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]

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]

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]

Mismatched socks

This week’s Riddler Classic is a problem familiar to many…

I have $n$ pairs of socks in a drawer. Each pair is distinct from another and consists of two matching socks. Alas, I’m negligent when it comes to folding my laundry, and so the socks are not folded into pairs. This morning, fumbling around in the dark, I pull the socks out of the drawer, randomly and one at a time, until I have a matching pair of socks among the ones I’ve removed from the drawer.

On average, how many socks will I pull out of the drawer in order to get my first matching pair?

Here is my solution:
[Show Solution]

Gift card puzzle

Here is a puzzle from the Riddler about gift cards:

You’ve won two gift cards, each loaded with 50 free drinks from your favorite coffee shop. The cards look identical, and because you’re not one for record-keeping, you randomly pick one of the cards to pay with each time you get a drink. One day, the clerk tells you that he can’t accept the card you presented to him because it doesn’t have any drink credits left on it.

What is the probability that the other card still has free drinks on it? How many free drinks can you expect are still available?

Here is my solution:
[Show Solution]

Elf music

This holiday-themed Riddler problem is about probability:

In Santa’s workshop, elves make toys during a shift each day. On the overhead radio, Christmas music plays, with a program randomly selecting songs from a large playlist.

During any given shift, the elves hear 100 songs. A cranky elf named Cranky has taken to throwing snowballs at everyone if he hears the same song twice. This has happened during about half of the shifts. One day, a mathematically inclined elf named Mathy tires of Cranky’s sodden outbursts. So Mathy decides to use what he knows to figure out how large Santa’s playlist actually is.

Help Mathy out: How large is Santa’s playlist?

Here is my solution:
[Show Solution]

Hand sort

A card-rearranging problem on the Riddler blog. Here it goes:

You play so many card games that you’ve developed a very specific organizational obsession. When you’re dealt your hand, you want to organize it such that the cards of a given suit are grouped together and, if possible, such that no suited groups of the same color are adjacent. (Numbers don’t matter to you.) Moreover, when you receive your randomly ordered hand, you want to achieve this organization with a single motion, moving only one adjacent block of cards to some other position in your hand, maintaining the original order of that block and other cards, except for that one move.

Suppose you’re playing pitch, in which a hand has six cards. What are the odds that you can accomplish your obsessive goal? What about for another game, where a hand has N cards, somewhere between 1 and 13?

Here is my solution:
[Show Solution]

Sniff out the spies

This interesting problem appeared on the Riddler blog. Here it goes:

There are N agents and K of them are spies. Your job is to identify all the spies. You can send a given number of agents to a “retreat” on a remote island. If all K spies are present at the retreat, they will meet to strategize. If even one spy is missing, this spy meeting will not take place. The only information you get from a retreat is whether or not the spy meeting happened. You can send as many agents as you like to the retreat, and the retreat can happen as many times as needed. You know the values of N and K.

What’s the minimum number of retreats needed to guarantee you can identify all K spies? If each retreat costs \$1,000 per person, what is the total cost to identify all K spies?

To begin with, let’s assume you know that N = 1,024 and K = 17.

Here is my solution for $K=1$:
[Show Solution]

And here is a partial solution for $K \gt 1$:
[Show Solution]

Nightmare Solitaire

This Riddler problem is a probability question about an old favorite: Solitaire!

While killing some time at your desk one afternoon, you fire up a new game of Solitaire (also called Klondike, and specifically the version where you deal out three cards from the deck at a time). But your boredom quickly turns to rage because your game is unplayable — you can flip through your deck, but you never have any legal moves! What are the odds?

Here is my solution:
[Show Solution]

Where will the seven dwarfs sleep tonight?

The following problem appeared in The Riddler. It’s an interesting recursive problem.

Each of the seven dwarfs sleeps in his own bed in a shared dormitory. Every night, they retire to bed one at a time, always in the same sequential order, with the youngest dwarf retiring first and the oldest retiring last. On a particular evening, the youngest dwarf is in a jolly mood. He decides not to go to his own bed but rather to choose one at random from among the other six beds. As each of the other dwarfs retires, he chooses his own bed if it is not occupied, and otherwise chooses another unoccupied bed at random.

  1. What is the probability that the oldest dwarf sleeps in his own bed?
  2. What is the expected number of dwarfs who do not sleep in their own beds?

Here is my solution.
[Show Solution]