Forums → Forum Games → Riddles
So, there was a thread on here a while back dedicated to riddles. Thaboss hasn't seen any new thread recently, so he's decided to start his own. Feel free to post and respond to any riddles.
Here's a classic one.
Bob hates his job. One day, while Bob is working on the 34th floor of an office building, he decides he's fed up with it. He rips open a window, and jumps through it. Although Bob is on the 34th floor, and there was nothing to stop, slow down, or cushion his fall, when he landed, he was perfectly fine. How?
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thaboss is exactly correct.
Matt and Amy were eating at a resteraunt on a cruise ship. Halfway through the meal, the ship entered a violent storm. Matt had been on a cruise ship before and had the stomach for it. Unfortunately for Amy, she wasn't used to the rocking, which had grown dramatically in the storm. She was sick for the rest of the evening.
And for the record, cruise ships DO have resteraunts with individual names. Check out Carnival's website.
I still can't spell "restaurant"...
Alright, thaboss heard this one, but forgot the exact wording. So he googled it and found it, but he's just gonna copy/paste. So please don't google this because you WILL find it.
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
Interesting...is it no one?
"There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know. A word of warning: The answer is not simple. This is an exercise in serious logic, not a lateral thinking riddle. There is not a quick-and-easy answer, and really understanding it takes some effort."
This is definitely gonna take some thinking...
*in dumb voice* the night of the blue moon!
Why do you keep posting these?! This isn't a riddle, it's a logic puzzle!
Regardless, I got this.
We'll start with Mr. 23, who has blue eyes. He says "There are 200 people here. Someone has blue eyes. I already see that 100 have brown eyes and 99 have blue eyes.
"If I have brown eyes, the other 199 know this. That takes me out of the equation. They will be working with only 99 people with blue eyes. Therefore, on the 99th night, someone else will leave the island. If no one leaves, that means that more than 100 people have blue eyes. Therefore, I have blue eyes."
On day 99, 23 realizes that, out of 200 people (with 100 having brown eyes), more than 99 have blue eyes. His eyes are therefore blue.
The SAME LOGIC takes place the other way around for Number 154, who has brown eyes. "I see that 100 have blue eyes and 99 have brown eyes. If I have blue eyes, that takes me out of the equation. On the 100th day, one of them will have determined that they have blue eyes. If no one leaves, then more than 100 have blue eyes. Since 99 have brown eyes, I have blue eyes."
On day 100, 154, and 99 other people, leave the island.
I may be slightly off, but I am more than close enough.
Why do you keep posting these?! This isn't a riddle, it's a logic puzzle!
True, but thaboss likes logic puzzles. Those were his only two anyway.
Matt is correct.
The answer is that on the 100th day, all 100 blue-eyed people will leave. It's pretty convoluted logic and it took me a while to believe the solution, but here's a rough guide to how to get there. Note -- while the text of the puzzle is very carefully worded to be as clear and unambiguous as possible (thanks to countless discussions with confused readers), this solution is pretty thrown-together. It's correct, but the explanation/wording might not be the best. If you're really confused by something, let me know.
If you consider the case of just one blue-eyed person on the island, you can show that he obviously leaves the first night, because he knows he's the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: [THEOREM 1] If there is one blue-eyed person, he leaves the first night.
If there are two blue-eyed people, they will each look at the other. They will each realize that "if I don't have blue eyes [HYPOTHESIS 1], then that guy is the only blue-eyed person. And if he's the only person, by THEOREM 1 he will leave tonight." They each wait and see, and when neither of them leave the first night, each realizes "My HYPOTHESIS 1 was incorrect. I must have blue eyes." And each leaves the second night.
So: [THEOREM 2]: If there are two blue-eyed people on the island, they will each leave the 2nd night.
If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities -- "I have blue eyes" or "I don't have blue eyes." He will know that if he doesn't have blue eyes, there are only two blue-eyed people on the island -- the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes -- THEOREM 2 says that if he didn't, the other guys would have left. When he sees that they didn't, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave.
This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they'll each wait 99 days, see that the rest of the group hasn't gone anywhere, and on the 100th night, they all leave.
Excellent. If you like logic puzzles so much, check this one out.
Matt is a contestant in a dealy game show. He has to choose between three doors. Behind one door is freedom. Behind the other two are instant death. He must choose one door. Matt chooses one of the doors, but, before it is opened, the Hostbot (who knows which door has freedom) says "Hang on! Let's make this a little more interesting!" The Hostbot points to a different door. "That one has instant death behind it!" The door is opened, revealing a wall of spikes, then closed again. "Here's the catch! You can now choose a different door!"
Matt can keep the current door, which has not been opened, or he can change to the third door. Here's the question: Does he have better odds by choosing to change doors?
Monty Hall problem. Always switch your door. That way you have 50/50 odds instead of 1/3.
Only half right! Nice try.
Yes, this is the Monty Hall problem. Yes, you switch doors. But you switch doors to change your odds to 2/3, not 1/2.
One door has been revealed, but not removed. There are still three doors. The odds of changing doors now changes from 1/3 to 2/3, as opposed to keeping the same door, where your odds remain at 1/3.
Yeah. You're right. 2/3.
How quickly can you find out what is so unusual about this paragraph? It looks so ordinary that you would think that nothing is wrong with it at all, and, in fact, nothing is. But it is unusual. Why? If you study it and think about it, you may find out, but I am not going to assist you in any way. You must do it without coaching. No doubt, if you work at it for long, it will dawn on you. Who knows? Go to work and try your skill. Par is about half an hour.
Another good one.
Thaboss has a 12 page file of riddles collected over the ages, including many from the old riddle thread. It would be cool if everyone could collaborate all their riddles into one massive file... Any idea how?
Well, it relates to the alphabet somehow. I first though it used every letter of the alphabet, but it doesn't. Upon closer inspection, it doesn't use the most common letter in the English language: "e."
How to collaborate the riddles? Open up Microsoft Word and start saving them as documents, each titled with the riddle's theme. Make a folder called "AG Riddles" and save them to it. Yeah, it'll take a while but that's how I would do it.
Make you sure you "53 Eagles" at the top. With my name on it. Because I stumped you guys.
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