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Who Will Win The Politicians’ Secret Vote?

Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here — math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world, including you, the readers. You’ll find this week’s puzzle below.

Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the form at the bottom. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible for the shoutout, I need to receive your correct answer before 11:59 p.m. EST on Sunday — have a great weekend!

Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏 Paul Holmes 👏 of Christchurch, New Zealand, our big winner. You can find a solution to the previous Riddler at the bottom of this post.

Now, here’s this week’s Riddler, which has quite the economic genealogy. It comes to us from Laura Feiveson, an economist at the Federal Reserve’s Board of Governors; it came to her from Yale economist Barry Nalebuff; and it came to him from Nobel-winning economist Thomas Schelling.


Suppose that five politicians, disgusted with the current two-party system, come together to choose a third-party candidate to run in the 2016 presidential election. The politicians’ names are Anders (A), Blinton (B), Cubio (C), Drump (D) and Eruz (E). Not wanting to spend all their time campaigning in Iowa and New Hampshire in winter, they decide instead to pick which of them will be the candidate at a secret meeting with just the five of them. The voting procedure is as follows: They will first hold a vote of A versus B. (The five politicians are the only voters.) The winner of that vote will then be paired against C. That winner will be paired against D, and finally that winner will be paired against E. They will declare the winner of that last matchup to be their candidate.

Each of A, B, C, D and E wants to be the presidential candidate themselves, but also has clear preferences over the others. Furthermore, the politicians’ preferences are common knowledge. Their preferences are as follows (“X > Y” means Candidate X is preferred to Candidate Y):

Candidate A: A > B > C > D > E

Candidate B: B > A > E > D > C

Candidate C: C > D > A > E > B

Candidate D: D > B > A > E > C

Candidate E: E > D > B > C > A

All of the politicians are forward-looking and vote strategically.

Question 1: Who will be chosen as the presidential candidate?

Now assume that A has the flu and is forced to miss the voting meeting. He is allowed to transfer his vote to someone else, but he can’t make that other person commit to vote against her own self-interest.

Question 2: To whom should he transfer his vote, given his candidate preference outlined above (A > B > C > D > E)?

Question 3: Who will win the candidacy now?

Question 4: A month before the meeting, Candidate A must decide whether or not to get the flu vaccine. Should he get it?


Need a hint? You can try asking me nicely. Want to submit a puzzle or problem? Email me.


And here’s the solution to last week’s Riddler, concerning trying to cross a river after a dangerous storm, which came to us from Chris Scambler. You all nailed this one — 75.4 percent of submitted solutions arrived at the correct answer: 50 percent.

There are a number of ways to attack this problem, but the simplest requires (gasp!) essentially no math at all. In the stated problem, you want to cross the river on foot via the bridges and group of islands. But imagine also the captain of a ship, on the river to the west of the group of islands, who wants to pass through completely to the east side, but can’t pass a bridge if it hasn’t been knocked out (his ship is too tall). The two of you are facing the exact same problem in reverse — imagine rotating the original problem 90 degrees. You can cross the river if and only if the captain cannot pass through. Since your problems are structurally identical, the probability each of you will get your wish must be equal, and it will always be the case that exactly one of you gets your wish. (Your wishes are mutually exclusive.) Therefore you both must have a 50 percent chance — you of crossing the river, and the captain of passing by the islands. This answer is correct regardless of the size of the N-by-N+1 island group.

The captain’s problem is illustrated by this picture, from reader Chris Roudiez:

captain

You could also attack this problem with brute force, going through each of the 8,192 (\(2^{13}\)) possible states of the set of 13 bridges. Indeed you will find that 4,096 of those (50 percent) are traversable, as reader Jon Wiesman showed. Or you could attack it via computer simulation, as friend-of-the-Riddler Zach Wissner-Gross demonstrated with this video, for the 10-by-11 islands case:

Also, shoutout to this fantastic post on the problem’s solution, replete with nifty interactive bridge simulations, from reader Nick Brown.

Finally, it’s time to present the inaugural 🏆 Coolest Riddler Extension Award 🏆. The trophy goes to [drumroll …] Ryan Soklaski, who extended the problem to show, via simulation and in one graph, the chances of crossing the river for various island-group sizes and bridge strengths. This is what he found:

extension

For bridges that survive the storm half of the time, you have a 50 percent chance of crossing, no matter the number of bridges, as we saw in the solution above. (That can be seen in the intersection of lines in the middle of this graph.) But larger groups of bridges are more sensitive to changes in their individual bridges’ strength. Lower the chance of any one bridge surviving the night just a bit in an N=100 group of bridges, for instance, and the chances of being able to cross quickly plummet to near zero. However, raise the chance of any one bridge surviving just a little in a big group, and you’re quickly almost guaranteed safe crossing.

Soklaski also writes, “The behavior of this system is pertinent in condensed matter physics, as it is a simple example of a system that exhibits a critical phase transition.”

Honorable mention in the extension category goes to Wissner-Gross, who suggested a three-dimensional extension; to Daniel Tello, who suggested, ever so artfully, “wicker tripod” bridges; and to the team of Zach Helming, Ira Fich and Jordan Coffman, who suggested that every network of bridges be treated, itself, as a bridge, like so. “YO, MAN. FRACTALS,” they wrote in their submission. They continued: “Actually, no, that’s super dumb. Sorry.”

Well, three readers’ super dumb idea is one puzzle editor’s honorable mention. Thanks for Riddling.

Oliver Roeder is a senior writer for FiveThirtyEight.

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