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Are You Clever Enough?

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 Brad Fortner comes a puzzle for the cleverest among us:

You are very clever when it comes to solving Riddler Express puzzles. You are so clever, in fact, that you are in the top 10 percent of solvers in Riddler Nation (which, as you know, has a very large population). You don’t know where in the top 10 percent you are — in fact, you realize that you are equally likely to be anywhere in the topmost decile. Also, no two people in Riddler Nation are equally clever.

One Friday morning, you walk into a room with nine members randomly selected from Riddler Nation. What is the probability that you are the cleverest solver in the room?

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

Riddler Classic

From Oliver Ruebenacker comes a “classic” indeed:

You have four standard dice, and your goal is simple: Maximize the sum of your rolls. So you roll all four dice at once, hoping to achieve a high score.

But wait, there’s more! If you’re not happy with your roll, you can choose to reroll zero, one, two or three of the dice. In other words, you must “freeze” one or more dice and set them aside, never to be rerolled.

You repeat this process with the remaining dice — you roll them all and then freeze at least one. You repeat this process until all the dice are frozen.

If you play strategically, what score can you expect to achieve on average?

Extra credit: Instead of four dice, what if you start with five dice? What if you start with six dice? What if you start with N dice?

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

Solution to last week’s Riddler Express

Congratulations to ЁЯСП Peter Ji ЁЯСП of Madison, Wisconsin, winner of last week’s Riddler Express.

Last week, I had four equilateral triangles. I placed one on the floor. Then, I picked a random edge of this first triangle and attached it to a side of a second triangle. Next, I randomly picked one of the four edges of the resulting rhombus and attached the third triangle. Finally, I randomly picked an edge along the perimeter of the resulting shape and attached the fourth triangle to it.

What was the probability that I could create a regular tetrahedron by folding the four triangles along their edges?

As noted by solver Joan Carlock, no matter which edges you selected, the first three triangles always formed the same figure:

Three equilateral triangles that form an isosceles trapezoid. The longer base has two triangular edges labeled A and B. The shorter base is labeled D. The two legs are labeled C and E.

At this point, there were five edges to choose from in placing the final triangle (labeled A through E in the above diagram). If you chose edges C, D or E, then the triangles folded into a tetrahedron. But if you chose edges A or B, the four triangles folded into something else: the lateral surface of a square pyramid. Solver John Drake of Sydney, Australia, demonstrated these cases using Mathigon’s Polypad. That meant the answer was three-fifths, or 60 percent.

Well, that wasn’t too bad at all! But things got much trickier with the extra credit, where the four equilateral triangles were replaced with six squares. Lining up their edges one at a time, what was the probability that they could be folded up to form a cube?

Some readers observed that there were 35 distinct ways to attach six squares into what are known as “hexominoes.” Meanwhile, only 11 of these could be folded up into a cube. Surely, the answer was then 11/35.

But by adding the squares one at a time, not all 35 hexominoes were equally likely to be formed. Solver тАЛтАЛPeter Exterkate (also of Sydney, Australia!) found their respective probabilities by carefully building up from the single domino, to the triominos, to the tetrominoes, to the pentominoes and finally to the hexominoes. According to Peter, the most likely hexomino was the one shown below, which was formed with a probability of 107/900.

Six squares arranged to form a hexomino. Four of them form a two by two square. There is another square to the left of the top-left square, and another one below the bottom-left square.

In the end, adding together the probabilities of the hexominoes that could form cubes gave an overall probability of 31/180, or about 17.2 percent.

And the extra credit didn’t stop there. You were also asked about the chances of being able to randomly fold the remaining Platonic solids: octahedrons, dodecahedrons and icosahedrons. This was not for the faint of heart — which is why no one was able to solve these. I will leave these as an exercise for the reader. (Translation: I haven’t figured these out yet, either.)

Solution to last week’s Riddler Classic

Congratulations to ЁЯСП Brian Snodgrass ЁЯСП of Fort Wayne, Indiana, winner of last week’s Riddler Classic

Last week, you raced against Riddler Nation in the inaugural Riddler-opolis 500! This was based on the game of RaceTrack, originally published by recreational mathematician Martin Gardner in 1973.

You began at the midpoint of the starting line at the bottom (the thicker border), and your goal was to circumnavigate the race track shown below in a single counterclockwise loop. You moved from point to point on the grid, without ever venturing into the surrounding square or the central circle.

A 15-by-15 race track of grid points, with a central circle barrier of radius 3. The vertical starting line is centered below the circle, and the starting point is at the midpoint of this starting line.

Each move took one second. For your first move, you could have chosen among the nine grid points surrounding and including the starting position.

The same race track, now with 8 points surrounding the starting point. These represent the points you can move to on your first turn.

From there, it was essential to remember that you had inertia. So for your second move, you again had nine possible destinations, but they were determined by your current velocity vector. So you could maintain your current direction and speed, or you could alter your destination by one point in any direction (horizontally, vertically or diagonally). For example, if your first move had been up and to the right, then your possible second moves were shown below, although two of them would have caused you to crash into the wall.

The first move was diagonally up and to the right. Now there are 9 more points shown as possible destinations of the second move, centered on the point that is diagonally up and to the right of the first move.

At no time were you allowed to venture into the wall, though you were permitted to be tangent to the wall or on a grid point along a wall.

Finally, when you crossed the finish line, fractional times were considered. For example, if the finish line bisected your final move, then that move counted as 0.5 seconds toward your total time.

So how quickly were you able to navigate the track?

Last week, I asked you to submit your answer in a very specific format, using the digits 1 through 9 to indicate each move. Why? So that I could animate the race! I received 48 solutions, 46 of which used valid notation. Here were the results:

The race is played out. Several solutions veer off to the right, but most work their way around the central circle without ever hitting it. Some solutions cross the finish line after exactly 12 moves, while others cross the finish line between 12 and 13 moves.

A few readers thought they were indicating position rather than velocity with each step, which is why you see some paths wandering off to the right. Also, at least one reader tried to cheat a little by clipping the bottom-left corner of the inner circle, which was not allowed.

Many solvers crossed the finish line in under 13 seconds, but the winning time was exactly 12 seconds.

Vince Vatter and Jay Pantone found 322 distinct winning routes using dynamic programming techniques. Here they are, in all their glory:

It’s worth noting that the fastest paths around the track were not the shortest paths. To hug the inner circle, you had to take shorter steps each turn to keep your turning radius low. But if you gunned your motor just a little, allowing for a wider turn at the top of the track, you wound up finishing sooner.

I hope you enjoyed the race! We’ll have to run another one of these — with a more complicated track — in the near future.

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


  1. Important small print: In order to 👏 win 👏, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.