How many times can you add up the digits?

This week’s Fiddler is a puzzle about adding digits over and over again.

For any positive, base-10 integer $n$, define $f(n)$ as the number of times you have to add up its digits until you get a one-digit number. For example, $f(23) = 1$ because $2+3 = 5$, a one-digit number. Meanwhile, $f(888) = 2$, since $8+8+8 = 24$, a two-digit number, and then adding up those digits gives you $2+4 = 6$, a one-digit number. Find the smallest whole number $n$ such that $f(n) = 4$.

Extra Credit: For how many whole numbers $n$ between $1$ and $10,000$ does $f(n) = 3$?

My solution:
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Chessboard race

This week’s Fiddler is a puzzle with a surprise connection to physics!

A tiny ant is racing across a 2-by-2 chessboard, as shown below, where each smaller square has a side length of 1 cm. The ant starts at the bottom-left corner of the bottom-left black square and is trying to reach the top-right corner of the top-right black square. The ant moves faster on the white squares than on the black squares. Speed on the white squares is 1 cm per minute, while speed on the black squares is 0.9 cm per minute. What’s the least amount of time it will take the ant to reach the finish?

Extra Credit: Instead, the board is now 8-by-8, as shown below.

My solution:
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Pancake race

This week’s Fiddler is a logic puzzle about getting home as fast as possible.

Alice, Bob, and Carey start together and each walk home at a different constant speed. Once all three get home, they can have pancakes! Alice can walk home in 10 minutes, Bob can do it in 20, and Carey in 30. Fortunately, any of them can carry any of the others on their back without reducing their own walking speed. Assume that they can pick someone up, set someone down, and change direction instantaneously. What is the fastest they can get to eat pancakes?

Extra Credit
There is now a fourth: Dee. Dee is the slowest, needing 60 minutes to walk home. As before, anyone can carry anyone else, and they won’t get pancakes until everyone gets home. What is the fastest this can happen?

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:
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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:
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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]