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Who Will Survive The Dodgeball Duel?

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 Andrew Young, a score-settling puzzle from the schoolyard:

Three expert dodgeballers engage in a duel — er, a “truel” — where they all pick up a ball simultaneously and attempt to hit one of the others. Any survivors then immediately start over, attempting to hit each other again. All of the combatants have perfect accuracy and will eliminate any target they hit (dodgeball this is not), without the possibility of a return volley. Two of the players are fast, and one is slow. If a fast player and the slow player target each other, the fast one always wins. If the two fast players target each other, each has a 50 percent chance of winning. A fast player is not fast enough to throw twice before the slow player gets a first throw off, so the slow player will win when targeting a fast player who isn’t targeting him or her. All players are aware of the abilities of the other players.

Assuming each dodgeballer pursues an optimal strategy, with the goal being survival, what are the odds of victory for one of the fast players and the slow player?

## Riddler Classic

Consider another truel — one where the three combatants are equally fast but unequally accurate. Name them Abbott, Bob and Costello. Each has an accuracy of a, b and c, respectively. That is, if Abbott aims at something, he hits it with probability a, Bob with probability b and Costello with probability c. The abilities of each player are known by the others.

Let’s say Abbott is a perfect shot: a = 1. Again, suppose the players follow an optimal strategy. Which player, for every possible combination of Bob and Costello’s abilities (b and c), is favored to survive this truel? (You are welcome to submit your answer as a diagram, if you’d like.)

## Solution to last week’s Riddler Express

Congratulations to 👏 Daniel Plants 👏 of Kirkwood, Missouri, winner of last week’s Riddler Express!

Last week’s problems were inspired by a paper from a team of mathematicians and computer scientists that was presented at a conference called “Fun With Algorithms” in 2012. They all had to do with hanging a picture frame by a cord in slightly odd and very specific ways such that you could guarantee, under certain circumstances, that the frame would crash to the ground. Why? Because Riddler Nation, as you know, is a strange, mysterious and arbitrary place.

For the Riddler Express, you were asked to hang your frame from two nails in such a way that the removal of either nail would cause the frame to fall. If you imagine hanging a picture from two nails in a “normal” way, the removal of one nail would simply leave the picture hanging from the other nail, although maybe a bit askew. So we’ll need to do something different.

There’s not much of a silver mathematical bullet here, and most solvers arrived at the hanging method via trial and error. The trick, of course, is to arrange the cord such that the frame’s reliance on one nail is dependent on the other, and vice versa. That way, when one nail goes, the whole shebang collapses. This can be accomplished with loops. Solver Thomas Epp explained: “Pulling out either nail will cause the loop around that nail to vanish, and the cord ends up no longer wrapped around the other nail.” Thomas illustrated his solution in the Picasso-esque illustration below:

Our winner, Daniel, arrived at his solution after “lots of random drawings in my notebook, followed by wrapping my headphone cord around two cups with straws in them to test my solution.” MacGyver, eat your heart out.

Others thought outside the box, er, the frame. Ryan Garwood proposed a simple, elegant and, frankly, illicit solution: Ignore the cord altogether. Simply pound two nails into the wall, a bit apart, level with each other, and balance the frame on top of them. Remove one nail and down tumbles the frame.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Kyle Hird 👏 of Cincinnati, winner of last week’s Riddler Classic!

Last week’s Riddler Classic asked you to extend the frame puzzle to three and four nails. It also asked about an option using two red nails and two blue nails, where removing both nails of one color caused the picture to fall, but removing one nail of each color left the picture hanging.

The math paper mentioned above introduces some mathematical notation to describe various methods of picture hanging, and it connects these puzzles to the mathematical fields of group theory and algebraic topology. Suppose there are N nails to contend with. For one given nail from those N, call it i, $$x_i$$ means wrapping the cord around that ith nail clockwise, while $$x_i^{-1}$$ means wrapping the cord around the ith nail counterclockwise. Stringing these symbols together will create our particular picture-hanging recipes.

For three nails, the solution is: $$x_1 x_2 x_3 x_2^{-1} x_3^{-1} x_1^{-1} x_3 x_2 x_3^{-1} x_2^{-1}$$.

For four nails, it’s: $$x_1 x_2 x_1^{-1} x_2^{-1} x_3 x_4 x_3^{-1} x_4^{-1} x_2 x_1 x_2^{-1} x_1^{-1} x_4 x_3 x_4^{-1} x_3^{-1}$$.

Laurent Lessard provided excellent illustrations of these solutions. In his diagrams, which show a progression of arrangements toward an end solution, Laurent used the mathematical symbols for “and” (∧) and for “or” (∨). So, for example, in the first image below, the frame falls if nails A and B and C are removed. In the second image, it falls if nail A or nails B and C are removed. The third image is our solution: The frame falls if nails A or B or C are removed.

A similar process works for four nails. In the first image below, all of the nails must be removed for the frame to fall. In the final image, however, removing any of the nails will do.

Finally, for the question about two red nails and two blue nails, the solution is: $$x_1 x_2 x_3 x_4 x_2^{-1} x_1^{-1} x_4^{-1} x_3^{-1}$$, where nails 1 and 2 are red and 3 and 4 are blue.

But we’ve actually already seen the solution above! It was an intermediate step in the final illustration — just imagine that A and B are red and C and D are blue. It is also essentially the same problem as the Riddler Express. Just image one of the two nails being two red nails and the other being two blue nails, and the same logic holds.

I, for one, am glad this solution is now finally published. My neighbors were not pleased with the incessant hammering coming from my apartment this past week.

## 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 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.