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Can You Spot The Black Hole?

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

Amare the ant is back! This week, he’s trying to escape from a spider web.

Starting at the very center of the web, he crawls 1 inch radially away in a random direction. From there, he randomly chooses between crawling in the radial or tangential directions, with a 50 percent chance of each. When he crawls radially, he moves another inch from his current position directly away from the center of the web. When he crawls tangentially, he moves another inch in one of the two directions perpendicular to the radial direction. (To be clear, when Amare moves tangentially, he is still moving farther away from the center of the web, at least a little bit.)

After Amare has crawled a total of 4 inches, how far away from the center of the web should he expect to be on average?

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

Riddler Classic

Data from the Event Horizon Telescope was recently used to generate a never-before-seen image of the black hole — Sagittarius A* — at the center of our galaxy. One of the most striking things about the image is how clearly we can make out the black hole’s shadow (as shown below). That’s because the plane of its accretion disk is nearly perpendicular to the vector between us and the black hole.

Was this likely to occur, or was it just a cosmic coincidence? Let’s find out. Assuming the accretion disk was equally likely to be in any plane, what is the probability of it being within 10 degrees of perpendicular to us, thereby resulting in a spectacular image?

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

Solution to last week’s Riddler Express

Congratulations to 👏 Mike Fuerstman  👏 of Arlington, Massachusetts, winner of last week’s Riddler Express.

Last week, Dr. Watson and Sherlock Holmes were tracking a criminal mastermind in the present-day United States. Watson listed out the states the mastermind had visited in alphabetical order. They were: Alabama, Arkansas, California, Colorado, Florida, Georgia, Indiana, Louisiana, Maryland, Minnesota, Missouri, Montana, Nebraska, New Hampshire, North Dakota, Pennsylvania and South Carolina.

Watson stared at the list, but he couldn’t make out any sort of pattern.

But Holmes’s eyes lit up. “Why, it’s elementary, my dear Watson!”

What pattern did Holmes notice?

At first, you might have wondered if there was some sort of geographical connection among these 17 states. Next, you might have carefully inspected the letters, looking for some sort of alphabetical or linguistic pattern. For example, all 17 states include the letter A! Well, except that Missouri doesn’t have an A … and Alaska (among others) is missing from the list.

Looking back at the wording of the riddle, Holmes specifically said it was “elementary.” And that was the biggest clue of all — or, as Mike, this week’s winner told me, “You can just go straight to hell with that ‘elementary’ pun.”

Now, every U.S. state has a two-letter postal abbreviation. At the same time, most elements also have two-letter abbreviations, with exceptions like potassium (K) and oxygen (O). There happen to be 17 two-letter abbreviations that represent both states and elements. And those were the states in the puzzle. Here’s a complete list, courtesy of solver Julia McCarthy:

  • Alabama → AL ← Aluminum
  • Arkansas → AR ← Argon
  • California → CA ← Calcium
  • Colorado → CO ← Cobalt
  • Florida → FL ← Flerovium
  • Georgia → GA ← Gallium
  • Indiana → IN ← Indium
  • Louisiana → LA ← Lanthanum
  • Maryland → MD ← Mendelevium
  • Minnesota → MN ← Manganese
  • Missouri → MO ← Molybdenum
  • Montana → MT ← Meitnerium
  • Nebraska → NE ← Neon
  • New Hampshire → NH ← Nihonium
  • North Dakota → ND ← Neodymium
  • Pennsylvania → PA ← Protactinium
  • South Carolina → SC ← Scandium

Solution to last week’s Riddler Classic

Congratulations to 👏 Ryan Lafitte 👏 of Tucker, Georgia, winner of last week’s Riddler Classic.

Last week, you had four fair tetrahedral dice whose four sides were numbered 1 through 4.

You played a game in which you rolled them all and divided them into two groups: those whose values were unique, and those which were duplicates. For example, if you had rolled a 1, 2, 2 and 4, then the 1 and 4 would have gone into the “unique” group, while the 2s would have gone into the “duplicate” group.

Next, you rerolled all the dice in the duplicate pool and sorted all the dice again. Continuing the previous example, that would have meant you rerolled the 2s. If the result happened to be 1 and 3, then the “unique” group would have now consisted of 3 and 4, while the “duplicate” group would have had two 1s.

You continued rerolling the duplicate pool and sorting all the dice until all the dice were members of the same group. If all four dice were in the “unique” group, you won. If all four were in the “duplicate” group, you lost.

If we let A, B, C and D represent the four rolls, there were five distinct states you could have found yourself in after each round: ABCD (all four rolls were unique), AABC (one duplicate), AAAB (one triplicate), AABB (two duplicates) and AAAA (all four rolls were the same). For each of these states, you could assign a probability of winning the game upon reaching that state. For example, p(ABCD) — the probability of winning when all four rolls were unique — was 1. Meanwhile, p(AABB) and p(AAAA) were both zero. At this point, you still had to solve for p(AABC), which we’ll call x, and p(AAAB), which we’ll call y.

If you ever came across AABC, then you had to reroll the two As. WIth some casework, you found that you had a 2-in-16 chance of then getting ABCD, a 10-in-16 chance of getting AABC once again (perhaps with different values of A, B and C), a 2-in-16 chance of getting AAAB and a 2-in-16 chance of getting AABB. Mathematically, that meant x = 1/8 + (5/8)x + (1/8)y. Solving for x gave you the equation 3x = y + 1.

Finally, if you ever came across AAAB, then you had to reroll all three As, which meant there were 64 cases to consider. This time around, there was a 6-in-64 chance of then getting ABCD, a 36-in-64 chance of getting AABC, a 12-in-64 chance of getting AAAB, a 9-in-64 chance of getting AABB and a 1-in-64 chance of getting AAAA. Mathematically, this meant y = 3/32 + (9/16)x + (3/16)y. Solving for y gave you the equation 26y = 18x + 3.

At this point, you had two equations with two unknowns, x and y. After solving these simultaneous equations, you found that x was 29/60 and y was 9/20.

So, what was your probability of winning the game? On your very first roll, the probability you rolled ABCD and won at the outset was 3/32. The probability that you rolled AABC (meaning the probability you won from that point was x, or 29/60) was 9/16. And the probability that you rolled AAAB (meaning the probability you won from that point was y, or 9/20) was 3/16. If you rolled AABB or AAAA on your first roll, you lost.

Putting all of this together, your probability of winning was 3/32 + (29/60)(9/16) + (9/20)(3/16), or 0.45. In other words, it was slightly worse than a coin toss.

As is often the case in Riddler Nation, several solvers created and solved their own extensions of this puzzle. Emily Boyajian used Markov chains to derive a general expression for your chances of winning with four N-sided dice. Tom Keith studied your chances of winning with N N-sided dice. Interestingly, your chances oscillated for values of N between 2 and 9 before decaying toward zero:

Should you ever find yourself playing this game in the real world, according to Tom’s results you should opt for an odd number of dice.

Want more puzzles?

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.


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