Hummings of a Happy Hurkle

This is a hat problem. The participants are supposed to be infinitely clever Mensa members, each wearing a hat bearing a number, such that they can see each other’s numbers but not their own. The hat numbers are integers, greater than zero. There are 3 participants (A, B and C, or Albert, Brian and Clarence if you prefer), and the number on one of the hats is the sum of the numbers on the other two. They take turns as follows:

A: I don’t know my number.

B: I don’t know my number.

C: I don’t know my number.

A: My number is 25.

What are the numbers on the other two hats?

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Here are two puzzles, superficially similar, but different answers.

A. Alf meets Bert, and asks him how many children he has, and of what sex.
The puzzle: if Bert has two children and at least one of them is a boy, what are the odds that he actually has two boys?

B. Alf now asks the same questions of several more people, until he gets to Charlie.
The puzzle: if Charlie has two children and at least one of them is a boy, what are the odds that he actually has two boys?

C. Alf goes back to Bert, and asks him what days any boys were born on.
The puzzle: if Bert has two children and at least one of them is a boy born on a Tuesday, what are the odds that he actually has two boys?

D. Alf now asks the same questions of several more people, until he gets to Dave.
The puzzle: if Dave has two children and at least one of them is a boy born on a Tuesday, what are the odds that he actually has two boys?

This strange puzzle and/or variants of it was originally set by Martin Gardner. The surprising thing is that the exact same question can have two different answers depending on the assumed context from which the odds should be calculated. There are enough clues here to make the answers fairly obvious.

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Anyone fancy a logic puzzle/game? I just ran across this one.

http://www.tomjubert.com/irrational

It’s a download, text-based propositional logic with a cute back story. Only 10 questions, but it certainly made me think.

Enjoy!

Multiple Choice Question:
If you choose an answer to this question at random, what is the chance you will be correct?
A: 25%
B: 50%
C: 0%
D: 25%

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Two witnesses provided evidence in court.

Tom is a somewhat reliable witness giving accurate testimony 70% of the time. Harry is less reliable, giving accurate testimony 60% of the time.

Both testified to the same effect.

What is the chance that their evidence was indeed accurate ?

Think carefully now.

[Thanks to Mark W for this one.]

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See “the puzzle of the two switches” for a seriously hard logic puzzle. This is a variation of the same puzzle to make it even harder.

The puzzle is exactly the same, except that you can no longer assume anything about the initial state of the two switches. Each switch could be off or on.

Your challenge again is to devise a strategy whereby all prisoners can escape.

Solution over the fold.
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You are one of ten clever people about to be locked in a strange prison controlled by a computer. You will each be sent to individual isolated cells with no possibility of communication. Randomly, one cell door will open and the prisoner will be allowed access to a central room that contains two on-off switches, a keyboard and a screen showing the exit code. Each door has the same probability of opening. You can assume that both switches are off to start with.

A prisoner has two choices. If they enter the exit code on the keyboard they will be allowed out of prison, but the screen will then go blank. Doors will continue to open, but only those prisoners who can remember the exit code from a previous visit will be able to escape. If they flip either one of the switches they will then be allowed back to the cell they came from and the door will close again. Flipping more than one switch or any attempt to communicate with other prisoners will cause the computer to shut down, and lock all prisoners in forever.

Your challenge is to devise a strategy whereby all prisoners can escape. This means that all prisoners must have visited the room before the first one enters the exit code, and that they can communicate only by setting and observing the switches and the screen.

Solution over the fold.

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Andy and Bert have worked out a neat card trick, but they need your help to do it. The puzzle for you is to figure out how the trick works.

The essence of the trick is that you shuffle a normal pack of 52 cards thoroughlyand give 5 cards to Andy. Andy keeps one card, puts the others into a special order and gives them to Bert. Bert looks at the 4 cards and correctly names Andy’s card.

You can assume that the trick involves no deception and no communication to Bert other than the sequence of the 4 cards. Can you figure out how it is done?

Solution over the fold.
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This statement of the puzzle comes from Wikipedia. There are subtle variations, but only one puzzle.

Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are ‘da’ and ‘ja’ but you do not know which word means which.

You are free to ask any questions of any gods in any order to a total maximum of 3 questions. You may only ask questions that True and False can answer. For the purposes of this puzzle the Random god should be thought of as answering entirely at random with no regard for the content of the question. Differing interpretations of Random are responsible for some variations in the puzzle.

Solution over the fold.
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Tom just bought 100 animals for a total of $100. The 100 animals consists of sheep, pigs and hens.

The sheep cost $10 each. The pigs cost $2 each. The hens cost 50c.

How many of each animal did he buy?
It’s not too hard to solve by trial and error, but we are looking for a logical solution so we can be sure it’s unique.

Solution over the fold.

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