Riddle me this?

The French version of the KFC website is great. I love the Super-Croque

Aww man! I thought this was going to be a Gravel Puzzleworth thread full of puzzles. :(
― marianna lcl (marianna lcl), Thursday, April 7, 2005 8:53 PM (5 years ago) Bookmark

Absolute fave recent puzzle: http://jig.joelpomerantz.com/fun/dogsmead.html

fantastic!

<3

thank fuck i've finished it so i can go to bed!

Two classics that were in the movie Fermat's Room

-----------------------

A candy-store owner receives three opaque boxes of mislabeled candy. One contains sweet candies, one contains mints, and the third contains a mix of sweets and mints. The three boxes are, correspondingly, labeled “Sweets,” “Mints,” and “Mix,” but none of the labels is on its correct box. How many pieces of candy must the shopkeeper remove to determine the actual contents of the three boxes and relabel them properly?

-----------------------

The professor has 3 daughters. He tells a student "the product of their ages is 36, and the sum of their ages is my house number."

The student says "I'm missing a piece of information."

The professor says, "you're right. The eldest plays piano."

What are their ages?

2nd one of those has multiple solutions?

supposedly, but the solution is the one that most makes sense
this riddle has been told in different ways but this way is good enough

36, 1, and 1, the younger two are twins

"A student asks his teacher, 'How old are your three daughters?' The teacher replies, 'If you multiply their ages you get 36. If you add their ages you get my house number.' 'I am missing a detail,' protests the student. 'Oh yes,' says the teacher, 'the older one plays piano.' How old are the 3 daughters?"

I don't get how the sum = house number helps in any way? A house number could be anything? There are several solutions, surely? (I played piano at a young age).

you have to assume that the eldest isn't 35 because the teacher isn't old or something...

72, 1/2, 1, the youngest one is adopted

typo 36*

also, the house number line is important

but that's all I can say

so does the student already know the house number...?

18, the square root of -2, i, two of them are fabrications invented to make mathematicians' sums neat and tidy

there is no piano

does the student know the house number?

yes and that is why you can figure there is only one piece of information missing

actually that is also why 36, 1, 1 doesn't work

lol I googled the answer, captainlorax you messed up in your retelling - you need to use the comparative & not the superlative form

puzzles whose solutions involve nitpicking the semantics of puzzle-people talking the way nobody actually does = dud

I told the problem twice dayo

yeah, there was semantics involved. you had to figure out that the student knew the teacher's home number because he only need one detail

xxxxxp that's surely important to mention then?!
9,2,2

Use of "older" was telling. It's not clear the student knows the house number, my o/h thought this was what the 'extra info' he needed was.

oh now I C - now I have one more trick up my sleeve when I'm at the bar

that must be some nerdy bar

apparently the real answer is that if the student knows the house number, then he is deciding between two possibilities (6, 6, 1 & 9, 2, 2,) that adds up to the house # and knowing that there is only one elder 'solves' it

assuming that the twins in the 6, 6, 2 were born by c-section at the same time so neither one is 'older' than the other

it's a classic riddle supposedly

I don't know why I'm trying to do that Dog's Mead one, but I am. On first look it doesn't seem right, particularly clue 9 Down, and also 7 Down being 5 squares doesn't seem to fit with 14 Across being 3 squares. Is there anything stupid like blank squares to make it work?

Surely with the crappy student puzzle 2, 3 & 6 is another solution?

9 down is elegant when it all fits together! No blank spaces.

10a/10d trouble me - they seem to have multiple possible solutions.

oh wait unless martha comes into it. damn, can't remember how old she was.

apparently the real answer is that if the student knows the house number, then he is deciding between two possibilities (6, 6, 1 & 9, 2, 2,) that adds up to the house # and knowing that there is only one elder 'solves' it

Why isn't it 4, 3 & 3 ???

xp yes it's martha. phew.

Yeah - apparently the year of the puzzle is sometimes given?

