Don’t flip out

This week’s Fiddler is a probability question about a coin-flipping game.

Kyle and Julien are playing a game in which they each toss their own fair coins. On each turn of the game, both players flip their own coin once. If, at any point, Kyle’s most recent three flips are Tails, Tails, and Heads (i.e., TTH), then he wins. If, at any point, Julien’s most recent three flips are Tails, Tails, and Tails (i.e, TTT), then he wins.

However, both players can’t win at the same time. If Kyle gets TTH at the same time Julien gets TTT, then no one wins, and they continue flipping. They don’t start over completely or erase their history, mind you—they merely continue flipping, so that one of them could conceivably win in the next flip or two.

What is the probability that Kyle wins this game?

Extra Credit
Kyle and Julien write down all eight possible sequences for three coin flips (HHH, HHT, HTH, THH, HTT, THT, TTH, and TTT) on eight different slips of paper. They place these slips into a hat and shake it.

They will each randomly draw slips of paper out of the hat, at which point they will play the same game as previously described, but looking for the sequence specified on the slip of paper they each selected. Kyle draws first and looks at his slip of paper. After doing some calculations, he says: “Well, at this point, it’s about as fair a match as it could possible be.”

Which slip or slips of paper might Kyle have drawn? And what are his chances of winning at this point (i.e., before Julien selects his own slip of paper)?

My solution:
[Show Solution]

The Likeliest Monopoly Square

This week’s Fiddler is about rolling dice in the board game Monopoly.

We have a square board with 40 individual spaces around it, numbered from 0 to 39. All players begin on space 0 (akin to the “Go” square in Monopoly) and roll a pair of dice to determine how many spaces they advance each turn. However, unlike Monopoly, there is no way to otherwise advance around the board (i.e., there’s no “Chance,” “Community Chest,” going to jail, etc.). In their first pass around the board, which space from 1 to 39 are players most likely to land on at some point (i.e., not necessarily on their first or last roll, but after any number of rolls)?

Extra Credit
The square board has 10 spaces on each side. The first side has spaces 0 through 9, the second side has spaces 10 through 19, the third side has spaces 20 through 29, and the fourth side has spaces 30 through 39. Because you’re rolling two dice, it’s impossible to land on space 1 in your first pass around the board. Several other spaces on the first side of the board are similarly unlikely. Putting that first side of the board aside, which space from 10 to 39 are players least likely to land on at some point during their first pass around the board? (Another question: What if you rolled three dice at a time instead of two?)

My solution:
[Show Solution]

Picking a speaker at random

This week’s Fiddler is about selecting a speaker of the house at random. How long will it take?

There are three candidates who want the job of Speaker. All 221 members of the party vote by picking randomly from among the candidates. If one candidate earns the majority of the votes, they become the next Speaker. Otherwise, the candidate with the fewest votes is eliminated and the process is repeated with one less candidate. If two or more candidates receive the same smallest number of votes, then exactly one of them is eliminated at random. What is the average number of rounds needed to select a new Speaker?

Extra Credit
What if there were 10 candidates running for Speaker?

My solution:
[Show Solution]

The slow car chase

This week’s Fiddler is a problem about probability and a slow car chase!

You and your pursuer are stopped at traffic lights one block away from each other, as shown below. For both cars, it takes 1 minute to get to each subsequent light, and there is a 50-50 chance any light will be red upon arrival (independent of what happened previously). If a light is red, you must wait one minute before it turns green. What is the probability you will make it to the city limits without being caught by your pursuer?

Extra Credit
In the same scenario as before, imagine there are infinitely many lights. On average, how many minutes will it be until you are ultimately caught?

My solution:
[Show Solution]

Pinball probability

This week’s Fiddler is a challenging probability question.

You’re playing a game of pinball that includes four lanes, each of which is initially unlit. Every time you flip the pinball, it passes through exactly one of the four lanes (chosen at random) and toggles that lane’s state. So if that lane is unlit, it becomes lit after the ball passes through. But if the lane is lit, it becomes unlit after the ball passes through. On average, how many times will you have to flip the pinball until all four lanes are lit?

