Feast your eyes!

## Want more riddles?

Well, arenâ€™t you lucky? Thereâ€™s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. Itâ€™s called â€śThe Riddler,â€ť and itâ€™s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.