The Three Hats Game

Three players enter a room, and a maroon or orange hat is placed on each one’s head. The color of each hat is determined by a coin toss, and the outcome of one toss has no effect on the others. Each player can see the other players’ hats but not his own.

The players can discuss strategy before the game begins, but after this they may not communicate. Each player considers the colors of the other players’ hats, and then simultaneously each player must either guess the color of his own hat or pass.

The group shares a $3 million prize if at least one player guesses correctly and no player guesses incorrectly. What strategy will raise their chance of winning above 50 percent?

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A player who sees two hats of the same color (say, orange) guesses the opposite color (maroon). A player who sees hats of two different colors passes. There are 8 possible arrangements of hats: OOO OOM OMM MOO MMO MMM MOM OMO In six of these, two of the players see hats of different colors and so will pass, and the third player sees two hats of the same color and guesses the opposite color, winning. In the other two cases, all three players are wearing hats of the same color, so all guess the wrong color and lose. This strategy produces a win in six of the eight cases, so the players will win 3/4 of the time. From “Applications of Recursive Operators to Randomness and Complexity,” the 1998 UCSB doctoral thesis of computer scientist Todd Ebert.

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February 17, 2015February 22, 2015 | Puzzles

Red and Black

I propose a card game. I’ll shuffle an ordinary deck of cards and turn up the cards in pairs. If both cards in a given pair are black, I’ll give them to you. If both are red, I’ll take them. And if one is black and one red, then we’ll put them aside, belonging to no one.

You pay a dollar for the privilege of playing the game, and then we’ll go through the whole deck. When the game is over, if you have no more cards than I do, you pay nothing. But for every card that you have more than I, I’ll pay you 3 dollars. Should you play this game?

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No. Because the cards that are put aside are paired by color, the remaining cards must be divided equally between red and black. So at the end of the game you and I will always have the same number of cards. You’ll lose a dollar every time you play. From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

February 14, 2015February 22, 2015 | Puzzles

Black and White

shinkman chess problem

By W.A. Shinkman. This is a self-mate in two moves: White makes a move, Black is allowed to make any legal reply, then White plays a second move that forces Black to checkmate him.

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1. Qd6! protects both white rooks and keeps Black’s bishop pinned. Black has only two options: 1. … Kc3 Rf3+ 2. Bxf3# 1. … Ke3 Rb3+ 2. Bxb3# From Benjamin Glover Laws, The Two-Move Chess Problem, 1890.

February 8, 2015February 15, 2015 | Puzzles

A Holiday Puzzle

New Year’s Day normally falls one week after Christmas: If Christmas falls on a Thursday, then New Year’s Day will fall on a Thursday as well. What is the most recent year in which Christmas and New Year’s Day fell on different days of the week?

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Trick question. Within any calendar year, Christmas and New Year’s Day always fall on different days:

February 6, 2015February 15, 2015 | Puzzles

Switching Visits

A prison warden greets 23 new prisoners with this challenge. They can meet now to plan a strategy, but then they’ll be placed in separate cells, with no means of communicating. Then the warden will take the prisoners one at a time to a room that contains two switches. Each switch has two positions, on and off, but they’re not connected to anything. The prisoners don’t know the initial positions of the switches. When a prisoner is led into the room, he must reverse the position of exactly one switch. Then he will be led back to his cell, and the switches will remain undisturbed until the warden brings the next prisoner. The warden chooses prisoners at his whim, and he may even choose one prisoner several times in a row, but at any time, each prisoner is guaranteed another visit to the switch room.

The warden continues doing this until a prisoner tells him, “We have all visited the switch room.” If this prisoner is right, then all the prisoner will be set free. But if he’s wrong then they’ll all remain prisoners for life.

What strategy can the prisoners use to ensure their freedom?

