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You’ve Been Marooned By Kidnappers. Can You Escape At Dawn?

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

From Marcus Farbstein and Mark Baird, a logical mystery story, with your very freedom hanging in the balance:

You’re super rich, and you often joke with your cadre of intelligent friends and family about getting kidnapped. You all agree that if you were ever kidnapped, the evildoers would knock you out so stealthily that you’d never feel the blow. Then, one snowy night, you step out of a restaurant and, just as predicted, never feel the blow.

When you stir back into consciousness, it’s night, but it’s not snowy. You find yourself sitting on a beach. The sky is clear, with no moon. In front of you stands a shadowy figure whose face you can’t make out. He throws a blocky rectangular object at your feet. “That’s a satellite phone,” the figure growls. “It’s got one minute of battery left in it. Use that to call your people to let ’em know you’re not dead — but not until daylight.” He tosses a paper bag next to the sat phone. “That’s some sandwiches and water, enough for a few days. That’s salt,” he explains, waving toward the surf. “If your people pay our ransom,” he continues, “We’ll come get you. Otherwise, there won’t be any more paper bags. Remember, wait until daylight to make that call.”

He then turns and climbs into a dinghy in the light surf, starts its outboard motor and zooms away. All this time you’ve been too groggy to do anything but listen. Now you watch as the dinghy disappears into the gloom, its wake a faint wash of phosphorescence that quickly fades. Later, there’s a bare wink of lights at the horizon, presumably the mothership getting underway and leaving.

Even though it’s a moonless night, there’s sufficient starlight to assess your surroundings. Your grogginess is gone and you walk about. You’re on a tiny island, which you estimate is a bit more than a mile by half a mile. There are no trees; it’s all flat sand. You taste the water rolled up by the surf, and it is indeed salt. The air is cool, but not cold. Your wallet, expensive chronometer, keys, cell phone, jewelry and small change are all gone; all you have are the clothes on your back — even your shoes and socks have been taken. The bag contains four sandwiches, all liverwurst with peanut butter on cheap rye bread, and four one-pint bottles of water. No napkins. Your knowledge of astronomy is too weak to try to estimate your location by the stars, but you’re not stupid. Before daybreak, you’ve worked out exactly how you’ll use that minute of time on the satellite phone so that your people, who are also not stupid, will be able to dispatch rescue.

What will you say?

Submit your answer

Riddler Classic

From Theodore James, some further international intrigue:

Mathematician Dr. Lana Gurtin has a problem to solve. She was hired by British intelligence for a top-secret assignment, but things have not gone according to plan. The year is 1942 and the Germans are rolling out a new and powerful tank, the Uberpanzer. Prominently displayed on the back of each Uberpanzer is its serial number, which is simply the number of tanks that had been built when it rolled off the line. So the first tank built has the serial number 1, the second one built has a 2, and so on.

Recently, a number of these new tanks were spotted by British scouts who recorded the serial numbers that they saw. They immediately sent this information to British intelligence, hoping that the serial number data could be used to estimate the total number of Uberpanzers the Germans have built. This is when Dr. Gurtin was brought on to head the project.

But then the unexpected happened. A German spy intercepted the dossier with the data before it could reach MI6. By the time British agents caught up with the spy, most of the data had been destroyed. However, two pieces of information were recovered from the debris. One: The lowest serial number recorded was 22. Two: The highest serial number recorded was 114.

Luckily, Dr. Gurtin knows exactly what to do. Assuming that the original data set was a random sample of serial numbers, what is Dr. Gurtin’s best estimate of the total number of Uberpanzers the Germans have built?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Tyler Silber 👏 of New York City, winner of last week’s Riddler Express!

Last week, you and your friend faced off in a friendly game of Lotería, a traditional Mexican game of chance akin to bingo. You each had a card with a four-by-four grid of images, drawn randomly from a deck of 54 possible images. Each image could appear at most once on a card. A caller randomly drew cards from a deck containing all 54 possible images, and you marked that image off on your card if it appeared there. The game ended when one of the players filled their entire card. What was the probability that either of you ended the game with an empty grid — that is, the odds that none of your images was called?

