Happy Thanksgiving! In honor of the holiday I have given this riddle a gratuitous and certainly unnecessary theme. Several of you may have seen this riddle before; I think it’s on the less difficult side.
You are a turkey slated to be eaten for Thanksgiving. However, it turns out you can talk and have pleaded for your release. Your captors are inclined to pardon you; on the other hand, they are still hungry, so they have proposed a game of chance:

You are provided with 100 balls, 50 white, 50 black; and two urns. You must put the balls into the urns. One of the urns will randomly be chosen, and then a ball will be randomly chosen from that urn. If a white ball is selected, you will be pardoned; but if a black ball is selected, you will be eaten. You can’t leave an urn empty.

How should you divide the balls to maximize your chance of survival?

A quick one to get back into the swing of things. Paraphrased from a riddle that is, I believe, from a book by Raymund Smullyan. The answer to this one is very simple but eluded me for a long time. It finally hit me in a flash of insight as I was walking around; intrepid reader John, who related it to me, ended up putting together a truth table.
In a certain mental hospital, there are doctors and patients, all mixed together. Some of each are insane and others are sane. Sane people always tell the truth, and insane people always lie.

There is an investigator in this hospital whose job it is to release sane patients. (He has another job, which is to admit insane doctors as patients. That isn’t relevant to this riddle, but it may come up in a sequel.) Sane doctors are not released (they are supposed to stay around and help the patients.)
You are a sane patient in this hospital, and you want to be released. The investigator does not know that you are sane, nor that you are a patient. Making only one statement, how can you convince the investigator that you are a sane patient?

Solve this one yourself and be smarter than me. In retrospect it seems so obvious…

You are blindfolded and put in front of a table with 100 quarters. I tell you that 30 of the quarters show heads, and the rest show tails. How can you put the quarters into two piles such that the two piles have the same number of quarters showing heads?

No, you can’t use any other sense to determine what side of a quarter is which.

Hint 0: It only looks impossible.

Hint 1: Try it with a very small number of quarters, like 3.

I have been calling the logic problems posted here “riddles.” In an offline discussion, reader MJB asked me why this was so. To him, the best examples of riddles were those posed by the Sphinx to Oedipus or those exchanged by Bilbo and Gollum; exercises in unraveling metaphor. He suggested that “puzzles” more accurately reflects the nature of these problems, which are based in logic.

I defended my choice, because the choice of “riddle” over “puzzle” was something I actually did think about, although it was really an intuitive judgment; nothing I thought about so deliberately as I did in this discussion. To my mind, the best examples of puzzles are exercises like Sudoku, crosswords, cryptograms, jigsaws, or any of the various puzzles you find in Dell newsprint magazines. These are a lot of fun and satisfying to work out, but they seem fundamentally different and the thrill of victory is not quite as strong. That’s why I went with “riddle,” because the problems I relate seem so different than these puzzles. Certainly no publisher could print a whole magazine full of the kind of riddles I enjoy every month!

Thinking about it some more, the difference seemed to be that puzzles come in similar kinds, and their solutions follow a common pattern. There are only so many techniques that are used for, say, a Sudoku or a cryptogram. Finding when to use these techniques and choosing the right one comes with experience, but the similarities are clear. But, MJB asked, aren’t some of the truth/liar style riddles like that?

I believe they are not. The easiest riddles of this form, the most classic ones, involve casting a question that a liar and truth-teller will give an acceptable answer to. About the strongest common thread is that you need to establish exactly what you will and won’t be able to know at the end of the riddle. For instance, you might ask 3 yes/no questions without ever having the luxury to know what the words for “yes” and “no” are!

Furthermore, coming up with a problem of a similar form is hard. Try it! It’s as hard as coming up with a satisfying metaphorical riddle, one that leaves you thinking “I should have known that!” rather than “how the heck was I supposed to know that?” at the end. By contrast, coming up with a puzzle is easy. Want a cryptogram? Take a quote, add the substitution cipher your computer gives you, and BOOM. Sudoku? Take a grid of number and subtract some of them. Jigsaw puzzle? Take a photograph, print it on cardboard, and cut it up.

But it was the crossword puzzle that really drove home the distinction for me. Creating a crossword is not a mechanical process. Creating the grid of words takes time and effort; then you need to come up with clues. But just like the other puzzles, you start with the answer and go from there. And that, I think, is the distinction between riddles and puzzles. In a puzzle, the answer comes first. In a riddle, the question comes first—the really hard part of making a riddle is finding an interesting question. I think MJB accepted this reasoning, and I’m actually quite proud of that.

(And if it’s not clear, the kind of riddles I enjoy are the “logic” riddles, as opposed to what might be called “metaphor” riddles.)

Do you have any other thoughts as to what makes a puzzle different than a riddle? If so, please share. I have some ordinary articles planned, but I’m sure further riddles will feature before too much longer.

From the XKCD forums. Recast here for clarity and because I am not confident that the poster posted with enough clarity. I think I have the solution but, unlike most days, I am not 100% confident.

For those who are unfamiliar, Texas Hold’em is a particular kind of poker that has become very popular in recent years. Here’s how it works: A single deck is used. Each player gets two cards face-down. There is a round of betting. Three cards are flipped face-up, and there is another round of betting. Another face-up card, another round of betting, then a final face-up card, and a final round of betting. Everyone who is left makes their best 5-card poker hand from their own two hidden cards plus the five common cards on the table, and the best hand wins. You are under no obligation to use any cards from your own hand; if the board contains AKQJT of one suit, all players who have not folded will tie with a royal flush.