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If the prisoners knew that both switches were off at the start, then the solution would be fairly straightforward. One switch (say, the left one) will be used for counting, and the other is a meaningless dummy. One prisoner is chosen as the Counter. Each other prisoner is instructed to flip the left switch from off to on one time, at the first opportunity, but otherwise to flip the dummy switch on each visit. When the Counter enters the room and finds the left switch on, he turns it off; otherwise he too flips the dummy switch. When the Counter has found the left switch turned on 22 times, he knows that all prisoners have visited the room and can safely tell the warden so. Because the prisoners don’t know the initial states of the switches, this plan must be enhanced (otherwise, if the left switch starts in the on position, then the Counter will make his announcement prematurely). The solution is to ask each prisoner to flip the left switch from off to on twice in total, rather than once. The Counter will wait until he’s seen the left switch turned on 44 times altogether, and then tell the warden that all prisoners have visited the room. (At first I wondered why they couldn’t just stick with the first scenario but add 1 to the expected count to allow for the possibility that the left switch had started in the on position. But this won’t work: If the Counter is waiting until he’s found the left switch turned on 23 times, and if it started in the off position, then he’ll wait forever, because each of the 22 other prisoners will turn it on only once.)

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February 4, 2015February 15, 2015 | Puzzles

Black and White

keeble chess problem

By J. Keeble. White to mate in two moves.

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1. Ng5!, and the other knight will discover mate. From O.A. Brownson Jr., Chess Problems, 1876.

January 28, 2015February 8, 2015 | Puzzles

Threewise

http://commons.wikimedia.org/wiki/File:Napoleon's_theorem.svg — Image: Wikimedia Commons

Draw an arbitrary triangle and build an equilateral triangle on each of its sides, as shown.

Now show that AP = BQ = CR.

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Image: Wikimedia Commons Rotating the figure 60° counterclockwise around B carries A to R and P to C. And rotating it 60° clockwise around A carries B to R and Q to C. So AP = BQ = CR. From Edward Barbeau, Murray Klamkin, and William Moser, Five Hundred Mathematical Challenges, 1995.

January 23, 2015February 1, 2015 | Puzzles

Black and White

By Theophilus A. Thompson. White to mate in two moves.

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1. Ra4+! Bxa4 2. b4# From Theophilus A. Thompson, Chess Problems, 1873.

January 19, 2015January 25, 2015 | Puzzles

Ice Work

Three hockey pucks, A, B, and C, lie in a plane. You make a move by hitting one puck so that it passes between the other two in a straight line. Is it possible to return all the pucks to their original positions with 1001 moves?

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No. When viewed from above the three pucks form triangle ABC. With each hit the orientation of the triangle changes, so that sometimes “ABC” reads clockwise, sometimes counterclockwise. The original position was reached after zero hits, so it can be restored only after an even number of hits. From Arthur Engel, Problem-Solving Strategies, 1998.

January 15, 2015January 25, 2015 | Puzzles

Knights and Scoundrels

A problem from the 1994 Italian Mathematical Olympiad:

Every inhabitant on the island of knights and scoundrels is either a knight (who always tells the truth) or a scoundrel (who always lies). A visiting journalist interviews each inhabitant exactly once and gets the following answers:

A₁: On this island there is at least one scoundrel.
A₂: On this island there are at least two scoundrels.
…
A_n-1: On this island there are at least n – 1 scoundrels.
A_n: On this island everyone is a scoundrel.

Can the journalist decide whether the knights outnumber the scoundrels?

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I don’t have the olympiad solution, but it seems to me that there must be an equal number of knights and scoundrels. The first half of the interviewees are knights, and the second half are scoundrels. If there were more than n/2 knights, then they’d crowd out the scoundrels and their greater claims could not be true. If there were more than n/2 scoundrels, then they’d crowd out the knights and their lesser claims could not be false. UPDATE: Also, n must be even. If it’s odd, then the middle interviewee can be neither a knight nor a scoundrel, which is impossible. Thanks to those who wrote in to point this out.

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January 10, 2015January 18, 2015 | Puzzles