Jamshrivelson

Question: One morning, Phil was playing with his daughter, who loves to cut paper with her safety scissors. She especially likes cutting paper into “strips,” which are rectangular pieces of paper whose shorter sides are at most 1 inch long. Whenever Phil gives her a piece of standard printer paper (8.5 inches by 11 inches), she picks one of the four sides at random and then cuts a 1-inch wide strip parallel to that side. Next, she discards the strip and repeats the process, picking another side at random and cutting the strip. Eventually, she is left with nothing but strips. On average, how many cuts will she make before she is left only with strips? Extra credit: Instead of 8.5 by 11-inch paper, what if the paper measures $m$ by $n$ inches? (And for a special case of this, what if the paper is square?) [Solution]

Question: The Robot Weightlifting World Championship was such a huge success that the organizers have hired you to help design its sequel: a Robot Tug-of-War Competition! In each one-on-one matchup, two robots are tied together with a rope. The center of the rope has a marker that begins above position 0 on the ground. The robots then alternate pulling on the rope. The first robot pulls in the positive direction towards 1; the second robot pulls in the negative direction towards $-1.$ Each pull moves the marker a uniformly random draw from $\left[0,1\right]$ towards the pulling robot. If the marker first leaves the interval $\left[-\frac12,\frac12\right]$ past $\frac12,$ the first robot wins. If instead it first leaves the interval past $-\frac12,$ the second robot wins. However, the organizers quickly noticed that the robot going second is at a disadvantage. They want to handicap the first robot by changing the initial position of the marker on the rope to be at some negative real number. Your job is to compute the position of the marker that makes each matchup a $50:50$ competition between the robots. Find this position to seven significant digits—the integrity of the Robot Tug-of-War Competition hangs in the balance! [Solution]

Question: Hames Jarrison has just intercepted a pass at one end zone of a football field, and begins running — at a constant speed of $15$ miles per hour — to the other end zone, $100$ yards away. At the moment he catches the ball, you are on the very same goal line, but on the other end of the field, $50$ yards away from Jarrison. Caught up in the moment, you decide you will always run directly toward Jarrison’s current position, rather than plan ahead to meet him downfield along a more strategic course. Assuming you run at a constant speed (i.e., don’t worry about any transient acceleration), how fast must you be in order to catch Jarrison before he scores a touchdown? [Solution]

Question: When you roll a pair of fair dice, the most likely outcome is $7$ (which occurs $1/6^\text{th}$ of the time) and the least likely outcomes are 2 and 12 (which each occur $1/36^\text{th}$ of the time). Annoyed by the variance of these probabilities, I set out to create a pair of “uniform dice.” These dice still have sides that are uniquely numbered from $1$ to $6,$ and they are identical to each other. However, they are weighted so that their sum is more uniformly distributed between $2$ and $12$ than that of fair dice. Unfortunately, it is impossible to create a pair of such dice so that the probabilities of all $11$ sums from $2$ to $12$ are identical (i.e., they are all $1/11$). But I bet we can get pretty close. The variance of the $11$ probabilities is the average value of the squared difference between each probability and the average probability (which is, again, $1/11$). One way to make my dice as uniform as possible is to minimize this variance. So how should I make my dice as uniform as possible? In other words, which specific weighting of the dice minimizes the variance among the $11$ probabilities? That is, what should the probabilities be for rolling $1, 2, 3, 4, 5$ or $6$ with one of the dice? [Solution]

Question: You have four standard dice, and your goal is simple: Maximize the sum of your rolls. So you roll all four dice at once, hoping to achieve a high score. But wait, there’s more! If you’re not happy with your roll, you can choose to reroll zero, one, two or three of the dice. In other words, you must “freeze” one or more dice and set them aside, never to be rerolled. You repeat this process with the remaining dice — you roll them all and then freeze at least one. You repeat this process until all the dice are frozen. If you play strategically, what score can you expect to achieve on average? Extra credit: Instead of four dice, what if you start with five dice? What if you start with six dice? What if you start with $N$ dice? [Solution]

Question: Suppose you have a chain with infinitely many flat (i.e., one-dimensional) links. The first link has length $1,$ and the length of each successive link is a fraction $f$ of the previous link’s length. As you might expect, $f$ is less than $1.$ You place the chain flat on a table and some ink at the very end of the chain (i.e., the end with the infinitesimal links). Initially, the chain forms a straight line segment, and the longest link is fixed in place. From there, the links are constrained to move in a very specific way: The angle between each chain and the next, smaller link is always the same throughout the chain. For example, if the $N^\text{th}$ link and the $N+1^\text{st}$ link form a $40$ degree clockwise angle, then so do the $N+1^\text{st}$ link and the $N+2^\text{nd}$ link. After you move the chain around as much as you can, what shape is drawn by the ink that was at the tail end of the chain? [Solution]

Question: Today marks the beginning of the Summer Olympics! One of the brand-new events this year is sport climbing, in which competitors try their hands (and feet) at lead climbing, speed climbing and bouldering. Suppose the event’s organizers accidentally forgot to place all the climbing holds on and had to do it last-minute for their 10-meter wall (the regulation height for the purposes of this riddle). Climbers won’t have any trouble moving horizontally along the wall. However, climbers can’t move between holds that are more than 1 meter apart vertically. In a rush, the organizers place climbing holds randomly until there are no vertical gaps between climbing holds (including the bottom and top of the wall). Once they are done placing the holds, how many will there be on average (not including the bottom and top of the wall)? Extra credit: Now suppose climbers find it just as difficult to move horizontally as vertically, meaning they can’t move between any two holds that are more than 1 meter apart in any direction. Suppose also that the climbing wall is a 10-by-10 meter square. If the organizers again place the holds randomly, how many have to be placed on average until it’s possible to climb the wall? [Solution]