To simplify the problem, since betting is not relevant to this riddle: In Texas Hold’em, the goal is to make the best poker hand possible out of your two hidden cards plus five that are face up and common to everyone. High pairs, especially aces and kings, are regarded as excellent hands.
The riddle is this: One player has a pair of kings–a strong hand! Adding as few other players as possible, construct a setup where that player has no chance to win any of the pot. That is, no matter what 5 cards end up being common to all players, he cannot hope to even tie for the best hand.

I think I have the solution that requires 6 other players, which is what the poster alludes to.

As promised, here is the followup riddle. This one is hard… very hard. There’s a solution, with commentary, behind the cut. Although intrepid reader John Rhoadhouse did find it independently!

You are faced with three creatures, numbered 1, 2, and 3. One of them always tells the truth, one always lies, and one may tell the truth or lie arbitrarily. You need to find out which is which in three yes-or-no questions. Here is the catch: Although they understand questions posed in English, they do not speak it. The words they use for “yes” and “no” are “ja” and “da,” but you don’t know which is which.

This is where Hint 0 comes in. If this riddle were posed by, say, some guy named NarutoFan69 on the offtopic forums for your favorite Webcomic, you would be justified in reasoning:

1. There are 12 different possibilities for a setup: 6 arrangements of the three creatures, multiplied by 2 for the meanings of “yes” and “no.”
2. With 3 yes-or-no questions, you can distinguish between a maximum of 8 (2³) different outcomes.
3. So there’s no good solution, and NarutoFan69 is posting a lame trick question (“read the name tags LOL”).

But hopefully you trust me not to give you a riddle without a good answer. The answer is behind the cut, although solving it yourself is a mark of great honor.

(more…)

A riddle for you today, and another one next time. There’s a method to my madness this time, though, because I want to talk about Hint 0 and why it’s so important for the kinds of riddles I like. (Hint 0: A solution exists.) Today’s riddle is much easier, I think, if you take Hint 0 into account. It’s even more important for the follow-up.

As a reward for your valiant service as a knight, the King offers you the hand of one of his daughters in marriage. The oldest daughter always tells the truth. The middle daughter may tell the truth or lie, arbitrarily. The youngest daughter always lies. Of course, you don’t know which daughter is which. You want to make sure you marry either the oldest daughter or the youngest. (If you marry the liar, at least you know where you stand.) You only get to ask one yes-or-no question of one daughter. How can you use this question to pick a daughter who meets your goal?

Clarifications: If it makes your questions more convenient, you may assume the daughters have names, say, Anne, Beth, and Carol. You cannot ask a question that the daughter might not know the answer to. For instance, you cannot ask “What would Beth say if I asked her if she is the oldest daughter?” because if Beth is the middle daughter, the other two wouldn’t know.

Hint 1: Strictly speaking, you don’t need the oldest/middle/youngest designations, but it makes the question you ask a lot more elegant.

This riddle was related to me by intrepid reader John Rhoadhouse. If nobody has posted the answer by the time the next article goes up, I will post the answer in a comment at that time.

John went to a party with his date, Marie. At the party were three other couples, for a total of 8 people. During the night a certain number of people shook hands with each other. Nobody shook hands with their own date, nor with themselves. Afterwards John asked each person how many hands he or she had shaken. Each person gave a different answer.

How many hands did Marie shake?

Clarifications: John doesn’t ask himself how many hands he has shaken, and he is allowed to duplicate one of the other guests; in fact, a little consideration reveals that his number of handshakes must duplicate one of the other guests’.

Hint 0: A solution exists. There is no need for guessing or trickery.

In the last article, reader Post commented that he ended up not getting second interview based on the fact that he couldn’t solve one of these riddles. That’s unfortunate. I dislike riddles as a test of mental ability (as opposed to an exercise of mental ability) because the one thing they really test really well is “have you heard this riddle before?”

That said:

A dragon and a knight agree to fight a duel. The duel will be “fought” in the following manner:

• There are six poisoned wells available to both the dragon and the knight, numbered 1-6.
• There is a seventh poisoned well, numbered 7, which is located high on a mountain and available only to the dragon.
• The poison works in the following way: If you drink water from a poisoned well, you will die in 24 hours, unless during that time you drink from a higher-numbered well. If this happens, the poisons cancel one another out and you are fine. There is no way to determine whether or not you have actually consumed poison or what kind, until you either die or fail to die.
• On the morning of the duel, the dragon and the knight each bring a glass of something. They consume each other’s glasses, and then go their separate ways, leaving plenty of time to drink from any wells available.
• The dragon’s behavior is unknown, but you may assume that he is not necessarily stupid or blinded by hubris.

Here is the riddle: Why is the duel not quite as unfair to the knight as it seems? What gives the knight hope that he can win? Just how unfair is the duel, anyway?
For people who enjoy ruining riddles with inane nitpicks: The dragon and knight won’t physically interfere with each other in any way. No poison will be used except that which comes from the wells. If poisons are mixed in the glass, assume they have no potency.

I’m moving, so riddles will be the order of the next couple of updates. That’s my excuse and I’m sticking to it.

You start at a particular spot on the Earth. You walk a mile south, then a mile east, then a mile north. You end up exactly where you started. Desctibe the set of all points where you could have started. Assume the Earth is a sphere.

Hint: The number of points is not 1.