NickB, if they were 4,3,3 (which totals 10), the student would be certain - there's no other way to get a total of ten, so he wouldn't say 'I need more information'.

(I think the student puzzle is a good one but you have to have done similar puzzles before imo)

Ah, I see that now - thanks!

tempted to break out the monty hall problem here...

Ledge I have another you might like...

go ahead!

A Neptunian alien has been cheated! He buys a set of gloves for 100 zkos, and gets 28 zkos back in change. Sadly, one of the gloves is full of fnags, and has to be thrown away! He now calculates he has paid 12 zkos for each warm finger.

How many fingers and hands did the alien have? You'll want to think, of course, about the basis of the human number system, and how they might do things differently on Neptune...

holy shit

How many pieces of candy must the shopkeeper remove to determine the actual contents of the three boxes and relabel them properly?

????

Am I missing something here or is this just the actual answer that you'd pick right away?

I guess that depends on whether the shopkeeper is aware that "none of the labels is on its correct box", but that would be a stupid thing to be aware of. Is the manufacturer going to ring up and go "hi, we mislabelled the boxes, and we definitely didn't get any of them right even by accident, but we can't tell you which is which"?

it's not the kind of thing an efficiet shopkeeper is likely to worry overmuch about, if we're going to be realistic about the whole thing

True - probably not going to see realist maths/logic puzzles get their own bookshop section any time soon

If you mix sweet sweets with minty sweets, they all end up tasting minty imo. That mixed jar is pure shopkeeper madness.

xp and THAT'S WHY COLLEGE AIN'T WORTH A DAMN

why don't you mix the contents of all three boxes and sell the whole thing as 'mixed'?

or, y'know, just open the jars and check. life is hard enough

i've got some equations for this neptune guy but i don't know what to do with them. one of them almost looks like a quadratic but it doesn't = 0.

4 hands with 2 fingers on each?

^ sex pistols, essentially

I wd suggest that Gravel's hint indicates that Uranians probly have a minimum total of 9 fingers

Neptunian rather. Brain addled by reading Burroughs this morning.

Is it true that aliens have two fingers in Uranus?

I hate riddles that involve math. Even if it's just arithmetic. I like better the ones about Artists, Bandleaders, and Cooks from Aldershot, Birmingham, and Cheltenham.

DM - no.

NV, yes, what I think you think I meant is indeed what I meant.

Too tired and distracted to do much other than test random bases tho which is not really working out.

and i'm stuck in simulatenous equation hell.

one of them almost looks like a quadratic but it doesn't = 0

I got here and then cheated by plugging (left hand side of equation - r.h.s. of eqn) into Excel for a giant grid of numbers and looking for the 0, so now I'm wondering what the right way of doing it is.

My mother is a maths teacher and would be ashamed :(

Trial and error is really not helping here.

I will say that imo Neptunians have more than 30 fingers in total.

imo 4fh-f(h^2)-2h+8 = 0. could be wrong tho.

i've taken an A4 page to work out that 6 = 6

but it's definitive

oh well i cheated. got pretty close i think, although mixed up an f and an h, and couldn't figure out the final rearranging trick. apparently there isn't a unique solution.

don't see why 4 hands of 2 fingers don't work tbh, but i mean i stopped studying maths at 15 for all intents and purposes, i missed out on aliengebra

4 hands 2 fingers = base 8, alien gets 28 in change, 28 isn't a number in base 8.

4 hands with 2 fingers does work I think, I missed a basic calculation. But Gravel's hint indicates that we shd expect the Neptunians to use a number system with the same base as their total number of fingers - in the same way that we use a base 10 system. We know that the Neptunian sytem must be at least base 9 because there's an 8 in the figures that the puzzle gives.

Agree that the whole base assumption is unwarranted tho, now I think about it.

when you need to resort to that level to tell me how i'm wrong, you've already lost imo

oops, xxpost.

