# Can You Roll the Dungeon Master’s Dice?

### You and a friend are choosing different kinds of dice at random. What’s the probability you’ll still roll the same number?

Welcome to Fiddler on the Proof! The Fiddler is the spiritual successor to FiveThirtyEight’s The Riddler column, which ran for eight years under the stewardship of myself and Ollie Roeder.

Each week, I present mathematical puzzles intended to both challenge and delight you. Beyond these, I also hope to share occasional writings about the broader mathematical and puzzle communities.

Puzzles come out Friday mornings (8 a.m. Eastern time). Most can be solved with careful thought, pencil and paper, and the aid of a calculator. Many include “extra credit,” where the analysis gets particularly hairy or where you might turn to a computer for assistance.

I’ll also give a shoutout to 🎻 one lucky winner 🎻 of the previous week’s puzzle, chosen randomly from among those who submit their solution before 11:59 p.m. the Monday after that puzzle was released. I’ll do my best to read through all the submissions and give additional shoutouts to creative approaches or awesome visualizations, the latter of which could receive 🎬 Best Picture Awards 🎬.

## This Week’s Fiddler

From Julien Beasley comes a coincidence of dice:

Two people are sitting at a table together, each with their own bag of six “DnD dice”: a d4, a d6, a d8, a d10, a d12, and a d20. Here, “d*X*” refers to a die with *X* faces, numbered from 1 to *X*, each with an equally likely probability of being rolled.

Both people randomly pick one die from their respective bags and then roll them at the same time. For example, suppose the two dice selected are a d4 and a d12. The players roll them, and let’s further suppose that both rolls come up as 3. What luck!

What’s the probability of something like this happening? That is, what is the probability that both players roll the same number, whether or not they happened to pick the same kind of die?

## Extra Credit

Julien’s puzzle idea didn’t stop there. Instead of two people sitting at the table, now suppose there are *three*.

Again, all three randomly pick one die from their respective bags and roll them at the same time. For example, suppose the three dice selected are a d4, a d20, and a d12. The players roll them, and let’s further suppose that the d4 comes out as 4, the d20 comes out as 13, and the d12 comes out as 4. In this case, there are *two* distinct numbers (4 and 13) among the three rolls.

On average, how many *distinct* numbers would you expect to see among the three rolls?

## Making the Rounds

There’s so much more puzzling goodness out there, I’d be remiss if I didn’t share some of it here. This week, I’m sharing a gem from Matt Enlow that I saw way back in, oh, 2023. Here’s the link to the original post, but be warned—the thread contains spoilers! For a spoiler-free experience, I’ll paraphrase the puzzle here:

In three-dimensional space, suppose you have six points, no four of which lie in the same plane. Furthermore, suppose that three of the points are colored red, while the other three are colored blue. Is it

alwayspossible—no matter which six points are picked and no matter which three are red and which three are blue—to find a plane such that all three red points are on one side of the plane and all three blue points are on the other side of the plane?

If you think the answer is yes, then prove it! If you think the answer is no, then under what sort of conditions is impossible to find such a dividing plane?

## Want to Submit a Puzzle Idea?

Then do it! Your puzzle could be the highlight of everyone’s weekend. If you have a puzzle idea, shoot me an email. I love it when ideas also come with solutions, but that’s not a requirement.

## Last Week’s Fiddler

Congratulations to the (randomly selected) winner from last week: 🎻 Q P Liu 🎻 from Santa Cruz, California. I received 15 timely submissions, of which 8 were correct—the smallest number of solvers in Fiddler history to date.

So let’s get one thing out of the way up front—this was arguably the hardest Fiddler I’ve ever run, and would have been more appropriate as Extra Credit, had it not been for the *even more challenging* version of this puzzle below. Special kudos to all of you who even *attempted* this one!

Last week, I started off by observing that there were countless ways to pick three points within the unit disk (i.e., the points on the unit circle and inside the unit circle). And for each of these ways, there were countless triangles you could draw within the disk that did not strictly contain any of the three points. By “strictly,” I meant that the three points—or “obstacles,” as I’ll refer to them—could be on the edges of a triangle, but not *inside* a triangle.

For any trio of obstacles in the unit disk, you were asked to consider the *largest* triangle (by area) within the disk that didn’t strictly contain any of the three obstacles.

As an illustration, I offered the diagram below:

The three red obstacles were all within the unit disk. Various green and red triangles that didn’t strictly contain the obstacles were shown. The largest of these by area was highlighted with a blue edge, and its area was approximately 1.1842.

By positioning the three obstacles more strategically, it was possible to reduce the area of the largest such triangle. What was the *minimum* possible area this largest triangle could have?

