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This one was already posted...

Ian the Mental-Ian wrote:This one was already posted...

I've just looked through all 11 pages again and I don't see it ............

This one is very different to the riddle set by Magnus on page 7 and another riddle by myself on page 10.

It sounded like the one you posted... My mistake.

moonbeam wrote:Seeing how no-one else is posting a riddle, here's another one for ya .......


Three gods,on a distant planet - one always tells the truth; one always lies and the other one lies and tells the truth at random.

Your task is to identify which god is which.

You are allowed to ask 3 questions which must be answerable with either a yes or a no - if you ask a question that causes one of them to answer with anything but a yes or a no (such as a maybe) - you will be incinerated on the spot.

The first question is asked to all 3 gods and they all answer in turn, with a yes or a no.

The 2nd question is asked to one (and only one) of the gods and he answers with a yes or a no.

The 3rd question must be directed to the same god that you asked the 2nd question to.

To make things awkward, the gods speak a language that you have never heard before and as such, you do not know what the words "yes" and "no" translate to in their language. After asking the first question, the gods may have all answered with the same answer (yes OR no), in which case you will only know one of their yes and no words; or they may have given different answers (yes AND no), so after the first question you may be lucky enough to know what both the words are (even though you won't know which word means yes and which word means no).
However, as you are a newcomer on this planet, use of their language is punishable by instant death, so you are not allowed to speak these words, or communicate them in any written form or whatever.

What questions do you ask in order to determine which god is which and to also enable you (without asking) what words mean yes and no??

Every time I've looked at this thread, someone had just answered a riddle, so I gave up checking!

I've not heard this particular variant before, so here goes:

The three gods are A, B and C.

Q1. Do all three of you always speak the truth?

The truthful god will say the equivalent to "No", the liar god will say "Yes" and the third god will say either, depending upon his mood.

There are other questions that would meet the same purpose, I think. What you are looking for is a question that will guarantee one god giving a different answer from the other two. This god will either be the liar or the truthteller. The switching god will always be one of the two uttering the same answer.

The gods reply as follows (the order doesn't actually matter, save for the purposes of illustration):

A: Hé
B: Ku
C: Ku

This means that A is either the constant fibber, or the ever-truthful god, and cannot be the 'switcher'.

Q2: (Addressed to A) Do both the other two gods tell lies at least some of the time?

If A says, "Ku," you have found the truthteller. You also know that ku means yes, and that hé means no.

If A says, "Hé," he is the lying god. Hé means yes, and ku means no.

This is because, after the first question, you know for certain that A either always lies, or always tells the truth. You have enough information in the question to know that, if you ask Q2 to the liar, you will get another lie, because either B or C is the one who always tells the truth. The liar will say that yes, both the other gods sometimes lie, and will use the same word he did in answer to the first question. The truthful god will also say yes to Q2, but will have said no to Q1.

So, you now know whether A is the truthful god, or the liar. You also know which word is which.

Q3: Is god B the one who sometimes tells the truth and sometimes lies?

If A is a liar, you could ask if B is the truthful god, and if A is the truthteller, then you could find out if B is the liar. However, the third question I suggested covers both possibilities.

There are four possible outcomes, but now you know whether you are speaking to the liar or the truthteller, and the words for yes and no, it's easy to work out the actual answer to Q3.

We got a winnerrrrrrrrrrrr .........

My solution is slightly different in that when asking question 2 - I ask a question that the liar and truth-teller would both give the same answer to, such as, "are you the liar?". They would both answer, "no" to this ..... apart from that it's pretty much the same:

Solution wrote:Firstly ask the question to each of them, "are you the random one?" this will always get either 2 yes's and a no, or 2 no's and a yes 'cos the liar will always reply yes and the truth-teller will always reply no, with the random one saying either yes or no.

