Ok–so we do this thing at work where we throw riddles and brain teasers at each other. I’m usually pretty good at them, or else I’ve just heard them all already. 
Anyway, one guy who is a horribly smug little prig just sent out this one today.
He’s already said it’s NOT a combination lock, and that you can’t send the key separately. Does anyone know the answer to this one? I can’t stand the little smirk he’s walking around wearing for one second longer!
bella–desperately
You could have your friend send you a lock, open, to which he has the key. Then you send the locked box back to him.
Drive it over to him yourself. Nowhere does it say you HAVE to use a third party for delivery.
or what cher3 said.
or a goofy, box-inside-a-box solution which i don’t care to type out.
<smug>Google is your friend!<smug>
Sorry about that. But I plugged one of the sentences into Google and this is a rather common riddle. A solution.
Which is pretty clever, since the box is locked the whole time. I think it’s actually similar to some of the ways that you can ensure that a message is authentic and stuff.
…or you could just admit to the smirking man that he has you stumped. Ego only stings for a minute.
Your problem is common in RW – and is quite tough, so I’ll throw you a few hints. Good luck!
Game theory hint: Why are they supplying multiple locks?
Lateral hint: Suppose, instead of a box, I want to send you an encrypted e-mail, so that you alone can read my e-mail to you, but not my e-mail to anyone else? We do this all the time, but how?
Damn. Here I was hoping it was “Name three words that end with -gry.” I know the answer to that one now 
Thanks everyone! SmackFu and beagledave had the answer he was looking for. I am saved from his smirk, at least temporarily.
Oh, and Ace–admit defeat? Never! Bwahahaha
bella–adding Google to her “friends” list
belladonna, you want some great ones to get back at them, check here:
http://www.srainc.com/people/brentr/riddles.html
My favorite riddle page ever because… there are no solutions at the site! There are still some I cannot figure out, unfortunately. And I don’t like the google method.
I actually managed to figure two of erislover’s riddles out (not counting the ones I’ve heard before)! The answer to Lying Defeeted should be:
The first native must have green soles. If he has purple soles, he would lie when he said that he had either green or purple soles (when he must have one or either). The second man also has green soles, since he truthfully says the first man has green soles. Since the third native says the second native has purple shoes, the third native must be a liar, so he has purple soles.
For Hat-Trick:
You are wearing a red hat. If the person in the back saw two green hats, he would know that he was wearing a red hat (because there are only two green hats). He doesn’t say anything, so there must be at least one red hat in front of him. The middle person, knowing this, must see a red hat in front of him. If he saw a green hat, he could guess that he must be wearing the red hat. However, since he doesn’t say anything, he must see a red hat on the person in front of him.
I go along with Cher3.
Why do so many solutions involve sending the box or ring to your friend first? Why shouldn’t your friend just send you the lock by itself?
jmiz; your solution to Lying Defeeted is not accurate, but blame the wording of the riddle:
[spoiler]The riddle is flawed. The question to the first native shouldn’t be “what colour are your soles?” but “are your soles green?” The given question could lead to the following solution:
The first native has purple soles. He answers the question by saying the native word for “yellow” and lies about what he heard, reporting the answer as “green”.
The second native also has purple soles. He heard the first native say “yellow” and lies about it, changing the answer to “green”.
The third native has green soles. He tell the truth about the second man.
This version runs counter to your solution, which would follow if the initial question was “are your soles green?”
If the first native has green soles, he says the native word for “yes”.
If the first native has purple soles, he lies about his answer and says the native word for “yes” instead of the truthful “no”. so in either case, his answer is “yes”.
The second native has green soles and when asked what the first native’s answer was, he truthfully says “he said he has green soles”. Whether the first native’s answer was truthful is irrelevant; the fact is he said he had green soles.
If the second native had purple soles, his answer would have been a lie, saying the first native said he didn’t have green soles.
Based on the response, the second native must have green soles.
