Can You Save The Drowning Swimmer?

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-sized 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 Humberto Barreto, a seaside problem:

You’re a lifeguard standing on the beach, right at the edge of the water, and gazing out over the ocean. You see someone drowning 100 meters to the right of you and 100 meters away from shore. You can run 100 meters in 15 seconds and swim 100 meters in 75 seconds. (The beach drops off steeply, meaning that you can’t run in the water.) What’s the fastest you can get to the victim?

Submit your answer

Riddler Classic

A few weeks ago, this column featured a simplified poker game called baby poker. Today, from Laurent Lessard, a fascinating twist on that game:

Toddler poker is played by two players. Each is dealt a “card,” which is actually a number randomly chosen uniformly from the interval [0,1]. (It could be 0.1, or 0.9234781, or 1/π, and so on.) The game starts with each player anteing $1. Player A can then either “call,” in which case both numbers are shown and the player with the higher number wins the $2 on the table, or “raise,” betting one more dollar. If A raises, B then has the option to either “call” by matching A’s second dollar, after which the higher number wins the $4 on the table, or “fold,” in which case A wins but B is out only his original $1. No other plays are made.

What is the optimal strategy for each player? Under those strategies, how much is a game of toddler poker worth to Player A?

Extra credit: What if the value of the raise is $k — i.e., players stand to profit $k instead of $2 after the raise?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Jeffrey Deipolyi 👏 of West Islip, New York, winner of last week’s Express puzzle!

You were asked to submit a whole number between 1 and 1,000,000,000, and the submission that was closest to ⅔ of the mean of all the submissions would be declared the winner.

The mean was 195,921,656, or nearly 20 percent of the upper limit of the range. Jeffrey, our winner and resident Nostradamus, chose 196,352,731.

What prediction does math have to offer about the outcome of this puzzle? This game — a number of guessers all aiming for ⅔ of their average — has a relatively straightforward Nash equilibrium, which you can find using something called the iterated deletion of dominated strategies. (Just flows off the tongue, I know.) In layman’s terms, that means: Identify the worthless strategies — those for which there is always a better alternative. Eliminate those strategies from the game’s rules. Then, consider the game anew, identify the worthless strategies once again and delete those, and so forth. For some games, just one strategy will remain. That’s your equilibrium.

In this game, it is never optimal to submit a number greater than 666,666,666, because those numbers can never win. (The highest the average could possibly be is 1,000,000,000, so the highest that ⅔ of the average could possibly be is about 666,666,666.) So eliminate all those possible submissions. Then, assuming that everyone is following this rational suit, it’s never optimal to submit a number greater than 444,444,444, because now those numbers can never win. So eliminate all those. Then it becomes never optimal to submit a number greater than 296,296,296, and then 197,530,864, and so on. Following that logic all the way down, the unique equilibrium is that every player submits the lowest possible number — in this case, the number 1. No one can get any closer to ⅔ of the average, and everyone ties for the win.

Suffice it to say, we did not arrive at that equilibrium. In this case, the trolls prevailed. Eighty-one “solvers” submitted an answer of 1,000,000,000 and an additional 25 submitted 999,999,999 — answers that can literally never win. Their submitted justifications ranged from “to throw off everyone else’s logical calculation” to “game theory can’t handle anarchy” to “I just want to watch the world burn.” And burn it did. Game theory is not well-equipped to deal with modern trolldom.

Jeffrey arrived at his winning submission this way: He started at a baseline average of 500,000,000. “On a hunch, most entries would therefore be ⅔ of that. The question really is, ‘How many times will people consider the ⅔-multiplier?’ I guessed somewhere between two and three, and chose at random the last six digits.”

Excluding all the submissions of 1,000,000,000 and 999,999,999, the mean dropped to 153,895,557.

Solution to last week’s Riddler Classic

Riddler Nation has a new king. All hail 👑 Cyrus Hettle 👑 of Lexington, Kentucky.

Last week, I put you each in command of an army of 100 soldiers. You waged a war over 10 castles, worth 1, 2, 3, …, 9, and 10 victory points. You could distribute your troops among the castles however you liked, but you didn’t know in advance how your archenemy — who was, in turn, each other submitter — would distribute his or hers. For each one-on-one war, whichever of the two combatants sent more troops to a castle conquered it and claimed its points, and whoever claimed the most points won. I adjudicated all the nearly 1 million one-on-one matchups from about 1,400 submitted battle plans, and the bloody results are in.

The table below summarizes the top five’s strategies — how many soldiers they sent to each castle — and what record in the round robin of one-on-one wars they wound up with.

Our new ruler Cyrus did not provide a rationale for his troop deployment, but far be it from me to question his highness. As Sun Tzu said, “All warfare is based on deception.”

Ken Nickerson, who finished in a very close second with a similar strategy, explained his thinking like this: “In general, I will win over a given opponent if I tend to win castles by having only slightly more soldiers than my opponent, but lose castles by having much less than my opponent. Each of my 10 castles will have armies designed to slightly beat one of the following opposing strategies for that castle: an essentially undefended castle (Castles 6, 9 and 10), an equally distributed castle (Castles 1, 2, 3, 4 and 5) or a focused attack castle (Castles 7 and 8).”

Brett Seymour, our bronze medalist, described his approach this way: “Winning the lower seven gives you more than half the points, so the top three values are largely ignored save for a scouting force of two to three to prevent a lone scout of the opponent from stealing. Then starting at Castle 7 we deploy in force and curve down from there.”

Here’s how all the people who submitted battle plans distributed their troops:

A few general patterns emerged. The most striking: not fighting hard for Castle 10. The median number of troops sent to fight for the highest-point castle was only two soldiers (the same as the median sent to Castles 2 and 3). Castle 9 was similarly relatively avoided. Indeed, everyone in the top five sent at most three soldiers to either of those castles. The rationale most often given was that others would probably fight hard for those high-value castles, so vying for them may be too costly. (As a result, though, you may have actually done well competing for Castle 10.) Castle 8 was the most hotly contested, on average.

In game theory, this sort of problem is known as a Colonel Blotto game. Blotto games are simple to describe but dastardly to solve. Nash equilibria exist in these games but are found using gnarly math and computation. Perhaps your fellow warlords’ strategies can shed some empirical light on the game, however. For the curious — or the microeconomics grad students scrabbling for a paper topic — we’ve posted all the battle plan data, stripped of identifying information, to GitHub.

Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.