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Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and 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 week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

Our first-ever Riddler Express centered on the “Sesame Street” character Count Von Count. He counts aloud by tweeting one number at a time, and with the numbers spelled out like so: “One!” “Two!” “Three!” … “Five hundred thirty eight!” etc. The original puzzle was to see how high the count could count before hitting Twitter’s 140-character limit. Riddler Nation found that it was all the way up to 1,111,373,373,372, or, as the count would write it, “One trillion one hundred eleven billion three hundred seventy three million three hundred seventy three thousand three hundred seventy two!”

But since then, Twitter has expanded its limit to 280 characters. That means Count Von Count’s possibilities have expanded as well. How high can Count Von Count count now? How much higher can the count count because of the new character limit?

Important note! The count is enthusiastic and must end all of his tweets with an exclamation point.

Riddler Classic

Speaking of compulsive counting counts, I, too, have a problem. Whenever I see a string of digits — a license plate, a ZIP code … a stranger’s PIN number — I try to turn them into a true mathematical equation by maintaining the digits’ order and inserting symbols. For example, the ZIP code of my office is 10023. I could turn that into 1+0+0+2=3 or 1×0=0×23 (or many other equations).

This game gets more complicated the more digits you have, and strings of four or five digits seem to be the sweet spot where there’s a lot of fun to be had.

Considering strings of all lengths and inserting only common mathematical symbols — ) ( + – × ÷ ^ = — what proportion of each string length has true mathematical equations lurking inside of it? (For example, for strings of length three, you’d consider the groups of digits 000, 001, …, 999, trying to insert symbols into each one to find a correct equation. You’d find that 000 has many possibilities, whereas 129 has none. For strings of length four, you’d consider 0000 through 9999, and so on.) As the strings get longer, there’s more you can do with them: Is there a string length where every possible string has a correct equation inside of it?

Extra credit: What if you were able to insert some more exotic symbols, such as the notation for factorial or modulus?

Solution to last week’s Riddler Express

Congratulations to ÑÑâÐ David Leman ÑÑâÐ of Indianapolis, winner of last week’s Riddler Express!

In competitive darts, a common game is called 501. A player starts with 501 points and subtracts the score of each throw. He or she must finish with exactly zero points. (Also, according to the rules, the final dart must land in either the bull’s-eye or the outer, doubled segments.) Finishing a game in the minimum number of throws is a rare feat, akin to a perfect 300 game in bowling. Last week I asked: What is the minimum number of throws? How many different ways are there to do it?

The minimum number of throws to count down from 501 is nine. The most valuable space on a dartboard is triple-20 — the piece of the 20 slice inside the inner ring — which is worth 60 points. 501 divided by 60 gives us a number between eight and nine, so we know we can’t do it in eight throws or fewer. Here’s one way to do it in nine:

60 + 60 + 60 + 60 + 60 + 60 + 60 + 57 + 24 = 501

That’s a string of seven triple-20s, then a triple-19 and then a double-12, for a total of 501. (Remember that your game must end on a doubled throw.)

There are 3,944 possible routes to nine-dart perfection — some of these include the same throws, but in different orders. The basic idea is to figure out how many 60s you might need, what throws could supplement those, and then how many ways you could mix up the order in which those throws came. Solver Mike Seifert went low-tech and shared his work:

And solver Amy Leblang took the high-tech route and was kind enough to share the code she used to calculate that total.

To see this math in action, here’s a video (with a hat tip to my colleague Daniel Levitt) of a man very nearly completing two perfect nine-dart finishes in a row:

Solution to last week’s Riddler Classic

Congratulations to ÑÑâÐ Ben Breadsell ÑÑâÐ of Perth, Australia, winner of last week’s Riddler Classic!

First, the maximum possible score is 7. This can be achieved with one dart at the very center of the board and six other darts evenly spaced along the board’s edge. Mahalingam Vaidhyanathan, the puzzle’s submitter, illustrated what this looks like:

Second, the probability of getting a score greater than 1 is about 0.41 — to be precise, it’s $$\frac{3\sqrt{3}}{4\pi}$$.

To see how this works, take a look at another of Mahalingam’s illustrations:

The blue circle represents the dartboard, the green dot represents where your first dart landed, and the blue dot is the center of the dartboard. The green circle represents the “bad” area for your second dart — throw it anywhere within the green circle and you’ll either be within 1 foot of your first dart (game over) or miss the board entirely (which you’re too skilled to do anyway).

So, to solve for the probability that you’ll get a score greater than one dartpoint, we need to solve for how often you’ll land on the dartboard but outside the “bad” radius around your first dart. That in part depends on where that first dart landed.

Let x be the distance between your dart and the center of the board. To get a score greater than 1, your second dart must fall outside of the green circle and in the blue area. That area of the overlap, call it A, equals:

\begin{equation*}A=2\cos^{-1}\left(\frac{x}{2}\right)-x\sqrt{1-\left(\frac{x}{2}\right)^2}\end{equation*}

To find the probability over all the possible x’s where your dart could land, we remember that our darts fall according to a uniform density and then integrate over x as it ranges from 0 (your dart lands at the center) to 1 (your dart lands on the edge). That looks like this:

\begin{equation*}1-\frac{1}{\pi}\int_0^1 (2x)(A)dx=\frac{3\sqrt{3}}{4\pi} \end{equation*}

That equals roughly 41 percent.

The answer to the third, and last, question: The expected value of your score is a fairly meager 1.47. The integral math to get that answer gets too complicated to handle with pencil and paper, so you’re best off building a computer dart-throwing simulation. Solver Samir Khan was kind enough to share the code he used to simulate 100,000 dart games.

We’ve already found the probability of your scoring exactly 1 point. It’s 1 minus the answer above, or about 59 percent. The probability of scoring 2 points turns about to be about 35 percent, 3 points about 6 percent, 4 points about 0.1 percent, and anything higher is negligible. Solver Steven W. made a histogram of the frequency of scores, over 1 million games:

Taking an average of these scores weighted by their probabilities gives an expected score of about 1.47.

Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.

Footnotes

1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. EST on Sunday. Have a great weekend!

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.