Extra credit: Instead of four lanes, now suppose your pinball game has $n$ lanes. And let’s say that $E(n)$ represents the average number of pinball flips it takes until all $n$ lanes are lit up. Now, each time you increase the number of lanes by one, you find that it takes you approximately twice as long to light up all the lanes. In other words, $E(n+1)$ seems to be about double $E(n)$. But upon closer examination, you find that it’s not quite double. Moreover, there’s a particular value of $n$ where the ratio $E(n+1)/E(n)$ is at a minimum. What is this value of $n$?

My solution:
[Show Solution]

The weaving loom problem

This week’s Fiddler is a classic problem.

A weaving loom consists of equally spaced hooks along the x and y axes. A string connects the farthest hook on the x-axis to the nearest hook on the y-axis, and continues back and forth between the axes, always taking up the next available hook. This leads to a picture that looks like this:

As the number of hooks goes to infinity, what does the shape trace out?

Extra credit: If four looms are rotated and superimposed as shown below, what is the area of the white region in the middle?

My solution:
[Show Solution]

Betting on football with future knowledge

This week’s Fiddler is a football-themed puzzle with a twist: you can see the future! Sort of.

You know ahead of time that your football team will win 8 of their 12 remaining games, but you don’t know which ones. You can place bets on every game, placing bets either for or against your team. You can bet any amount up to how much you currently have. You want to implement a betting strategy that guarantees you’ll have as much money as possible after the 12 games are complete. If you did so, then after the 12 games how much money would you be guaranteed to have if you started with $100?

My solution:
[Show Solution]

A more elegant alternative solution, due in large part to a clever observation by Vince Vatter.
[Show Solution]

Optimal baseball lineup

This week’s Fiddler is a problem about how to set the optimal baseball lineup.

Eight of your nine batters are “pure contact” hitters. One-third of the time, each of them gets a single, advancing any runners already on base by exactly one base. (The only way to score is with a single with a runner on 3rd). The other two-thirds of the time, they record an out, and no runners advance to the next base. Your ninth batter is the slugger. One-tenth of the time, he hits a home run. But the remaining nine-tenths of the time, he strikes out. Your goal is to score as many runs as possible, on average, in the first inning. Where in your lineup (first, second, third, etc.) should you place your home run slugger?

Extra Credit: Instead of scoring as many runs as possible in the first inning, you now want to score as many runs as possible over the course of nine innings. What’s more, instead of just having one home run slugger, you now have two sluggers in your lineup. The other seven batters remain pure contact hitters. Where in the lineup should you place your two sluggers to maximize the average number of runs scored over nine innings?

My solution:
[Show Solution]

Braille puzzle

This week’s Fiddler is a counting problem about the Braille system.

Braille characters are formed by raised dots arranged in a braille cell, a three-by-two array. With six locations for dots, each of which is raised or unraised, there are 26, or 64, potential braille characters.

Each of the 26 letters of the basic Latin alphabet (shown below) has its own distinct arrangement of dots in the braille cell. What’s more, while some arrangements of raised dots are rotations or reflections of each other (e.g., E and I, R and W, etc.), no two letters are translations of each other. For example, only one letter (A) consists of a single dot – if there were a second such letter, A and this letter would be translations of each other.

Of the 64 total potential braille characters, how many are in the largest set such that no two characters consist of raised dot patterns that are translations of each other?

Extra Credit In addition to six-dot braille, there’s also an eight-dot version. But what if there were even more dots? Let’s quadruple the challenge by doubling the size of the cell in each dimension. Consider a six-by-four array, where a raised dot could appear at each location in the array. Of the 224 total potential characters, how many are in the largest set such that no two characters are translations of each other?

My solution:
[Show Solution]

Making something out of nothing

This week’s Fiddler is a problem about composing functions. Here it goes:

Consider $f(n) = 2n+1$ and $g(n) = 4n$. It’s possible to produce different whole numbers by applying combinations of $f$ and $g$ to $0$. How many whole numbers between $1$ and $1024$ (including $1$ and $1024$) can you produce by applying some combination of $f$’s and $g$’s to the number $0$?

Extra Credit: Now consider the functions $g(n) = 4n$ and $h(n) = 1−2n$. How many integers between $-1024$ and $1024$ (including $-1024$ and $1024$) can you produce by applying some combination of $g$’s and $h$’s to the number $0$?

My solution:
[Show Solution]