The third native says the second native has purple soles. Since you know this to be incorrect, the third native must be a purple-soled liar.
[/spoiler]
Three’s a crowd is interesting:
I predict one person will get $99.99, one person will get one cent, the third will get nothing, UNLESS the three scoundrels all possess perfect logic, in which case they’ll be able to predict the above result and will instead settle immediately for a 3-way even split.
Open Door Policy is pretty basic:
Ask “what would the other guy say is the dangerous door?” and take that door.
Apairantly Not is amusing:
Break up each pair one at a time and give each person one sock.
Timing is Everything:
Joe was walking for 50 minutes.
These riddles are rough. I thought I’d figured out River City… but I was wrong… grr.
OK, I’ve got Asphynxiation:
If the husband is lying, then the sphynx with the knife always lies, and therefore he has the knife. If the husband is telling the truth, then the sphynx with the knife always tells the truth, and he still has the knife. Ergo, the husband has the knife.
I got Trouble in River City. I have to solve every riddle I see just to prove to myself I’m not getting dumber.
I figured it out using pennies as normal balls, dimes as potentially light balls, and nickels as potentially heavy balls.
Step 1: Put 4 balls on each side of the scale.
Step 1 possible outcome: The sides are even. Therefore if there is an odd ball it must be one of the 4 balls you haven’t balanced yet.
Step 2: Replace one of the 4 untested balls with a normal ball. Put 2 untested balls on one side of the balance, and one normal and one untested ball on the other.
Step 3A: If the 2 sides balance, then the only remaining untested ball is the only possible odd ball. Put it against a normal ball and you have your final answer.
Step 3B: If the 2 sides are not even, then take the 2 balls from the side that did not include a known normal ball. If this side was lighter in step 2 the balls are potentially light, if it was heavier they are potentially heavy. Balance them against each other and whichever one fulfills its potential is the odd ball. If they are even then you know the ball that was with the normal ball is the odd ball. You also know if it is heavy or light based on step 2.
Step 1 possible outcome: The sides are not even. Therefore you now have 4 potentially heavy balls and 4 potentially light balls, and only 2 steps left.
Step 2: Take 2 of the potentially light balls, and 1 of the potentially heavy balls and set them aside. Of the 5 remaining balls put 2 potentially heavy balls and a potentially light ball on 1 side of the balance, and 1 potentially heavy and 1 potentially light on the other. Add 1 normal ball so there are 3 balls on each side.
Step 3A: If the 2 sides are even, then 1 of the 3 balls you set aside must be the odd ball. Put 1 potentially light ball and 1 potentially heavy ball on 1 side of the balance against 2 normal balls. If the side goes down the potentially heavy ball was indeed heavy, if it goes up the potentially light ball was light. If the sides are even then the potentially light ball that is still set aside is indeed light.
Step 3B: If the side with 2 potentially heavy balls and a potentially light ball was heavier in step 2, then you know that either 1 of those 2 potentially heavy balls is the odd ball, or the potentially light ball on the other side of the balance is the odd ball. Therefore you are down to 3 balls, 2 potentially heavy and 1 potentially light, and you can do the same thing you would have done in step 3A.
Step 3C: If the side with 1 potentially heavy ball and 1 potentially light ball was heavier then you know that either the potentially heavy ball is the odd ball, or the potentially light ball on the other side is the odd ball. Therefore you are down to only 2 balls, 1 potentially heavy, 1 potentially light. Put these 2 balls on 1 side of the balance against 2 normal balls and if they are heavier the potentially heavy ball is heavy, and if they are lighter the potentially light ball is light.
Three Dots:
Everybody raised their hand, so everybody knows there must be at least 2 green dots. The reason for this is that if there was only one green dot then whoever had it would not see any green dots. Therefore if anybody saw a red dot they would immediately know they had a green dot in order for there to be 2. But nobody immediately knew they had a green dot, so nobody saw a red dot, and you must have a green dot.
So that’s what PHP does.
(And I thought I was proving how smart I am…)