## 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:
[Show Solution]

## Ellipse packing

You’ve heard of circle packing… Well this week’s Riddler Classic is about ellipse packing!

This week, try packing three ellipses with a major axis of length 2 and a minor axis of length 1 into a larger ellipse with the same aspect ratio. What is the smallest such larger ellipse you can find? Specifically, how long is its major axis?

Extra credit: Instead of three smaller ellipses, what about other numbers of ellipses?

My solution:
[Show Solution]

## Loteria

This week’s Riddler Classic is about Lotería, also known as Mexican bingo!

A thousand people are playing Lotería, also known as Mexican bingo. The game consists of a deck of 54 cards, each with a unique picture. Each player has a board with 16 of the 54 pictures, arranged in a 4-by-4 grid. The boards are randomly generated, such that each board has 16 distinct pictures that are equally likely to be any of the 54.

During the game, one card from the deck is drawn at a time, and anyone whose board includes that card’s picture marks it on their board. A player wins by marking four pictures that form one of four patterns, as exemplified below: any entire row, any entire column, the four corners of the grid and any 2-by-2 square.

After the fourth card has been drawn, there are no winners. What is the probability that there will be exactly one winner when the fifth card is drawn?

My solution:
[Show Solution]

## Perfect pizza sharing

This week’s Riddler Classic is about how to cut a pizza to achieve precise area ratios between the slices.

Dean made a pizza to share with his three friends. Among the four of them, they each wanted a different amount of pizza. In particular, the ratio of their appetites was 1:2:3:4. Therefore, Dean wants to make two complete, straight cuts (i.e., chords) across the pizza, resulting in four pieces whose areas have a 1:2:3:4 ratio.

Where should Dean make the two slices?

Extra credit: Suppose Dean splits the pizza with more friends. If six people are sharing the pizza and Dean cuts along three chords that intersect at a single point, how close to a 1:2:3:4:5:6 ratio among the areas can he achieve? What if there are eight people sharing the pizza?

My solution:
[Show Solution]

To jump straight to the results:
[Show Solution]

## Optimal Wordle

This week’s Riddler Classic is about the viral word game Wordle.

Find a strategy that maximizes your probability of winning Wordle in at most three guesses.

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]

## Flipping your way to victory

This week’s Riddler Classic concerns a paradoxical coin-flipping game:

You have two fair coins, labeled A and B. When you flip coin A, you get 1 point if it comes up heads, but you lose 1 point if it comes up tails. Coin B is worth twice as much — when you flip coin B, you get 2 points if it comes up heads, but you lose 2 points if it comes up tails.

To play the game, you make a total of 100 flips. For each flip, you can choose either coin, and you know the outcomes of all the previous flips. In order to win, you must finish with a positive total score. In your eyes, finishing with 2 points is just as good as finishing with 200 points — any positive score is a win. (By the same token, finishing with 0 or −2 points is just as bad as finishing with −200 points.)

If you optimize your strategy, what percentage of games will you win? (Remember, one game consists of 100 coin flips.)

Extra credit: What if coin A isn’t fair (but coin B is still fair)? That is, if coin A comes up heads with probability p and you optimize your strategy, what percentage of games will you win?

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]

## The number line game

This Riddler puzzle is a game theory problem: how should each player play the game to maximize their own winnings?

Ariel, Beatrice and Cassandra — three brilliant game theorists — were bored at a game theory conference (shocking, we know) and devised the following game to pass the time. They drew a number line and placed \$1 on the 1, \$2 on the 2, \$3 on the 3 and so on to \$10 on the 10.

Each player has a personalized token. They take turns — Ariel first, Beatrice second and Cassandra third — placing their tokens on one of the money stacks (only one token is allowed per space). Once the tokens are all placed, each player gets to take every stack that her token is on or is closest to. If a stack is midway between two tokens, the players split that cash.

How will this game play out? How much is it worth to go first?

A grab bag of extra credits: What if the game were played not on a number line but on a clock, with values of \$1 to \$12? What if Desdemona, Eleanor and so on joined the original game? What if the tokens could be placed anywhere on the number line, not just the stacks?

Here are the details of how I approached the problem:
[Show Solution]