## Outthink the Sphinx

This week’s Riddler Classic is a tricky puzzle that combines logic and game theory.

You will be asked four seemingly arbitrary true-or-false questions by the Sphinx on a topic about which you know absolutely nothing. Before the first question is asked, you have exactly $1. For each question, you can bet any non-negative amount of money that you will answer correctly. That is, you can bet any real number (including fractions of pennies) between zero and the current amount of money you have. After each of your answers, the Sphinx reveals the correct answer. If you are right, you gain the amount of money you bet; if you are wrong, you lose the money you bet. However, there’s a catch. (Isn’t there always, with the Sphinx?) The answer will never be the same for three questions in a row. With this information in hand, what is the maximum amount of money you can be sure that you’ll win, no matter what the answers wind up being? Extra credit: This riddle can be generalized so that the Sphinx asks N questions, such that the answer is never the same for Q questions in a row. What are your maximum guaranteed winnings in terms of N and Q? If you’re just looking for the answer, here it is: [Show Solution] Here is a more detailed write-up of the solution: [Show Solution] ## Baking the biggest pie This week’s Riddler Classic is about baking the biggest pie. Just in time for π day! You have a sheet of crust laid out in front of you. After baking, your pie crust will be a cylinder of uniform thickness (or rather, thinness) with delicious filling inside. To maximize the volume of your pie, what fraction of your crust should you use to make the circular base (i.e., the bottom) of the pie? Here is my solution: [Show Solution] ## Turning radius This week’s Riddler Classic is a simple-looking question about the turning radius of a truck. Suppose I’m driving a very long truck (with length L) with two front wheels and two rear wheels. (The truck is so long compared to its width that I can consider the two front wheels as being a single wheel, and the two rear wheels as being a single wheel.) Suppose I can also rotate the front wheels by$\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!