Anyway, base = number of fingers shd really be made an explicit fact of the puzzle otherwise it's legitimate to use base 10 imo.

i get the base argument, i got as far as introduction to hex in my comp sys degree. but i didn't see the need.

admittedly i haven't run a thorough test on 4hands two fingers though. just not convinced it doesn't work.

I got a unique solution at 3 hands of 2 fingers, with 4 fingers tending towards 12zko per gloved finger as the number of hands approached infinity, but if dude has an infinite number of hands then he should expect to be ripped off by glove merchants, and anyway losing a glove wouldn't be a problem as he'd still have an infinite number of gloves for his infinite number of hands

but I got really confused at various points despite cheating, so, considerably not confident in that answer

perhaps I should go on the thread where Indians ask for exam paper solutions and see who can fix me up with a model answer, showing working

Anyway, base = number of fingers shd really be made an explicit fact of the puzzle

figuring that out is the point of the puzzle i thought!

alien's got four hands with two fingers each, paid 12 kzos per finger warmed = 6 fingers leaving one hand ungloved as per puzzle

Totally depends where the glove shop is. If it's on Earth, we can assume it's base 10. I don't think we can assume it's on Neptune, cos he wouldn't be considered an alien on Neptune amirite.

you think they don't have the same immigration problems we do?

every fucking thread

haha!

hey i fully support the rights of migrant workers anywhere in the solar system.

Sol Economic Area

I have just realised what NV's hint got at in particular (was doing the base thing but had not spotted that) which obviously makes me very wrong indeed

feh

when you need to resort to that level to tell me how i'm wrong, you've already lost imo

― k¸ (darraghmac), Monday, September 13, 2010 3:07 PM

this has just struck me as a good ILM/ILE site description tbh

I find the answer to the Neptunian one *really* satisfying btw!

Hint in ROT13: snpgbevfr

('Base = number of fingers' is the early breakthrough that makes the following maze seem worth exploring imo)

Oops, I was subtracting the 20 of 28 and then adding the 8. That's not going to help. Back to square 1.5...

^ first nerd problems

ROT13 nerdiness comes to the fore.

I like rot13 answers on the puzzle threads, it means that even though I can never do the puzzles I can convince myself my brain has not completely atrophied by mentally rot13-ing

imo 4fh-f(h^2)-2h+8 = 0. could be wrong tho.

Just been looking at this with a bit of paper and I got:

8h-4fh+2f+(f^2)=0

...which isn't giving any answers > 6 fingers, goddammit

^ scratch that, should be 8h+4fh-2f-(f^2)=0 which gives lots of answers eg. f=12 & h=3

should deffo be 8 not 8h... anyway here is the page i cheated from, correct equations in second post:

http://www.groupsrv.com/hobby/about687459.html

i did like the puzzle, just didn't have the maths chops to quite make it all the way (prob didn't help that i was trying to do it at work).

Oh, I was using 'f = total number of fingers' (= B in that solution), so no wonder our equations look different.

Here's how I did it (rot13)

Gnxr s nf gur ahzore bs svatref (juvpu vf nyfb gur onfr)

Gur ovt abgr: s fdhnerq

Gur cevpr: s fdhnerq zvahf 2s zvahf 8. (Orpnhfr bs gur 28)

Snpgbevmr guvf: (s cyhf gjb)(s zvahf sbhe).

Jung'f gur cevpr cre jnez svatre? S cyhf gjb! Fb vg jvyy NYJNLF tb arngyl vagb gur cevpr, naq vg'yy tb va s-zvahf-4 gvzrf.

Fb gur ahzore bs jnez svatref vf s zvahf sbhe.

Fb nal nafjre jvgu sbhe svatref ba n unaq jbexf.

nice reasoning!

Orpnhfr bs indeed

2 hands: 6 fingers on one, 4 on the other. 10 fingers total. dude loses the four-fingered glove. 72 z spent / 6 fingers = 12 z per finger.

unless we're meant to assume (as i did, at first) that the alien really IS spending the full 100, at which point i give up.

aw math riddles. snore

dusty bin to thread

ha this is all gone a bit complicated but still don't see why my solution doesn't work! (aside from unneccessary assumptions tbh)

i used occam's razor to cut off all his other arms, in other words

yeah, but like others have said, if we're working in base 8 (total number of fingers), then our alien consumer could never have gotten 28 z in change. in base 8, what we'd call 28 is just 30.

