Ants on a circle
You have a circle with a number of ants scattered around it at distinct points. Each ant starts walking at the same speed but in possibly different directions, either clockwise or anticlockwise. When two ants meet they immediately change directions, and then continue with the same speed as before. Will the ants ever, simultaneously, be in the same positions as when they started out?
(Thanks to OMD.)
What are the chances of that happening? Surely all that bouncing around is going to shuffle them all up. Well, the answer, which you've probably now guessed, is that, yes, they do all end up at the starting point. And the time at which this happens (although there may be earlier times as well) is just the time it would take for one ant to go around the entire circle unhindered. The trick is to start by ignoring the collisions, just think of the ants walking through each other. Clearly there will then be a time at which the ants are in the starting positions. But are the ants in their own starting positions? This is slightly harder to see, but you can easily convince yourself, and furthermore at that time they will also be moving in the same direction they were to start with (this is not necessarily true of earlier times at which they may all be in the starting positions).
Four switches and a lightbulb
Outside a room there are four switches, and in the room there is a lightbulb. One of the switches controls the light. Your task is to find out which one. You cannot see the bulb or whether it is on or off from outside the room. You may turn any number of switches on or off, any number of times you want. But you may only enter the room once. (Thanks to Tomfr.)
The trick is to realize that there is more to the bulb than light.
Step one: turn on switches 1 and 2, and go and have some coffee. Step two: turn off 1 and turn on 3, then go quickly into the room and touch the lamp.
It is controlled by switch 1 if it is hot and dark, 2 if it is hot and light, 3 if it is cold and light, 4 if it is cold and dark.
In a dark room there is a table, and on this table there are 52 cards, 19 face up, 33 face down. Your task is to divide the cards into two groups, such that in each group there must be the same number of face-up cards. You can t switch on a light, ask a friend for help, all the usual disalloweds. Is this even possible?
(Thanks to golftango and Bruno Dupire.)
An elegant lateral thinking puzzle, with a simple solution.
Move any 19 cards to one side and turn them all over. Think about it!
The use of an odd number, 19 in this case, can be seen as either a clue or as a red herring suggesting that the task is impossible.
A group of children are playing and some of them get mud on their foreheads. A child cannot tell if he has mud on his own forehead, although he can see the mud on the foreheads of any other muddy children. An adult comes to collect the children and announces that at least one of the children has a dirty forehead, and then asks the group to put up their hand if they know that they have mud on their forehead. How can each child determine whether or not their forehead is muddy without communicating with anyone else? (Thanks to weaves.)
If there is only one child with mud on his forehead he will immediately know it because all of the other children are clean. He will therefore immediately raise his hand.
If there are two children with muddy foreheads they will not immediately raise their hands because they will each think that perhaps the adult is referring to the other child. But when neither raises his hand both will realize that the other is thinking the same as he and therefore both will raise their hands.
Now if their are three muddy children they will follow a similar line of thinking but now it will take longer for them all to realize they are muddy. And so on for an arbitrary number of muddy children.
To make this work we really need something to divide time up into intervals, a bell perhaps, because no doubt not all children will be thinking at quite the same speed!