This passage, from Left Hand of Dorkness’ link above, did the trick for me:
mmm
20 1 18 4 13 6 10 15 …?
There is a compilation of Medieval riddles called the Exeter Book of Riddles*, filled with the kind of things that show up in the “Riddle Game” between Gollum and Bilbo in The Hobbit. Tolkien undoubtedly had this book in mind when he wrote that chapter.
There are three similar riddles in it that Tolkien didn’t dare use, because the answer is too obscure. Here’s one of them:
Give up? The answer is:
[spoiler]
Creation.
That is, everything that God made.
If you think that this kind of answer is “cheating”, well, so do I.
I imagine some jester or lore-master, after having stumped the Lord of the Castle with one of these and grinningly revealing the answer being hauled away to the Dungeon or the Block by the irate Lord’s guards , and desperately crying out “I have another one, Lord! It’s a fair one, Lord! I SWEAR the answer’s not “Creation”!!!”
https://en.wikipedia.org/wiki/De_creatura[/spoiler]
- dating from the tenth century. There’s a Penguin edition translation of the hole thing, with answers and commentary. Sometimes the answers are only tentative.
Exeter Book - Wikipedia
Don’t make me call Manwë on you.
Walking down a path on a very important journey, you come to a Y-shaped fork in the path. One direction is where you need to go, the other leads to certain death.
There are 2 sentries, one standing next to the start of each of the 2 new paths. To learn which is the right path, you can ask one of the sentries one question. You know one of the sentries always tells the truth and one of the sentries always lies, but you don’t know which is which. What question do you ask?
Ask either sentry “which direction would the other sentry tell me to go?”. Whatever they say, take the other direction.
The Blue-eyed islander one has to be up there as a ball-breaker of a riddle - sure it’s been discussed here before and is pretty well known on the internet.
https://xkcd.com/blue_eyes.html
If it was just completely impenetrable then it wouldn’t be very interesting, but you read the solution and it makes sense - still impressed if anyone gets / got it straight up though.
[quote=“septimus, post:20, topic:817593”]
Here’s a classic that stumps most.
A prison warden is ready to retire and to spend his final years on Tahiti. There are only four inmates left anyway, so he decides to just shut the prison down. But he can’t decide whether to let all the prisoners go free, or send them all to the firing squad. Instead of flipping a coin he devises one final sadistic game. Addressing all the prisoners together he says,
"Tomorrow I’ll put a blob of paint on each of your foreheads, randomly choosing red or blue paint. Each of you will see the paint on the other three foreheads, but not your own. Feel free to discuss strategy now, but once the game starts tomorrow, if there’s any communication whatsoever among you, you all die.
Tomorrow you will each have paper and pencil and will have a chance to guess your paint color. On the paper, secretly write ‘Red’, ‘Blue’, or ‘Pass.’
[ul][li] If all four of you Pass, all four will die.[/li][li] If anyone with Blue paint writes ‘Red’, all four will die.[/li][li] If anyone with Red paint writes ‘Blue’, all four will die.[/li][li] If at least one of you guesses correctly, and none of you guesses wrong, then all four of you will live.[/ul]The prisoners appoint you to devise their best strategy.[/li][/QUOTE]
Are you sure you haven’t left a step out, like, knowing whether the person before you passed, or something like that? Because, as stated-- assuming the prisoners all write simultaneously-- i can’t see any way to communicate among the prisoners (even by something as simple as answering or not), which means I can’t see anything better than having prisoner #1 randomly write red or blue and the others passing. That gives a 50/50 shot at living or dying.
Does the warden read and respond to the first prisoner before the second one has to write, maybe?
Seems to me that the best you can do is a 50/50 chance. Tell 3 to pass and the other takes a guess.
My guess was that the gold treasure was in both caskets. But that answer depends on how you interpret the ambiguously worded (1).
(1) There is gold treasure in one of the caskets. (FALSE, if you take it to mean only one. But it could also mean at least one, in which case it would be TRUE and blow up my solution.)
(2) There is gold treasure in the red casket. (TRUE)
(3) Exactly two of these four inscriptions are true. (TRUE)
(4) None of these inscriptions is true. (FALSE)
What builds up castles
And tears down mountains
Makes some people blind
And helps others to see?
Sand. As in sand castles, erosion, sandstorms, and eyeglasses, respectively. I remember this one from a Super Friends cartoon.