...25, 26, 27, 30. so he's gotta have at least 9 fingers total.

in base 8, base 10 28 = 34

Point agreed with, tho.

if we're working in base 8

unneccessary assumptions!

yeah, that's another way to look at it. was concerning myself only with the (2 finger sets) + 8 aspect.

if they've evolved to complicated financial transactions then they're at a stage where they can conceive a decimal system :)

xpost

It's not so much an unnecessary assumption, I just think the wording of the puzzle shd make clear that it is a necessary condition.

Like there are non-decimal human counting systems so biology doesn't equal mathematical destiny, obv

i'm being all challopuzzly tbh

aliens don't exist you nerds

it's true, we don't have to assume base 8 just because the alien has 8 fingers (in your proposed answer), but the riddle does direct us to look in that direction.

think that bothers me most about it is the clear statement that the alien "buys a set of gloves for 100 zkos..." doesn't matter how much he got in change, really (aside from it's potential utility as a clue towards base-whatever, it's a red herring).

the actual purchase price seems to have been 100.
and 100 = (total number of fingers) x (total number of fingers).
and 12 = (total number of fingers) + 2.

at which point the puzzle becomes rather difficult to solve.

eeks: "direct" and "direction", "think that bothers me...", fucking "it's".

^ yeah i kinda though there was something in that part too before i got reductionist on the whole thing

in coming up with my lazy answer above i just ignored that bit and pretended that the purchase price was supposed to be 72 after all, cuz 12 into 72 is way easier to manage

and you only have to try one candy right? that one's a hell of a lot easier to work out...

Bah, I got the right fingers-per-hand last night after realising my stupid sign error, but couldn't work out number-of-hands, so I thought I'd gone wrong again. Turns out from GP's answer and ledge's link that number-of-hands doesn't matter after all. Oops.

Yeah, sorry for the lack of clarity! I mean to state something like: "He pays the trader with a 100 zko note, and gets 28 zko back in change..."

what the hell is a fnag?

vapmrie's totoh

I guess everyone is familiar with four fours?

You must use exactly four fours, and the operations plus, minus, times, divide, square, and factorial (and as many brackets as you want) to make all the numbers from 1 to 100.

So one is (4÷4) x (4÷4), two is (4÷4) + (4÷4) etc...

Ooh, that'll keep me occupied for a Friday!

(some quick clarifications: you don't have access to 'square root', but '44'/'444' is allowed)

square but not square root seems odd. especially as square is indicated with a '2'!

They're the original buttons on some ancient Texas instruments calculator iirc?

http://www.interactivemaths.net/system/files/newlanguagepuzzle.pdf - good!

still not convinced this isn't a hoax

Haven't finished it yet - group effort?

do you know if there's answers anywhere?

I haven't looked - I found it on a maths site, apparently scanned from a weekly puzzle column in a Portland local paper? I'm pretty sure it's not a scam though!

What I've got so far, although I'm convinced I could just have any old nonsense with any justification for it:
================================

3 = bring the television here, please
5 = go through the next broken window
6 = past, present and future
8 = yes, but which?
9 = from Good Friday to easter
10 = first the bad news (really not sure about this one)
11 = pardon?
15 = send a little map
16 = say it again
17 = carry a magazine or radio into work
18 = happy anniversary
21 = where?
22 = act jealous, love (not sure, and no idea what this sentence even means)
23 = attention workers

symbols at start of 3, 15, 16 could be take/bring/send
empty circle could be 'say' or similar
a lot of the symbols don't get re-used, which is annoying.

if 8 is correct then 2 = no, but maybe tomorrow

and therefore 20 = i saw that yesterday

think i have all of them now btw

three of us managed to finish this. got 'sunday, during church' for 10 (1st day, during, house of god)

didn't really like this, just felt like arbitrary symbol matching.

