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]
Maximum Dice
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]
Things Have Chained
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]
Random Rock Climb
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]
Robot Weightlifters
Question: The Robot Weightlifting World Championship’s final round is about to begin! Three robots, seeded 1, 2, and 3, remain in contention. They take turns from the 3rd seed to the 1st seed publicly declaring exactly how much weight (any nonnegative real number) they will attempt to lift, and no robot can choose exactly the same amount as a previous robot. Once the three weights have been announced, the robots attempt their lifts, and the robot that successfully lifts the most weight is the winner. If all robots fail, they just repeat the same lift amounts until at least one succeeds.
Assume the following:
all the robots have the same probability $p(w)$ of successfully lifting a given weight $w$;
$p(w)$ is exactly known by all competitors, continuous, strictly decreasing as the w increases, $p(0) = 1,$ and $p(w) \rightarrow 0$ as $w \rightarrow \infty$; and
all competitors want to maximize their chance of winning the RWWC.
If $w$ is the amount of weight the 3rd seed should request, find $p(w).$ Give your answer to an accuracy of six decimal places.
[Solution]
Survivor
Question: Players A, B, C, and D are playing a game which starts with A saying the number $1.$ From that point on, each player says a number between $1$ and $4$ greater than the number spoken by the preceding player. The player who says $20$ wins the round, the player after them is removed from the game, the player after them starts the next round. Any other players survive to the next round. Each player’s top aim is to win the whole game but, if they realize that’s impossible, they’ll prioritize surviving to the next round. Who will survive the $4$ players game?
[Solution]
Cornhole Connoisseur
Question: it’s the end of a close game of cornhole, and your team is $3$ points away from the win. By house rules, you have to hit the $3$ points exactly, or you lose. On your team are three athletes. The first is The Aggressor who has a $40%$ chance to get the cornhole, a $30%$ chance to hit the board and a $30%$ chance to miss entirely. The second is The Conservative who has a $10%$ chance to get the cornhole, a whopping $80%$ chance to hit the board, and a $10%$ chance to miss entirely. Finally, there’s The Waste who’s blind drunk at every game and will always miss the board. Given your roster and the free choice to use any of them in any situation, and you play optimally, what is the chance that you win the game?
[Solution]
Election Comeback
Question: it’s election night and your candidate is behind in the count. However, a significant fraction of the vote is still out in uncounted mail-in ballots. What are the chances that your candidates come back for the stunning victory if a whole bunch of people vote? What happens when the polls are tilted in one direction?
[Solution]
Crescent Observatory
Question: you’re watching the Moon from your room, alone, like every night. Seeking a higher purpose, you decide to track the projected area of the illuminated portion of the Moon throughout the month. If your data is accurate, how much faster will the illuminated Moon’s area be growing at half Moon as compared to a crescent Moon of $1/6^\text{th}$ area?
[Solution]
A Permutation Most Foul
Question: You’re playing a game of three on three basketball in a weird league. The rule on foul shots on a three pointer is that all three players from the team get to take turns shooting three foul shots. However, to get the second foul shot the team must make the first, and to get the second foul shot the team must make the second. As soon as a player misses, the foul shooting is over. If the players on your team have foul shot percentages $a,$ $b,$ and $c$ (no two equal), then what is the maximum number of shooting orders that would produce the same expected number of points?
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