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.

## Riddler Express

From James Marek comes some madness in the month of March:

According to James, there was a scavenger hunt this year to determine which lucky individuals received tickets to the Duke’s home men’s basketball game against rival North Carolina. (Spoiler alert: Underdog North Carolina won the game.)

The original scavenger hunt had five tasks. For the purposes of this riddle, let’s say it has three tasks. For each task, whoever finishes first gets 1 point, whoever finishes second gets 2 points and so on. Your score is the total number of points you earn across all three tasks, and lower scores are better. Knowing how popular basketball is at Duke, it’s safe to say that many people are participating. But only the top-10 finishers in the scavenger hunt will get tickets.

Without knowing how anyone else did on the scavenger hunt, what is the highest score that *guarantees* you are in the top 10? (Being tied for 10th is acceptable.)

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

## Riddler Classic

From Ed Carl comes a surprising game of dice:

We’re playing a game where you have to pick four whole numbers. Then I will roll four fair dice. If any *two* of the dice add up to any *one* of the numbers you picked, then you win! Otherwise, you lose.

For example, suppose you picked the numbers 2, 3, 4 and 12, and the four dice came up 1, 2, 4 and 5. Then you’d win, because two of the dice (1 and 2) add up to at least one of the numbers you picked (3).

To maximize your chances of winning, which four numbers should you pick? And what are your chances of winning?

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

## Solution to last week’s Riddler Express

Congratulations to 👏 Michael DeLyser 👏 of State College, Pennsylvania, winner of last week’s Riddler Express.

Last week, you had three fair coins, three fair dice and a full deck of cards in your possession. First, you flipped all three coins and noted the number of heads. Next, you tossed all three dice and noted the number of ones or sixes. Finally, you drew three random cards from the deck of 52 and noted the number of hearts.

What was the probability that all three numbers were the same?

Across all three sets of items — coins, dice and cards — the number of each had to be zero, one, two or three. So if you worked out the probabilities of each of those numbers for each set of items, then you had everything you needed to find the answer.

First, let’s take a look at the coins. With three flips, the probability of getting zero heads was 1/8, one heads was 3/8 (since the heads could have come on any of the three flips), two heads was 3/8 and three heads was 1/8.

Now for the dice. The probability you rolled a one or six was 2/6, or 1/3. That meant the probability of rolling none of them was (2/3)^{3}, or 8/27. The probability of rolling exactly a single one or six was 3(1/3)(2/3)^{2}, or 12/27. The probability of rolling two of them was 3(1/3)^{2}(2/3), or 6/27. And the probability of rolling three of them was (1/3)^{3}, or 1/27.

And last, the cards. In last week’s puzzle, I neglected to say whether the cards were picked with or without replacement — that is, after you pick a card, whether you place it back in the deck and reshuffle or put it aside. If you were picking *with* replacement, the probability of drawing zero hearts was (3/4)^{3}, or 27/64. The probability of drawing one heart was 3(1/4)(3/4)^{2}, or 27/64 (again). The probability of drawing two hearts was 3(1/4)^{2}(3/4), or 9/64. And the probability of drawing three hearts was (1/4)^{3}, or 1/64.

Finally, if you were drawing without replacement, like solver Cordelia Hu of Knoxville, Tennessee, the numbers were a little more involved. The probability of drawing zero hearts was (39/52)(38/51)(37/50). The probability of drawing one heart was 3(13/52)(39/51)(38/50). The probability of drawing two hearts was 3(13/52)(12/51)(39/50). And the probability of drawing three hearts was (13/52)(12/51)(11/50).

With all these probabilities in hand, your final task was to multiply across, finding the probability for the same numerical result from the coins, dice and cards, and then add up the probabilities for the different numbers. For example, the probability of getting zero across the board was 1/8 (for the coins) times 8/27 (for the dice) times 27/64 (for the cards). (And if you drew the cards without replacement, that final probability would have been 703/1700.) The product of these three probabilities was 1/64.

After doing similar computations for getting one, two and three, the answer was **1,351/13,824**, or about 9.8 percent, when the cards were drawn with replacement. When the cards were drawn *without* replacement, the answer was just a tad higher: **6,089/61,200**, or about 10 percent.

So while each of the numbers had only a small handful of possible values, getting all of them to align was a relatively rare occurrence.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Guy D. Moore 👏 of Darmstadt, Germany, winner of last week’s Riddler Classic.

Last week, two ants named Geo and Desik were racing along the surface of a cone. The circular base of the cone had a radius of 2 meters and a slant height of 4 meters. Geo and Desik both started the race on the base, a distance of 1 meter away from its center.

The race’s finish was halfway up the cone, 90 degrees around the cone’s central axis from the start, as shown in the following diagram:

Geo and Desik both wanted your help in strategizing for the race. What was the length of the shortest path from the start to the finish?

Finding the shortest distance between two points on the surface of a three-dimensional shape is no picnic. But in *two*-dimensional space, the shortest distance is a straight line. So, if we could “flatten” the cone’s lateral surface down to two dimensions, that would greatly simplify matters.

As it turns out, you *can* flatten the lateral surface of a cone. If you’ve ever unrolled a conic cup by the water cooler, you’d know that if you cut along a slant height, the cone will flatten into a circular sector (i.e., a portion of a circle). The radius of this circle was the slant height of the cone, 4 meters. And its distance around was the same as the circumference of the cone’s circular base, which was 4𝜋 meters. That meant the cone unraveled into a semicircle of radius 4.

And so, the ants would start by traveling in a straight line on the circular base of radius 2. Then, once they reached its circumference, they’d travel along the curved surface in such a way that corresponded to a straight line on the flattened cone. Your task now was to choose the optimal point on the circumference of the circular base that connected these two paths.

Suppose the ants went to a point at an angle *θ* around the circumference from the radius they started on, as shown in the diagram below. By the law of cosines, that distance would be √(5−4cos(*θ*)).

But where on the semicircle’s circumference did the ants wind up? The start and finish were separated by 90 degrees, or 𝜋/2 radians, around the cone’s axis. Furthermore, when you flattened the cone, what was once 360 degrees around became just 180 degrees around (it was a *semi*circle, after all), meaning all angles were effectively halved. So the ants were 𝜋/4 radians away, plus another *θ*/2 due to their wandering along the base.

Applying the law of cosines once again, you found that the distance within the semicircle was 2√(5−4cos(*θ*/2+𝜋/4)).

Finally, you had to add the lengths of these two segments and find which value of *θ* minimized the sum.

Before revealing the answer, let’s see how the distance changed as a function of *θ*:

There was indeed a unique minimum distance, shown in red. As far as the ants were concerned, this minimum appeared to occur when they traveled in a straight line and didn’t have to change direction upon switching from the circular base to the lateral surface.

If you graphed the function f(*θ*) = f(√(5−4cos(*θ*)) + 2√(5−4cos(*θ*/2+𝜋/4)), you’d find the minimum occurred at a very nice value of *θ*: -𝜋/6 radians, or -30 degrees. (The minus sign showed up because the ants moved *clockwise* relative to their starting position on the circular base.) Meanwhile, the minimal distance was about **3.718 meters**.

Several solvers, including Laurent Lessard and Jenny Mitchell, were even able to find an exact solution: 3√(5−2√3).

In any case, thank you to everyone for helping the ants Geo and Desik identify the geodesic!

## 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.