Sorry about the last one.

Hope I can restore some credit with this: https://6408676999133382487-a-1802744773732722657-s-sites.googlegroups.com/site/jlnwebster/mathematics/puzzles-1/Number%20Square.png?attachauth=ANoY7cqBR3sXllolK07AUe3x7SCeM4n0YcFYtFmqjWbS5SZ4KCbxy8fuzX0HXNW_OXKMeUcdDRkcXEZ562wmcmbtlD3rhKYWQUrX9A6dMogauE0he1LCENPoVH1T9TisY7YypAs8_GrrS8JacPXzSSBxR_WpAvnMYTSbKuK6-3CAcohVDS0etPIuMlS1slCmxeeO-bVa-IxSJKjj2--aznXgvLR0s9VLs61AaokmFprsuP2NP2T_D9Y%3D&attredirects=1

Oops! I mean this: https://sites.google.com/site/jlnwebster/mathematics/puzzles-1/Number%20Square.png - don't download any attachments or whatever.

NB: It is bastard hard and I personally am stuck.

ambiguous in places. also waht is parity?

not that that will make it any easier.

Parity = odd or even?

This is the kind of thing that I am v bad at but will probably still look at later until I tie myself in knots with contradictory guesses and become angry

I think parity means 1 is odd, 0 is even? Which wouldn't make sense, because there aren't any zeroes.

Oh right, it just means odd/even in this context. I am dumb.

(There a clarification on the website that 'distance' means 1,2 or 3, not 0,1 or 2.)

Rows/columns also numbered 1-4 rather than 0-3? (lol spot the computer nerd)

Yeah, I'd imagine so.

i think i've proved to my satisfaction that the squares (^2 i mean) aren't 4 and 8. so they must be 1 and 1.

now instantly disabuse me of this notion.

what? 8 isn't 4^2. idiot.

Is there an answer online? I may have reached one, but I don't trust it. (ilxmail works if you don't want to spoil it onthread)

I took a leap of faith at one 2-choice junction and don't believe I hit a contradiction, so I suppose I should go back to the other option and make sure it doesn't work...

(a solution is here, if you want it... http://jaxwebster.wordpress.com/2010/06/19/self-referential-number-square/)

Interesting. I have that answer and I did (more or less) those steps but in a completely different order. I like a puzzle which doesn't have to be done in the one true order - you might think that something as self-referential as this would be pretty rigid about ordering, but it seems not.

Thanks, GP!

Haha I finally finished this and really liked it.

Weirdly once I had what ledge posted I could do the rest!

I didn't realise that the "new language puzzle" upthread uses (or maybe is just loosely based on, I haven't looked into it for more than 20 seconds here) a real "conlang", if there can be such a thing as a "real" "artificial language".

But anyway, this is a thing, which actually gets used in the world outside puzzles, apparently:
http://en.wikipedia.org/wiki/Blissymbolics
http://www.blissymbolics.org/pfw/

Blissymbols have become popular as a method of augmentative and alternative communication (AAC) for non-speaking people with cerebral palsy or other disorders, for whom it can be impossible to otherwise communicate with spoken language. However, Bliss did not approve of his language being used as an AAC; he sued the celebral palsy centers who employed it in this way, and they settled out of court

what a twat!

