The blue-eyed islanders

Today’s Riddler problem is another classic. The current incarnation of the puzzle is about error-prone mathematicians, while the classic version is about blue-eyed islanders.

A university has 10 mathematicians, each one so proud that, if she learns that she made a mistake in a paper, no matter how long ago the mistake was made, she resigns the next Friday. To avoid resignations, when one of them detects a mistake in the work of another, she tells everyone else but doesn’t inform the mistake-maker. All of them have made mistakes, so each one thinks only she is perfect. One Wednesday, a super-mathematician, whom all respect and believe, comes to visit. She looks at all the papers and says: “Someone here has made a mistake.”

What happens then? Why?

Here is the solution:
[Show Solution]

The puzzle of the pirate booty

Today’s puzzle was posed on the Riddler blog, but it’s actually a classic among problem-solving enthusiasts, and is commonly known as the pirate game. Here is the formulation used in the Riddler:

Ten Perfectly Rational Pirate Logicians (PRPLs) find 10 (indivisible) gold pieces and wish to distribute the booty among themselves.

The pirates each have a unique rank, from the captain on down. The captain puts forth the first plan to divide up the gold, whereupon the pirates (including the captain) vote. If at least half the pirates vote for the plan, it is enacted, and the gold is distributed accordingly. If the plan gets fewer than half the votes, however, the captain is killed, the second-in-command is promoted, and the process starts over. (They’re mutinous, these PRPLs.)

Pirates always vote by the following rules, with the earliest rule taking precedence in a conflict:

  1. Self-preservation: A pirate values his life above all else.
  2. Greed: A pirate seeks as much gold as possible.
  3. Bloodthirst: Failing a threat to his life or bounty, a pirate always votes to kill.

Under this system, how do the PRPLs divide up their gold?

Extra credit: Solve the generalized problem where there are P pirates and G gold pieces.

Here is the solution to the main problem:
[Show Solution]

And here is a solution to the general case:
[Show Solution]

If this sort of problem interests you, I recommend taking a crack at the riddle of the blue-eyed islanders, or the unfaithful husbands. More information here as well.

Baseball division champs

Today’s post is a Riddler problem about baseball, and it goes like this:

Assume you have a sport (let’s call it “baseball”) in which each team plays 162 games in a season. Assume you have a division of five teams (call it the “AL East”) where each team is of exact equal ability. Specifically, each team has a 50 percent chance of winning each game. What is the expected value of the number of wins for the team that finishes in first place?

The problem statement is a bit vague, so we must make some assumptions in order to solve the problem. Our first solution assumes that the win-loss records of the teams are mutually independent.
[Show Solution]

This next solution explains how to tackle the more realistic case where the games are not played independently, and explains why the independence assumption is a good one for the AL East.
[Show Solution]

Monsters’ gems

Once again, The Riddler does not disappoint! This puzzle is about slaying monsters and collecting gems.

A video game requires you to slay monsters to collect gems. Every time you slay a monster, it drops one of three types of gems: a common gem, an uncommon gem or a rare gem. The probabilities of these gems being dropped are in the ratio of 3:2:1 — three common gems for every two uncommon gems for every one rare gem, on average. If you slay monsters until you have at least one of each of the three types of gems, how many of the most common gems will you end up with, on average?

Here is my solution:
[Show Solution]

A more brute-force approach:
[Show Solution]

Yet another solution approach with very nice write-up can be found on Andrew Mascioli’s blog

Overflowing martini glass

This Riddler puzzle is all about conic sections.

You’ve kicked your feet up and have drunk enough of your martini that, when the conical glass (?) is upright, the drink reaches some fraction p of the way up its side. When tipped down on one side, just to the point of overflowing, how far does the drink reach up the opposite side?

Here is my solution:
[Show Solution]

Proud partygoers puzzle

Another great problem from the Riddler blog.

A group of N people are in attendance at your shindig, some of whom are friends with each other. (Let’s assume friendship is symmetric — if person A is friends with person B, then B is friends with A.) Suppose that everyone has at least one friend at the party, and that a person is “proud” if her number of friends is strictly larger than the average number of friends that her own friends have. (A competitive lot, your guests.)

Importantly, more than one person can be proud. How large can the share of proud people at the party be?

The solution:
[Show Solution]

Elevator button puzzle

This problem was originally posted on the Riddler blog. Here it goes:

In a building’s lobby, some number (N) of people get on an elevator that goes to some number (M) of floors. There may be more people than floors, or more floors than people. Each person is equally likely to choose any floor, independently of one another. When a floor button is pushed, it will light up.

What is the expected number of lit buttons when the elevator begins its ascent?

My solution:
[Show Solution]

A much more elegant solution, courtesy of Ross Boczar
[Show Solution]