Turning radius

This week’s Riddler Classic is a simple-looking question about the turning radius of a truck.

Suppose I’m driving a very long truck (with length L) with two front wheels and two rear wheels. (The truck is so long compared to its width that I can consider the two front wheels as being a single wheel, and the two rear wheels as being a single wheel.)

Suppose I can also rotate the front wheels by $\alpha$ and the back wheels — independently from the front wheels — by $\beta$. What is the truck’s turning radius?

Here is my solution:
[Show Solution]

Can you eat all the chocolates?

This week’s Riddler Classic is a neat puzzle about eating chocolates.

I have 10 chocolates in a bag: Two are milk chocolate, while the other eight are dark chocolate. One at a time, I randomly pull chocolates from the bag and eat them — that is, until I pick a chocolate of the other kind. When I get to the other type of chocolate, I put it back in the bag and start drawing again with the remaining chocolates. I keep going until I have eaten all 10 chocolates.

For example, if I first pull out a dark chocolate, I will eat it. (I’ll always eat the first chocolate I pull out.) If I pull out a second dark chocolate, I will eat that as well. If the third one is milk chocolate, I will not eat it (yet), and instead place it back in the bag. Then I will start again, eating the first chocolate I pull out.

What are the chances that the last chocolate I eat is milk chocolate?

Here is my original solution:
[Show Solution]

And here is a far more elegant solution, courtesy of @rahmdphd on Twitter.
[Show Solution]

Flawless war

his week’s Riddler Classic has to do with the card game “War”. Here is the problem, paraphrased:

War is a two-player game in which a standard deck of cards is first shuffled and then divided into two piles with 26 cards each; one pile for each player. In every turn of the game, both players flip over and reveal the top card of their deck. The player whose card has a higher rank wins the turn and places both cards on the bottom of their pile. Assuming a deck is randomly shuffled before every game, how many games of War would you expect to play until you had a game that lasted just 26 turns (with no ties; a flawless victory)?

Here is my solution:
[Show Solution]

Polygons with perimeter and vertex budgets

his week’s Riddler Classic involves designing maximum-area polygons with a fixed budget on the length of the perimeter and the number of vertices. The original problem involved designing enclosures for hamsters, but I have paraphrased the problem to make it more concise.

You want to build a polygonal enclosure consisting of posts connected by walls. Each post weighs $k$ kg. The walls weigh $1$ kg per meter. You are allowed a maximum budget of $1$ kg for the posts and walls.

What’s the greatest value of $k$ for which you should use four posts rather than three?

Extra credit: For which values of $k$ should you use five posts, six posts, seven posts, and so on?

Here is my solution:
[Show Solution]

Cutting a ruler into pieces

This week’s Riddler Classic is a paradoxical question about cutting a ruler into smaller pieces.

Recently, there was an issue with the production of foot-long rulers. It seems that each ruler was accidentally sliced at three random points along the ruler, resulting in four pieces. Looking on the bright side, that means there are now four times as many rulers — they just happen to have different lengths. On average, how long are the pieces that contain the 6-inch mark?

With four cuts, each piece will be on average 3 inches long, but that can’t be the answer, can it?

Here is my solution:
[Show Solution]

Squaring the pentagon

This week’s Riddler Classic is a number theory problem, which I will paraphrase:

The number 50 can be represented using a set of interleaved dots where the number of columns is one greater than the number of rows; the same way the stars in the US flag are arranged. If we add 1 to this number, we obtain 51, which can be represented using concentric pentagons. Here are diagrams showing both arrangements:

What is the next number with the property that it can be represented as interleaved dots but when you add 1 to it it can be represented using concentric pentagons?

Here is my solution:
[Show Solution]

Connect the dots

This week’s Riddler Classic is a problem about connecting dots to create as many non-intersecting polygons as possible. Here is the problem:

Polly Gawn loves to play “connect the dots.” Today, she’s playing a particularly challenging version of the game, which has six unlabeled dots on the page. She would like to connect them so that they form the vertices of a hexagon. To her surprise, she finds that there are many different hexagons she can draw, each with the same six vertices.

What is the greatest possible number of unique hexagons Polly can draw using six points?

(Hint: With four points, that answer is three. That is, Polly can draw up to three quadrilaterals, as long as one of the points lies inside the triangle formed by the other three. Otherwise, Polly would only be able to draw one quadrilateral.)

Extra Credit: What is the greatest possible number of unique heptagons Polly can draw using seven points?

Here is my solution:
[Show Solution]

Dungeons & Dragons

This week’s Riddler Classic is a probability problem about the game Dungeons & Dragons. Here it goes:

When you roll a die “with advantage,” you roll the die twice and keep the higher result. Rolling “with disadvantage” is similar, except you keep the lower result instead. The rules further specify that when a player rolls with both advantage and disadvantage, they cancel out, and the player rolls a single die. Yawn!

There are two other, more mathematically interesting ways that advantage and disadvantage could be combined. First, you could have “advantage of disadvantage,” meaning you roll twice with disadvantage and then keep the higher result. Or, you could have “disadvantage of advantage,” meaning you roll twice with advantage and then keep the lower result. With a fair 20-sided die, which situation produces the highest expected roll: advantage of disadvantage, disadvantage of advantage or rolling a single die?

Extra Credit: Instead of maximizing your expected roll, suppose you need to roll N or better with your 20-sided die. For each value of N, is it better to use advantage of disadvantage, disadvantage of advantage or rolling a single die?

Here is a detailed derivation of the relevant probabilities:
[Show Solution]

And here are the results:
[Show Solution]

Flip to freedom

This week’s Riddler Classic is a problem about coin flipping. The text of the original problem is quite long, so I will paraphrase it here:

There are $n$ prisoners, each with access to a random number generator (generates uniform random numbers in $[0,1]$) and a fair coin. Each prisoner is given the opportunity to flip their coin once if they so choose. If all of the flipped coins come up Heads, all prisoners are released. But if any of the flipped coins come up Tails, or if no coins are flipped at all, the prisoners are not released. If the prisoners cannot communicate in any way and do not know who is flipping their coin or not, how can they maximize their chances of being released?

Here is my solution:
[Show Solution]

When did the snow start?

This week’s Riddler Classic is a neat calculus problem:

One morning, it starts snowing. The snow falls at a constant rate, and it continues the rest of the day.

At noon, a snowplow begins to clear the road. The more snow there is on the ground, the slower the plow moves. In fact, the plow’s speed is inversely proportional to the depth of the snow — if you were to double the amount of snow on the ground, the plow would move half as fast.

In its first hour on the road, the plow travels 2 miles. In the second hour, the plow travels only 1 mile.

When did it start snowing?

Here is my solution:
[Show Solution]