Most difficult riddle you know(and the answer)?

[glee]

*Originally posted by DMark *
“Speak, friend, and the door will open.”

What magical phrase will you say to get in?

(Apologies to Tolkien fans, I sorta paraphrased it above.
It’s a riddle in the Hobbit.
Gotta read the book to get the answer or see Lord of the Rings first 10 minutes of the film.)

The runes on the doors of Moria are in Elvish, so one says “Mellon” (Elvish for friend).

This happens in the Fellowship of the Ring, not the Hobbit.
It’s not the first 10 minutes of the film either.
We may have to send the Watcher in the Water to haunt your dreams…

[Revtim]

Even if you ask “How many of each animal did Noah take on the ark?” the answer isn’t so clear cut. One place in the bible says two of each, another says something like seven of each.

[Yo_La_Tengo]

Ok, nice job, gatopescado. Let’s see if anyone can get this one…
First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me what’s always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
And if you read Harry Potter, you can’t answer this question. I’d like to see if anyone can actually answer this w/o ever hearing it before…

[Snooooopy]

*Originally posted by Yo La Tengo *
**Ok, nice job, gatopescado. Let’s see if anyone can get this one…
First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me what’s always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?
**

Ass. I don’t want to kiss anyone’s ass.

[TheOtherOne]

The man who builds it doesn’t want it
The man who buys it doesn’t need it
and the man who needs it doesn’t know it…

So what is it?

[Bryan_Ekers]

*Originally posted by Yo La Tengo *
First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me what’s always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?

Most of the lines suggest the answer is “D”, but the “unwilling to kiss” is unclear.

[DoperChic]

*Originally posted by TheOtherOne *
**The man who builds it doesn’t want it
The man who buys it doesn’t need it
and the man who needs it doesn’t know it…

So what is it? **

A coffin.

[rjk]

*Originally posted by ultrafilter *
**

Exactly what it says. You keep rolling the die until it comes up six. Maybe you roll it the first time and it comes up six; maybe you have to wait for a hundred times. Keep doing this as an experiment, and the average number of times you roll it will be six.

**

How about:

How many people do you need in a room before there’s a 50% chance that two of them have the same birthday?

(The answer is somewhere in the 20’s, but I don’t recall it exactly.) **
[/QUOTE]

23. **
    [/QUOTE]

I haven’t done the arithmetic, but 23 is about right, IIRC. On the other hand, I did work out the die question, and six isn’t right.

The problems have very similar solutions. The general idea is to take the probability of failing on a single trial, and use that to get the odds for each successive try. For the whole series, we multiply the individual probabilities. So in the second problem we look for unique birthdays. The first person is ceretain to be unique, probability 1. The second is unique when he has a different b’day, 364/365 (ignoring leap years). The third is unique with probability 363/365, and so on. Multiplying as we go, the chance of all being unique drops below 50% between trial 23 and 24 (again, IIRC).

Similarly, we look at the odds of not rolling a six on a single trial, 5/6. Since the trials are independent, unlike the birthday problem, the odds don’t change for successive trials, and we keep multiplying by 5/6. Working it out, the chance of not rolling a six drops below 50% on the fourth roll, so we expect a six by then.

[Papamurf]

First think of the person who lives in disguise,
Who deals in secrets and tells naught but lies.
Next, tell me what’s always the last thing to mend,
The middle of middle and end of the end?
And finally give me the sound often heard
During the search for a hard-to-find word.
Now string them together, and answer me this,
Which creature would you be unwilling to kiss?

spider or perhaps spyder (I think)

[knock_knock]

originall posted by rjk:
Similarly, we look at the odds of not rolling a six on a single trial, 5/6. Since the trials are independent, unlike the birthday problem, the odds don’t change for successive trials, and we keep multiplying by 5/6. Working it out, the chance of not rolling a six drops below 50% on the fourth roll, so we expect a six by then.

ultrafilter is asking a different question - he wants to know the mean number of times you have to roll the die to get a six, not the median. The expected number of rolls can be expressed as

1p(6 on first roll) + 2p(6 on 2nd roll) + 3*p(6 on 3rd roll) + …

where p(6 on 2nd roll) represents the probability of getting a 6 on the 2nd roll and not before.

You could sum the series, but it’s not as simple as it looks. Instead, think of it this way: if you roll the die many times, you expect to get a six one sixth of the time. So on average, every sixth roll will be a six.

(btw, rjk, odds does not mean the same thing as probability. If the chance or the probability of something is 5/6, the odds of it happening are 1:5, and the odds against are 5:1)

[berdollos]

and never leave and other leave who never entered?

A hospital
(the dead and the newborn)

[rjk]

You have a standard six-sided die (i.e., equal probability of each number coming up). How many times do you expect to have to roll it before you get a six?

Although it’s certainly a long time since I took stats (and I didn’t even come close to acing it), I still expect four rolls. Maybe I just don’t understand, or maybe I’m misreading the problem. I’ll go dig up a stats book someplace.