The chances were minuscule — about \(3.508 \times 10^{-12}\), or on the order of one in a trillion.

Taylor Firman, this puzzle’s submitter, writes: “The inspiration for this problem actually comes from reality: While on vacation, a friend of mine ended a game with zero matching images and I seemed to be the only one flabbergasted by the odds.” And he discusses how to arrive at these flabbergastingly minuscule odds on his blog.

The solution flows from two simple facts. First, the players’ two grids can’t have any overlap — that is, they can’t share any images. If they did, when one player won, the other player’s grid couldn’t be empty. Second, the deck of images must be ordered such that all of one player’s images come up before any of the other player’s images. The mathematical trick is to combine the probabilities of these two events into our final answer.

For example, we can begin by calculating that, given the first player’s grid, there are 38!/(16!22!) — or about 22 billion — non-overlapping grids that the second player could have, out of a raw total of 54!/(16!38!) — or about 21 trillion — possible grids. These are the formulas for choosing 16 cards from sets of 38 (the 54-image deck minus the 16 images on the first player’s grid) and 54 cards, respectively. We can calculate the number of orderings of the images similarly, using factorials and the tools of combinatorics. I refer you to Taylor’s solution for the gritty details.

You were also asked how the probability of having an empty card at the end of the game changed as the rules of the game changed — for example, if there were more or fewer unique images in the deck. Taylor provided that data, too, in interactive chart form. If the size of the grid was fixed at four-by-four, here’s how the zero-match probability increased as the number of unique images increased:

The version of the chart on Taylor’s site also lets you see how the odds change for different grid sizes.

One in a trillion, you say? Never tell me the odds. Unless, of course, I ask you to in a math puzzle column, in which case, please do tell me the odds.

Solution to last week’s Riddler Classic

Congratulations to 👏 Eric Mann-Hielscher 👏 of Brooklyn, New York, winner of last week’s Riddler Classic!

On the excellent British game show “Countdown,” there is a segment called the Numbers Game. You ask for six numbered cards in total — up to four “large” cards, with “small” cards making up the balance. Large cards are drawn at random from a deck containing the numbers 25, 50, 75 and 100. Small cards are drawn at random from a deck containing two each of the numbers 1 through 10. Then a random number generator spits out a three-digit target number, and you have 30 seconds to use addition, subtraction, multiplication and division to combine your six numbers into a total as close to the three-digit number as you can. (You can only use a number as many times as it comes up in the six-number set. You can only use the mathematical operations given. At no point in your calculations can you end on something that isn’t a counting number. And you don’t have to use all of the numbers in your set.) If you can nail the target exactly, we’ll call that game solved.

This riddle was twofold. First, what number of large cards is most likely to produce a solvable game and what number of large cards is least likely to be solvable? Second, what three-digit numbers are most or least likely to be solvable?

Two large cards is best — you can solve about 98 percent of all target numbers. Zero large cards is the worst — you can solve only about 84 percent of the targets. In general, large target numbers tend to be the hardest to solve. Assuming you’ve chosen the optimal two large numbers, 967 is the most difficult target number — you can only solve it about 89 percent of the time. There are a few smaller target numbers that you can always solve.

Given the huge number of combinations of cards and target numbers, this riddle is a problem for a computer to solve. Well, for you and your computer. Our winner Eric, along with solvers Benjamin Phillabaum, Ryan Vilim and Ben Weiss, were kind enough to share their code.

Solver Jason Ash plotted the results of his programmatic solution for each number of large cards. You can see that the chances of solving tend to go down as the target number gets bigger, and that two large cards is the best choice — its chances of solving hover closest to 100 percent.

And solver Austin Buscher provided another look at the problem, charting the frequency of combinations capable of yielding each three-digit number. “The histogram shows what we would intuitively expect,” he writes. “Smaller numbers can be computed more often. There are also peaks at multiples of 25, 50 and especially 100, another intuitive byproduct.”

Now if only I could do all of this on TV in less than 30 seconds …

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 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. Eastern time on Sunday. Have a great weekend!

Oliver Roeder is a senior writer for FiveThirtyEight.

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