I kinda feel I lost some credit on this thread for the bad conlang puzzle upthread - anyway here's a conlang puzzle I think is absolutely brilliant, I ended up ordering a photocopy of the original book from the French national library.

http://twitpic.com/3vmtc/full

(not my desktop fwiw)

think i fail at the first hurdle for trying to click on the scrollbar/extra pages on the right of that pic.

and for saying "fail at the first hurdle"

I was looking for a languages thread but there are some articifial language puzzles itt so I'll put it here:
http://www.lrb.co.uk/blog/2015/09/03/helen-dewitt/tashu-duset-sekar/

and if you don't want to do the puzzles I agree with the article underneath too fwiw (modulo whiny caveats about opening things up w/o letting the new incomers be worn down by 3 years of being sneered at); hopefully the article got circulated round some teenager-populated nerd tumblr communities and such

I'd be curious to know how kids who haven't done Latin etc get on with the above but short of kidnapping some I will probably never know

surely you want someone who has done Latin as a control, right

(I could do them)

(I sometimes run a course that involves inventing a language - it's good fun)

Ooh, those were fun. Really not too hard, either.

xp I guess I was the control, though I am not a kid

(I could do them unless we were supposed to detect any nuances to the word order - I took the "unfixed" from the rubric to mean it didn't matter, though I spent a short time checking for patterns. But then I did Latin long ago and have spent time this very week swearing at German adjective declension, though studying German doesn't seem to rule applicants out of taking Test 1)

Tell me about your course, imago! Who is the target audience, how long is it, what do they do?

It's a day-long thing that forms part of either a weekend or a summer/easter week, generally aimed at 13-18 year-olds. I'll show them basic Esperanto and its mechanisms, discuss why it succeeds or fails as a language, and then run an exercise by which we write down letters in piles of vowels and consonants, turn them over, and then go around the table selecting patterns of vowels and consonants which are then turned over (by me) to make randomised new words. (For example, consonant-consonant-vowel-consonant-vowel-consonant could give us fbopad, or equally mjisev). Then I split up the group into four: nouns, verbs, adjectives and 'other' - each new group gets to claim in turn the words we've created and give them meaning. Finally all the groups collaborate to write a formative constitution for the new language. As I said, great fun :D

The names of 100 prisoners are placed in 100 wooden boxes, one name to a box, and the boxes are lined up on a table in a room. One by one, the prisoners are led into the room; each may look in at most 50 boxes, but must leave the room exactly as he found it and is permitted no further communication with the others.

The prisoners have a chance to plot their strategy in advance, and they are going to need it, because unless every single prisoner finds his own name all will subsequently be executed. Find a strategy for them which which has probability of success exceeding 30%.

I assume once names are found the boxes/names are not removed and are left exactly as they were?

Yep!

This is somewhat harder than the monty hall problem. I thought the old 'simplify!' would help, got a solution for two people but I can't generalise it. Hint needed or I will have to cheat.

Having a ... glimmering... of what to do, but can't turn it into a concrete plan of action yet.

Slowly I turned, step by step

Okay, I think I know what the strategy is in the case of four prisoners.

This is a hard puzzle derived from a computer science topic. Reading the solution, I first thought "why would THAT work?" then had an epiphany. I like it.

Yeah, I figured it was related to something like that, which is where I came up with my sketch of a strategy. Can't quite get the calculation to go through yet though.

Something I don't get about this. Seems like the very first guy has a 50% chance of finding his own name and then the second guy has a tiny bit higher than 50%(third guy a little bit more than that etc.). Already based on the first two 1/2 * (1/2 +epsilon) = 1/4 + delta = 25% + something small is already less than 30%.

Oh wait. Maybe not

heh, the prisoners one must be making the rounds cuz someone just posed it to me the other day. still haven't looked it up but i have the impression it's not reasonable to solve and impossible to calculate

some feasible riddles

1a. (easy, elegant solution) 9 gold coins, 8 of equal weight, 1 dud coins that weighs less. you have 2 weighs of a scale to find the dud coin
1b. (harder, inelegant solution) same as 1a but 12 gold coins, 3 weighs, and you don't know if the dud is lighter of heavier than the rest

2. you have 25 horses, and you want to pick the fastest 3 horses out of those 25. In each race, only 5 horses can run at the same time because there are only 5 tracks. What is the minimum number of races required to find the 3 fastest horses without using a stopwatch?