And here’s one I read in a book, but I don’t remember the author.
It’s at the beginning of eternity,
The end of time and space,
The beginning of every end,
And the end of every race.
The letter E
Not again…
Do they all answer at the same time?
If not, first two pass immediately. Third person passes immediately if the 4th has red and passes later if the 4th person has blue. 4th person answers corresponding to what the 3rd person did.
That was surprisingly difficult, but I finally got something.
If you see three of the same color, say the other color.
If you see two red and one blue, say blue.
If you see two blue and one red, pass.
You’ll win 10 out of 16.
[spoiler]First rule: if they see a pattern that could be RRRR or BRBB (using some prearranged order), then vote the opposite of that. The prisoners will obviously die on a RRRR or BRBB, but live on a RRRB, RRBR, RBRR, BRRR, RRBB, BBBB, BRRB, and BRBR.
Second rule, which only applies to prisoners 3-4: if they see a pattern that could be RBBB or BBRR, vote the opposite. RBBB and BBRR combinations die, but RBRB, RBBR, BBRB, and BBBR live.
That’s 12/16 chances to live.[/spoiler]
If you assume that the sentries are rigorous rules-followers, there’s another question you can ask, and you can even ask it of a lone rigorously rule-following sentry who randomly will answer truthfully or falsely. I read this one as a kid and was fascinated by its devious answer. See if you can figure it out before clicking on the spoiler!
[spoiler]If you were to answer the question “Is the left path the way to certain death” as truthfully as you’re answering this question, would your answer be “yes”?
Imagine the left path leads to certain death. A truth-teller would answer, “Is the left path the way to certain death?” with “Yes.” A liar would answer, “No.”
You’re asking if they’d answer “yes” to that question. A truth-teller would say “yes,” because they would. A liar would answer that question “no,” so when you ask if they’d say yes, they’ll lie and say “yes.”
Similar logic applies if the left path doesn’t lead to certain death.
This nested question allows you to cancel out lies with a lie-about-the-lie, getting a truthful answer every time.[/spoiler]
Isn’t this a gambler’s fallacy or reverse Monty Hall problem? IOW, it does matter if I see 3 red or 3 blue (or for that matter, 100 red or blue in a scenario with 100 prisoners). If it is a random process, the chances of me having either red or blue remain 50/50…right?
[spoiler]I don’t think that’s the right way of looking at it. Well, at least the analogy doesn’t make sense to me.
The first thing is that you want to concentrate badness. It’s true that for any individual’s guess they’ll be right only half the time. But what if you could make it so that all four prisoners were wrong at the same time, but when they were right it was at different times? And barring that, at least have more than one prisoner be bad simultaneously.
When they guess the opposite of RRRR, if RRRR comes up, then they’ll all guess and all be wrong. But if BRRR comes up, then only one individual makes a guess and she’ll be right. If RBRR comes up, then a different prisoner will be right. Etc. We get a 4x improvement because the good guesses are spread out.
Beyond that, there are similarities to error coding. Looking up Hamming distance would be a start. Basically, you want to find several “loci” and the set of combinations exactly one color-flip away. They can’t overlap or there will be ambiguity.
The “clean” setup would have three 4-color combos and the 12 additional combos that are a step away. This still only gets you 12 wins but doesn’t require different rules for different prisoners.
However, three combos like this don’t exist. What we can do instead is have two 4-color combos, with their 8 attendant color-flipped versions, and then two additional combos but only with two flipped versions for each. One possible set (in binary, since that’s easiest for me) is:
0000 -> 0001, 0010, 0100, 1000
1011 -> 1010, 1001, 1111, 0011
0111 -> 0110, 0101
1100 -> 1101, 1110
There are lots of possible combinations; that was just the first I found.[/spoiler]
No, it has no answer. It went on for years and I finally called the store and spoke with them.
BTW:[spoiler]If you have exactly 2[sup]n[/sup]-1 prisoners, you can always come up with an algorithm that saves you (1 - 1/2[sup]n[/sup]) of the time. So for 7 prisoners, you can live 87.5% of the time (and can’t do any better).
I’m not sure of a general algorithm for other numbers but it will be at most floor(2[sup]N[/sup] / (N+1)) * N / 2[sup]N[/sup].[/spoiler]