## Tower of goats

This week’s Riddler classic is a counting problem. Can the goats fit in the tower?

A tower has 10 floors, each of which can accommodate a single goat. Ten goats approach the tower, and each goat has its own (random) preference of floor. Multiple goats can prefer the same floor. One by one, each goat walks up the tower to its preferred room. If the floor is empty, the goat will make itself at home. But if the floor is already occupied by another goat, then it will keep going up until it finds the next empty floor, which it will occupy. But if it does not find any empty floors, the goat will be stuck on the roof of the tower. What is the probability that all 10 goats will have their own floor, meaning no goat is left stranded on the roof of the tower?

My solution:
[Show Solution]

## Squid game

This week’s Riddler Classic is Squid Game-themed!

There are 16 competitors who must cross a bridge made up of 18 pairs of separated glass squares. Here is what the bridge looks like from above:

To cross the bridge, each competitor jumps from one pair of squares to the next. However, they must choose one of the two squares in a pair to land on. Within each pair, one square is made of tempered glass, while the other is made of normal glass. If you jump onto tempered glass, all is well, and you can continue on to the next pair of squares. But if you jump onto normal glass, it will break, and you will be eliminated from the competition.

The competitors have no knowledge of which square within each pair is made of tempered glass. The only way to figure it out is to take a leap of faith and jump onto a square. Once a pair is revealed — either when someone lands on a tempered square or a normal square — all remaining competitors take notice and will choose the tempered glass when they arrive at that pair.

On average, how many of the 16 competitors will make it across the bridge?

Here is my solution.
[Show Solution]

And here is a much better solution!
[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]

## 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]