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.
**
[/quote]
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]
Yes, I remember this one from my elementary school days. I won’t ruin it for anyone who hasn’t heard it before, though Asylum’s already given you a HUGE clue.
Thats the key to the riddle: I direct your thoughts at the end when I do the math for you.
Remember ResIpsaLoquitor’s riddle said there are 3 red hats and 2 blue hats, for a total of 5 hats, that are then drawn from unseen by three men. Nocturne pointed out that M1 and M2 each saw one red and one blue, thus could not know what they were wearing. But M3 saw two blue hats and therefore knew that he had a red.
However, Asylum threw in a twist that M3 was blind; and said that M3 doesn’t need to see two blue hats, he only needs to know that M1 and M2 are unable to decide their own color from what they see. This assumption by M3 would be correct in the scenario: M1 = blue, M2 = blue, M3 = red. In this case M3 knows both blue hats are gone leaving him with only red.
But there is another possibility that fits the original riddle but that would make a blind M3 incorrect in his assumption that he is wearing red. That is if : M1 = red, M2 = red, M3 = blue. In this permutation both M1 and M2 each see one red and one blue hat, but M3 cannot assume that this means M1 and M2 are both wearing blue thus leaving him with red.
Then there is also the possibility that they all three could be wearing red (remember we started with 3 red and 2 blue).
So how would a blind M3 be able to deduce what he’s wearing?
Hmmm…well, if he hears that neither M1 nor M2 knows what they’re wearing, that MUST mean that M1 sees both M2 and M3 in either 2 reds, or a red and a blue. M2 has to have the same problem (although he’s not necessarily seeing the same combination as M1, just the same conflict).
In other words, M1 doesn’t know what he’s wearing, so he has to see either two reds (meaning that there’s one red left in the closet and 2 blue, and he could have either color), or that he sees a red and a blue (leaving him with a blue and 2 reds, again meaning that he’s wearing either color). M2 has to have the exact same dilemma, since he also doesn’t know what he’s wearing. M3 doesn’t need to SEE any of this; he can just piece it together from what he’s heard.
If M1 and M2 are wearing blue, then M3 must be wearing a red. Thus, M1 and M2 would see a red-blue combo, and this would account for why neither knows what he’s wearing.
If M1 is in red and M2 is in blue, (or vice-versa), we can’t tell what M3 is wearing. BUT if M3 is in blue, M1 would have said “I know what hat I’m wearing” (red) because the two blues are taken, so the problem is contradicted (unless M1 is a liar). If M3 is in red here, then M1 will see a red and a blue, and still realize that he could have EITHER a red or a blue (because one red and one blue are left). M2 will see two reds, and also still be uncertain.
Bingo. The only way either scenario can work is if M3 is wearing a red. From pure deduction (not sight), he knows that’s what he’s wearing.
Now, throw in an orange hat and M4… 
Dang, I don’t know. Does it have anything to do with the extra “in” (The third word of the quote)?
Oh perhaps he could just take the table and smash through a wall? They could be made out of cardboard…
Huh? Huh?
He looks in the mirror, sees what he saw, takes the saw, cuts the table in halves, puts the halves together to make a whole, and exits through the hole.
Sorry Res and Yi, close but not quite.
Remember I said that it doesn’t matter what color hats the first two men are wearing. They could both be wearing red or blue hats or one could be wearing red and the other blue. You’re right that the third man is wearing a red hat though.
As has been pointed out the first man would know what color hat he was wearing if he saw two blue hats. Since he said he didn’t know the color of his hat he obviously did not see two blue hats, he either saw two reds or one red and one blue .
The second man doesn’t even need to look at the first man’s hat. He knows the first man would have known what color hat he was wearing if the second and third man were both wearing blue . This allows him to deduce that he (the second man) and the third man are either both wearing red hats or one is wearing red and the other is blue. Now if the third man is wearing a blue hat what color would be the only possibility for the second man to be wearing (remember there are only two possible combinations left)? Red , otherwise the first man would have known his hat color (the second and third can’t both be wearing blue). However since the second man said he did not know he obviously did not see a blue hat on the third man.
Which leaves the third man with the conclusion that he must be wearing a red hat. And he never had to figure out what the two other men saw, only what they didn’t see.
And kudos to Nocturne and Evo.
And I’ve got one last riddle, although I think it goes over better verbally:
Bob’s mother has four children. The youngest name is Penny, the second youngest’s name is Nickel and the second oldest’s name is Dime. What’s the oldest child’s name? Yes this may seem really easy but be careful. I’ve amused myself many times by repeating this riddle over and over only to see the look of frustration when I finally tell the answer.
Q: What walks on four legs in the morning, two legs at noon, three legs in the evening, is green and goes 100 miles an hour?
A: A frog-man in a blender.
Bob.
You’re right, it would be more difficult verbally 
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 of them which door the other one would say was safe. Then take the other door.
(Applause)
Now put three people in the room.
One always lies, one always tells the truth and the other one varies randomly between truth and lies.
You have one question.
Answer follows shortly…
“Thwack - Ahhh!”
(The sound of someone slapping their forehead and groaning with mixed delight and embarassment as realization breaks through at last.)
Thanks Asylum.
Here’s another one that also works better verbally:
How many of each animal did Moses take on the ark?
It’s fun to watch people scramble through various numbers before they figure it out. And just in case you’re still puzzled, the answer isn’t “All of them.”
Well done, glee.
But your new twist throws me.
Oh, and the answer is zero. It wasn’t Moses, it was Noah.
(Sound of door opening)
Say to all three of them “Did you know they’re throwing a party behind the safe door? The beer is free!”, then follow them.
(sound of running feet and door closing)
“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.)
Oh what they hey. I’ll contribute…
This thing all things devours:
Birds, beasts, trees, flowers;
Gnaws iron, bites steel;
Grinds hard stones to meal;
Slays king, ruins town,
And beats high mountain down.
What is it?
time
of which i waste far too much on this damn internet!