Can you hop to the lily pad?

This week’s Fiddler is about hopping back and forth.

You are a frog in a pond with an infinite number of lily pads in a line, marked “1,” “2,” “3,” etc. You are currently on pad 2, and your goal is to make it to pad 1. From any given pad, there are specific probabilities that you’ll jump to another pad: Whenever you are on pad $k$, you will hop to pad $k−1$ with probability $1/k$, and you will hop to pad $k+1$ with probability $(k−1)/k$.

What is the probability that you will ultimately make it to pad 1?

My solution:
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Halloween Puzzle

This week’s Fiddler is about rounding!

You are presented with a bag of treats, which contains $n \geq 3$ peanut butter cups and some unknown quantity of candy corn kernels (with any amount being equally likely). You reach into the bag $k$ times, with $3 \leq k \leq n$, and pull out a candy at random. Each time, it’s a peanut butter cup! How many candy kernels do you expect to be in the bag?

My solution:
[Show Solution]

Round, round, get a round

This week’s Fiddler is about rounding!

Let $\text{round}(x)$ be the value of $x$ rounded to the nearest integer. Suppose $x_1,\dots,x_n$ are independent uniformly distributed random variables in $[0,1]$. Find the probability that
\[
\text{round}(x_1+\cdots+x_n) = \text{round}(x_1)+\cdots+\text{round}(x_n)
\]

My solution:
[Show Solution]

Showcase Showdown

This week’s Fiddler is based on “Showcase Showdown” on the game show “The Price is Right”.

Suppose we have some number of players. Player A is the first to spin a giant wheel, which spits out a real number chosen randomly and uniformly between 0 and 1. All spins are independent of each other. After spinning, A can either stick with the number they just got or spin the wheel one more time. If they spin again, their assigned number is the sum of the two spins, as long as that sum is less than or equal to 1. If the sum exceeds 1, A is immediately declared a loser.

After A is done spinning (whether once or twice), B steps up to the wheel. Like A, they can choose to spin once or twice. If they spin twice and the sum exceeds 1, they are similarly declared the loser. This continues until all players are done. Whoever has the greater value (that does not exceed 1) is declared the winner.

Assuming all players play the game optimally, what are player A’s chances of winning?

My solution:
[Show Solution]

Dungeon Master’s Dice

This week’s Fiddler is a probability question about a dice-rolling game.

Two people are sitting at a table together, each with their own bag of six “DnD dice”: a d4, a d6, a d8, a d10, a d12, and a d20. Here, “dX” refers to a die with X faces, numbered from 1 to X, each with an equally likely probability of being rolled. Both people randomly pick one die from their respective bags and then roll them at the same time. For example, suppose the two dice selected are a d4 and a d12. The players roll them, and let’s further suppose that both rolls come up as 3. What luck! What’s the probability of something like this happening? That is, what is the probability that both players roll the same number, whether or not they happened to pick the same kind of die?

Extra Credit
Instead of two people sitting at the table, now suppose there are three. Again, all three randomly pick one die from their respective bags and roll them at the same time. For example, suppose the three dice selected are a d4, a d20, and a d12. The players roll them, and let’s further suppose that the d4 comes out as 4, the d20 comes out as 13, and the d12 comes out as 4. In this case, there are two distinct numbers (4 and 13) among the three rolls. On average, how many distinct numbers would you expect to see among the three rolls?

My solution:
[Show Solution]

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]

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]

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]