3. (fun algebraic* spin on a classic balance puzzle) i pick an arbitrary polynomial in one variable p(x) = a + bx + cx^2 + dx^3 + ... yyou can give me values of x and i will evaluate them, p(x). our goal is to figure out what it is; i.e, to tell me a, b, c, d, ... in the fewest number of evaluations. which numbers do you evaluate at, and how many evaluations does it take? (note: solution has to hold for ANY polynomial, of arbitrary (finite) degree)

*don't let that scare you! anyone can solve this one ;-)

Thanks!

Back to prisoners. Feel like the comp sci approach has to do with binary numbers and checksums maybe, but not sure about the latter. Studying the case of 4 prisoners now. The only thing you know going in is that the boxes are ordered and that other guys have all chosen correctly, presumably. (But is this last actually of any use? Looks like not) There are 4 choose 2 ways equals 6 for anyone to pick. Which 4 of the six should they use? You don't want any two guys to use the same exact choice, based on some kind of symmetry argument, and furthermore you don't want to favor one particular box, so each box should be opened by exactly two prisoners. One possible set would be
P1 chooses B1,B2
P2 chooses B1,B3
P3 chooses B2,B4
P4 chooses B3, B4

The probability of any one prisoner's success is 1/4 but you can't just multiply the probabilities since the successes are not independent. Need to get a pencil and do the calculation. (Also note that the chances of overall success with this approach should be the same as that of any "conjugate" approach, such as one that has P1: B1,B4

Probability still way too low, unless there is some form of communication inside or outside the room I am missing, like they do something to the paper with the names on them to leave a binary encoded message.

I cheated. V elegant, still tough to intuit *how* it works, even for only four prisoners. Wouldn't have come up with it myself in a million years.

Yeah this is advanced set theory stuff, right? I feel like I get the principles at work but no way I could prove one method of search could have a better probability of success than another. Permutations is an album title to me.

Okay. Just took a look, almost seems like some kind of diagonal argument variant.

No, not quite.

advanced set theory stuff, right?

Psst, Tombot, over here

Anyway, really cool that a probability problem is solved with a group theoretic approach.

What I like best about this problem is that even when you solve it correctly all 100 prisoners are executed more often than not. Imagine them putting in all that work and planning, then the first prisoner sent in doesn't find his name.

B-b-but they have a moral victory. They converted a vanishingly small probability of survival into a 30% chance! Socrates would have blushed.

For the case of four prisoners I got a probability 1/3 of failure, 2/3 of success. Furthermore, after the first two prisoners make it, it is over, they have succeeded. The same as after the first 50 prisoners in the 100 prisoner case.

Also seems like the first prisoner reduces his own personal probability of failure from 1/2 to 1/4.

No, that's not quite right, I left something out.

No, first guys probability by his lonesome does not change.

But simply by surviving, he has already eliminated most of the cases the strategy doesn't work for.

By my count, the 4 prisoner problem has 14 bad cases and 10 good cases, so overall probability of success is 10/24 = 5/12. The first prisoner doesn't not find himself in 12 of the 14 bad cases, leaving two bad cases for the second prisoner to either encounter or to miss in order to survive, in which case victory is assured. Calculating the two step probability gives 1/2 * 10/12 = 5/12 once again.

Ah no, still not right, first guy does not eliminate as many cases as that.

i have the impression it's not reasonable to solve and impossible to calculate

I solved it! I couldn't calculate it until I understood it a bit better, but:

(spoilers)
Lbh'er rffragvnyyl orggvat gung gurer'f ab 'punva' nobir yratgu 50. Vs gurer vf, lbh ybfr*, vs abg, lbh jva. Lbh pna pnyphyngr gur punaprf gung n punva bs yratgu a rkvfgf naq vg'f fhecevfvatyl arng - sbe vafgnapr gur punaprf bs gurer orvat n punva bs yratgu 100 ner.... 1/100!

*: oneevat n fhcre jrveq beqrevat rssrpg

Excited to try flopson's tonight!

solved the prisoners one last night... god damn it that's a fantastic riddle