We don't know which word means what but whichever answer is only given once, we know that he can't be the random one - we just don't know yet whether he is the liar or the truth-teller, so we now ask this one a question that he must answer no to - the question is, "are you the liar?". The truth-teller will reply no to this and the liar will reply no to this. So whatever word he replies with - this translates as "no". Now look at the answer that he gave to the first question. If he replied with the same word, then he said no to the first question so he must be the truth-teller. If he replied with a different word, then that word must mean yes and he must be the liar. So we now know the translation of the words and we have the identity of either the liar or the truth-teller (depending on what answers were given).
Lastly, ask the truth-teller/liar, whilst pointing to one of the other 2 men, "is that the random one?" You can deduce the final identities, as you know the identity of the one you're asking the question to.

That was a bit shorter than my explanation!

Happy to forego my turn at setting a riddle in favour of someone who has a good question lined up (although I have a fairly easy one that I can use, if you insist).

I insist.

Just to keep things moving, here's another one for ya:

Twenty prisoners are scheduled for execution, but the sadistic warden has decided they will play a game first. They are to be lined up facing in the same direction, such that each prisoner can only see those other prisoners in front of him. Each of them will be given a hat, which can either be red or blue, chosen randomly. Starting at the back of the line and moving forward, each prisoner will be given a chance to guess the colour of his own hat. If he guesses right, he will be pardoned; wrong and he will be shot immediately.

The prisoners are given several minutes to talk with each other ahead of time. After the game starts and the hats are on, any attempt to signal each other or saying anything other than "red" or "blue" will result in them all being shot.

What is the optimal strategy for them to follow, i.e. what will guarantee the greatest number of prisoners surviving?


Let's start off with you can get at least 10 correct by every even numbered prisoner guessing that his colour, is the colour of the prisoner in front of him. So, for example prisoner 20 looks at prisoner 19's hat and says red (which may be right or wrong). Prisoner 19 now knows that his is red so he says red. Prisoner 18 looks at prisoner 17's hat and says that colour .... etc. This way guarantees that at least 10 are freed, but can you employ a better strategy?

The chap at the back calls out the colour of the majority of the hats in front of him (let's say red). Then the person directly in front of him calls red only if the hat of the person in front of him is blue. If he says nothing, then the next person ahead goes through the same process.

If the person does call red, then the next person ahead calls blue only if the next person in front is wearing a read hat. If he says nothing, then the next person ahead goes through the same process.

By the time we get to the front, the majority of people who haven't called will know what colour hat they're wearing.

This won't save everyone, and it would take a fair amount of skill and judgement.

Can't they just cast their eyes upward and look at the couler of the brim?

if you cant see the brim...

surely if they can chat beforehand, they can say "if i say a colour within 5 seconds, your hat is red, if i say a colour after 5 seconds, your hat is blue"

So the first guy guesses his own colour, and he may be right or wrong, but from that the guy in front of him knows his own colour due to the speed in which the guy behind him said his guess. Due to the fact that the next guy knows what his colour is, he then says his own colour, allowing him to survive, in the appropriate time span concerning the guy in front of him. This continues towards the end.

This results in 19 people living with 100% certainty, with a 50/50 chance as to the first guy living. SO potentially, all 20 may survive, with a guarenteed 19.

Would that work?

Sorry for forgetting about this thread.

Thanks for stepping in, moonbeam.

May I ask if there are exactly 10 of each colour of hat, or if the random choice is such that there could be 20 people with red hats?

ah yes! If there is 10 of each colour the blokey who says it first can count how many of each he can see, thus knowing what colour his hat is. Therefore, utilising the above methodology, each of the 20 will survive!

Okay, just to clarify the "rules":

The prisoners cannot see the brim (or any part) of their hat.

There are a random (unknown) amount of each colour of hats, so the prisoners may all end up with 20 blue hats, 20 red hats or any combination in between - they do not know how many of each colour of hats there are.

The prisoners are not allowed to comunicate in any way whatsoever. I supose your method works Beardy so I'm gonna impose an extra rule (heh heh ):
When the prisoner makes his guess, he writes it on a slip of paper and hands it to the warden, who then reads it out loud for everyone to hear. Each prisoner is given 10 seconds to write on his slip of paper - the warden then waits another 10 seconds before reading their guess out loud, so each announcement made by the warden takes the same amount of time each time.

Cheater

Just because I had the perfect method

Hmmmmmm.......now I need to get rethinking!