Wolfram Research has worked with the CBS/Paramount show NUMB3RS since its first season. Now in the fifth season, it remains the most popular show of Friday nights. “The Math behind NUMB3RS” gives a more in-depth look at some of the mathematics in each episode. With season 5, we’ve added a math puzzle to go with each episode. Fifteen episodes into season 5, there are fifteen puzzles available.



For each puzzle we make a web page with an episode file that contains the puzzle itself, a hint section that discusses the related mathematics, a quote from the episode, and a detailed solution. For example, in episode 509, “Conspiracy Theory,” the puzzle is “Nonrandom Matrix“—a type of Latin square also known as a Sudoku puzzle. In a Sudoku puzzle, you must fill in the matrix so that every row, column, and 3×3 box contains the digits 1 through 9.

The puzzle for episode 504 is called “Two Trains.” Two trains 100 km apart each travel towards each other at 10 km per hour. A bumblebee starts flying back and forth between the two trains at 60 km per hour, until the two trains crash into each other. How far does the bumblebee travel? As the site reveals, the two trains will collide in… well, try to solve it. The set-up for the infinite series can be seen at MathWorld‘s “Two Trains Puzzle” page. The NUMB3RS puzzle page also offers another classic puzzle, a train-shunting one by Sam Loyd:

The puzzle for episode 510, “Last Person Standing,” is based on the “Truel World” Demonstration. In this truel (a three-person duel), A is weakest, C is strongest. A shoots first, then B, then C. What is the best strategy for A, and what are the chances that A will win? To approximate this, a die is chosen, and each player must roll under a certain value to hit. Odds of hitting are given in the first line of each circle.

Originally, I suggested that in the best strategy, A should deliberately miss, while the strongest trade shots. When a player is eliminated, A should take aim at the survivor. Some math geniuses at Berkeley let me know that I was wrong. When players A B C have accuracies of a b c, player A should shoot at C when For example, if C always hits, and B hits a third of the time, there are two possible best strategies for A. When A should shoot at C. When A should deliberately miss at the start. When it doesn’t matter which strategy A picks.

The weakest player in a truel often has a nonintuitively high chance of winning. For example, when everyone has exactly a one-third chance to win. In this case, A should deliberately miss. In a game where A shoots C, the simplest set of odds where everyone has exactly a one-third chance to win is the following:

a = 1780225612071980440487233690 / 42514698180175799467200302197

b = 890112806035990220243616845 / 12666774444361618608640463004

c = 326459223625633611840607911167657350 / 3023918895788355583589056125975619389

(a,b,c) = (0.0418732, 0.0702715, 0.107959)

If the die is a 6-sided cube, and A hits on 1 and 2, B hits on 1-4, and C hits on 1-5, here is a visualization of an intermediary stage.

In homage to the next episode, “Cover Me,” you will need to manuever two agents around a dangerous estate. There are zones of safety, and an agent can cover for another as they rush through other areas. With these simple rules, how complicated could it be to safely get the agents to their goals? Tune in next week and visit http://www.cbs.com/puzzle or http://numb3rs.wolfram.com to find out.