Now before we get to optimizing the positions of the three obstacles, let’s first take a look at what the candidates for the largest triangle *could have* been. If all three obstacles were close to the edge of the disk, then the largest triangle you could make was an inscribed equilateral triangle, with an area of 3√(3)/4, or about 1.299. In fact, this was the largest triangle you could possibly fit in the disk.

But when the obstacles were closer to the middle, the largest possible triangle got smaller. In particular, there were two kinds of triangles to consider. First, you could have a triangle that was squeezed between the three obstacles. To make such a triangle, you picked any two of the three obstacles (noting that there were three ways to do this), and you identified the chord that contained them. This chord was one edge of your triangle. Extending one of the chord’s two endpoints (noting that there were two endpoints to choose from, as shown below) through the third obstacle gave you a second edge of the triangle.

The third edge closed off the triangle. In total, you had six of these “internal” triangles to consider.

Alternatively, depending on the three obstacle locations you chose, it was possible for the largest triangle to never even come close to one of your obstacles. Again, one edge passed through two obstacles, but this time the other two edges both veered away from the third obstacle. In this case, the third vertex of the triangle was at the midpoint of the arc subtended by the chord with two obstacles on it, since this maximized the height (and therefore area) of the triangle. There were three such “external” triangles.

That gave you a grand total of six internal triangles plus three external triangles, or nine total triangles to consider. For your enjoyment, I created an interactive graph in Desmos where you can drag the three obstacles around and see these nine triangles change in real time. The largest of these triangles is highlighted.

With these nine triangles constructed, what sort of arrangement minimized the area of the largest triangle? Well, when the obstacles were too far apart, the internal triangles wound up being larger than the external triangles. But when the obstacles were too close together, the external triangles wound up being larger than the internal triangles.

When optimization problems (like this one) ask for the “minimum of the maximum” or the “maximum of the minimum,” the solution often comes when the tradeoff between different choices is made moot by setting these options equal to each other. In this case, that meant finding an arrangement of obstacles for which all nine triangles—the six internal and the three external—all had precisely the same area. After all, when these nine areas *weren’t* all the same, then you could presumably make the largest triangle a little smaller by making the smaller triangles a little larger.

When the three obstacles were symmetrically spaced around the disk’s center so that they formed a concentric equilateral triangle, that guaranteed that the six internal triangles all had the same area as each other, *and* that the three external triangles all had the same area as each other. However, you needed a bit more algebra to ensure the areas of the internal triangles equaled the areas of the external triangles.

Consider the two triangles shown below (one internal and one external) when the three obstacles were each a distance *a* from the disk’s center. Since both triangles shared a common vertical base, having the same area meant they needed to have the same height, which was (1−*a*/2).

Notice that the points *A*, *B*, and *C* were all collinear, which allowed you to determine the coordinates of point *C*:

The final loose thread was making sure that point *C* was on the unit circle, i.e., that its distance from the origin was 1. Applying the Pythagorean theorem to its coordinates and then doing some more algebra ultimately gave you the quartic equation, 5*a*^{4} − 14*a*^{3} + 15*a*^{2} − 16*a* + 4 = 0.

I don’t know about you, but I never calculate the roots of quartics by hand. Fortunately, this quartic had a factor of (*a*−2). Dividing both sides by this factor gave you the slightly friendlier cubic equation, 5*a*^{3} − 4*a*^{2} + 7*a* − 2 = 0. I still don’t know about you, but I never calculate the roots of cubics by hand, either. In any case, the remaining real root was roughly 0.321.

Again, that was the distance between each of the three obstacles and the disk’s center. In terms of *a*, the area of each of the nine triangles was (1−*a*/2)√(1−*a*^{2}/4), which turned out to be roughly **0.8286**. And that was the answer! This was the area of the smallest largest triangle you could make by positioning the three obstacles within the disk. (If you were in the ballpark of 0.829, I marked your answer as correct.)

Solver Michael Schubmehl wasn’t satisfied with this approximation, and instead submitted an exact answer. I’m not going to transcribe it here, but you can find it in Michael’s write-up. Consider yourself warned: There are large, relatively prime coefficients and nested radicals.

## Last Week’s Extra Credit

Congratulations to the (randomly selected) winner from last week: 🎻 Peter Ji 🎻 from Madison, Wisconsin. I received 7 timely submissions, of which 5 were correct. As challenging as last week’s Fiddler was, this was even trickier. Five solvers was indeed a record low for any puzzle here.

For Extra Credit, instead of placing three obstacles, you had to place *four* obstacles within the unit disk to minimize the area of the largest triangle (by area) that didn’t strictly contain any of the obstacles.

What was the *minimum* possible area this largest triangle could have?

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