(And I do know the difference between odds and probability; I just let myself get sloppy. And I should have previewed!)

[ultrafilter]

The easiest way to solve the die problem is by conditional expectation. The trials are independent, so at each one, you can forget what happened before. Let p be the probability of rolling a six (p = 1/6, but it’s less typing if I use p), and let m be the mean number of times you have to roll the die to get a six. If you get it on the first try (with probability p), m = 1. Otherwise, you now expect to roll the die m + 1 times before you see a six. So by the formula for conditional expectation, m = p + (m + 1)(1 - p). Solve that, and you’ll get 1/p, which is equal to six.

It’s possible to do this problem using the geometric distribution and some fancy manipulations of series, but that’s hard, and not worth the time.

[Left_Hand_of_Dorkness]

I remember when I was a kid reading this riddle somewhere. Is it attributed to Samson? I can’t remember: supposedly, some famous schmoe who was imprisoned asked this riddle and was freed based on the inability of his captors to answer it. Anyway:

As I walked and as I ran
Out of the dead the living came.
Four there were, and five to be;
Answer this riddle or set me free.
ANSWER:
As the narrator jogged along, he saw a cow’s skull lying by the road. A bird had nested in it and flew out, leaving behind a nest with four chicks and one egg.

My favorite riddle as a kid:
Q. How do you spell “hard butter” with four letters?
A. G-O-A-T.

My favorite riddles as an adult:
Q. What’s so delicate that even saying its name can break it?
A. Silence.

Q. How many legs does a dog have, if you call its tail a leg?
A. Four (calling a tail a leg doesn’t make it a leg)

Daniel

[Yo_La_Tengo]

Papamurf got the answer to my riddle. Nice job. (Spy+d+er = Spider)

[askol]

Ah hell. I just wrote a post twice dealing with the expected value of dice question, using the expected value of a geometric random value with p=1/6, and it turned out to be six.

I would show you the excruitiating deatail, but I’ve already written it twice.

askol

[pravnik]

Originally posted by TNhippie *
**You are in a room that is inescapable, except for two identical doors. One door leads to instant death, the other leads to freedom. You don’t know which is which. There are two other people in the room. They know all the answers. One of them always lies; the other always tells the truth. You don’t know which is which. You may ask one question. You may direct it to one person.

What do you do?

I’ll give you a few minutes. **

Ask either one “If I asked that other guy over there which door was the safe door, what would he say”? Then walk through the door other than the one he said.

Another, from Lewis Carrol: on the island of knights and knaves, knights always tell the truth and knaves always lie. You walk up on Al, Bill and Charlie. Al says “All of us are knaves”. Bill says “Exactly one of us is a knight”. What are Al, Bill, and Charlie?

[Monocracy]

posted by ultrafilter:
You have a standard six-sided die (i.e., equal probability of each number coming up). How many times do you expect to have to roll it before you get a six?

posted by knock knock:

originall posted by rjk:
Similarly, we look at the odds of not rolling a six on a single trial, 5/6. Since the trials are independent, unlike the birthday problem, the odds don’t change for successive trials, and we keep multiplying by 5/6. Working it out, the chance of not rolling a six drops below 50% on the fourth roll, so we expect a six by then.

ultrafilter is asking a different question - he wants to know the mean number of times you have to roll the die to get a six, not the median.

Actually, to me, the question is pretty unclear. By it’s wording, it sounds like he’s asking for the mode, not the mean or median. And to that, the answer would be 1.

[glee]

*Originally posted by pravnik *
Ask either one “If I asked that other guy over there which door was the safe door, what would he say”? Then walk through the door other than the one he said.

Coooee! I already answered that!

*Originally posted by pravnik *
Another, from Lewis Carrol: on the island of knights and knaves, knights always tell the truth and knaves always lie. You walk up on Al, Bill and Charlie. Al says “All of us are knaves”. Bill says “Exactly one of us is a knight”. What are Al, Bill, and Charlie?

Al must be a knave - because if he’s a knight, then his statement makes him a knave.

Bill must be a knight - because if he’s a knave, then Charlie must be a knight (otherwise Al is telling the truth), but then Bill would be telling the truth.

Since Bill is a knight, his statement means Charlie is a knave.

[Left_Hand_of_Dorkness]

Another answer to the two doors, one-person-tells-the-truth-and-one-person-lies question:

If you were to answer the question, “Does the left door lead to safety,” as honestly as you’re answering this question, would you answer affirmatively?

If the person is a truth-teller, then their answer to this question will be truthful, and you can trust it.

If the person is a liar, then their answer to this question will contain a double negative (basically, they’ll have to lie about what the lying answer would be). Again, their answer will be accurate.

This convoluted solution works for any logic puzzle in which you ask yes/no questions that are answered either truthfully or nontruthfully, as long as the answerers have a strong grasp of their logical duties.

Daniel