split the middle piece thus allowing the others to be taken out

Eesh... you got me on that one Tantris.

Yes, well done, you all solved it very quickly.
On a role there Big Doak

Ah Iceray youre on the right track!

Split one of the pieces in half by splitting one loop. Thats $1.

Connect 2 pieces with that loop. Thats $2.

Then we have one long piece, and one sole loop.

Split the final loop, thats $3.

Use it to connect the ends of the long piece, thats $4 and a whole bracelet!

Correct.

Gotta give Iceray some credit on that one. I'm logging off now, happy riddling!

Cop-outs like Iceray's attempt to cheat are unnecessary. Oversimplifications like Iceray's "everyone tells the person in front of them" don't work at all. Bishop's solution to save 50% is really nice, and a great starting point to understand the whole problem. But probability should not play a role, and I will concede only 50% for this solution (not 75% as claimed). Big Doak is the only one to get something correct and complete, but I wish he hadn't phrased it as an example. Good show!

I prefer to use modular arithmetic, calling the colors of the hats "0 and 1" instead of "red and white". Then it's straightforward to generalize to a problem with any number of colors.

As for these balls...why not generalize to 3^n balls, with 1 slightly heavier, using only n weighings? Or what if, instead of 1 slightly heavier, we have a given number k < 3^n where k balls all weigh a little extra (by the same amount).

Good riddle. I got another

What is the solution PatDragon? I don't see where Big Doak phrased something as an example.

The first guy says either what he sees in front of him or what color of hats there is a majority of in front of him. Everyone after that says what the person behind him said unless he sees two more of the color hat opposite of what the person behind him said.

Am I correct?

And yes, PatDragon, you could rephrase the riddle so as to have 3 weighings with 27 balls. I don't know if you could solve it with more than one ball heavier though...

ZaZa, no.
The first guy says "red" if there's an even number of red hats and "white" if there's an odd number of red hats.
The others use this information to figure out their colour.

Try this one on for size: a plane with capacity n > 1 is being filled with passengers. The first passenger has lost his ticket, so chooses a seat at random. Every other passenger has his ticket, and knows what his assigned seat is. They board the plane 1 at a time, in the following way: if their assigned seat is empty, they take it. If it is full, they choose one at random. What is the chance that the last person boarding the plane gets his assigned seat?

1/n

meaning that the first one (the one that lost the ticket) has to randomly takes the seat that corresponded to his ticket, so that all next passengers get their correct places as well.

one half

No, there is still a chance that his spot will be open if someone takes the person who lost his ticket's seat.

if the last person's seat is sat in, then he can not sit there.
If the original passenger's seat is sat in before this happens, then the the last passenger will definitely sit on his seat

50% chance guys.

50%

I think the equation is...I forgot what ! is called other than !.

n!

Forget that..

1-1/n!

no, its 50%.
Every time someone can't sit in their own seat they have 3 options:
1) Take the original (first person's seat)
2) Take the last person's seat
3) Take the seat of another passenger.

If they pick option 3, then the process repeats, as another passenger chooses a random seat
if they pick option 2, then the last person won't sit on their seat
if he picks option 1, then everyone will be able to sit on their assigned seat again, so the last passenger will sit on his assigned seat.

Option 2 is irrelevant, as it merely repeats the process, but options 1 and 3 have equal chances of happening.
Hence 50%

Once option 1 occurs it will be 100% chance of the last person taking his own spot. Thus, with more people, the more chances of the first person's spot being taken. That's where ! comes into play.

Oh, forget that. You're right.

There are no factorials involved. 50% is correct, but it doesn't seem to me that anyone has tried to explain why. bishop, you sound like a smart guy - think you could solve it with induction?

It would seem we've got a good amount of riddle-hounds...1 more:

You're playing on a game show, where a prize is hidden behind one of three doors. The prize is randomly hidden at the start of the show, and will not move after that. You are asked to choose a door. After your choice is made, the game-show-host opens a different door to show you that there is nothing behind it (i.e. he opens a door which you did not pick, and is not the winning door). Clearly, the prize is either behind the door you chose already, or the other one that isn't open. The game show host asks you if you would like to change your answer.

Most people phrase the riddle thusly: should you change your answer, or would it make no difference? I prefer the following:

You should always change your answer - you'll be twice as likely to win. While it might seem intuitive that the doors are equally likely to win, this is not the case. Why?

An excellent one. I think "You should always change your answer - you'll be twice as likely to win. While it might seem intuitive that the doors are equally likely to win, this is not the case. Why?" gives away to much, but that's just me.