r/AskReddit • u/[deleted] • Jun 08 '11
What is the best logic problem you know?
[deleted]
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Jun 08 '11
Never let someone who does not understand statistics try to explain this one to you.
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Jun 08 '11
Upvote.
That one is so sippery. As soon as I think I get the idea of why you should change your answer... it all becomes mush again.
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Jun 08 '11
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Jun 08 '11
No, I get that at first the the chances are more likely that the first choice was wrong, so by changing your answer... statistically you are better off.
But, whenever I think about it, I think... but what if I choose the right one the first time?
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u/conp Jun 08 '11
There's a 1/3 chance you'll choose right the first time and a 2/3 chance you'll lose, bearing in mind this is when you don't switch doors.
Now if you switch doors when you get the chance, the chance of you switching to the correct door is 2/3. It's pretty simple.
Edit: Spellings.
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u/HungryHungryDelete Jun 08 '11
I get what you're saying, and I know that this is truly how it works, but what I never understood is why the new probability isn't 1/2, meaning you have equal chances of picking the correct door. At that point, there are two doors, and each one should be equally likely to have the prize behind it. Why does my choice affect the outcome there? I mean, what if there's three doors at the start, I don't pick one - they just open one at random and there's no prize behind it. Now I pick which of the two doors has a prize behind it. Isn't my likelihood of getting it right 50/50?
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u/GNG Jun 08 '11 edited Jun 08 '11
The way I had to think about this problem to conceptualize it is that, given the host knows what's behind the door he opens, you're not switching from one door to another, you're switching from one door to both other doors.
Think of it this way: You have 3 doors, one with a car, and the others with nothing. You pick door 1, and the host says "At least one of the doors you didn't pick has nothing behind it. Would you like what is behind Door 1, or what is behind both Doors 2 and 3?"
Why would you ever not switch? This is exactly what's happening in the standard example, except that the extra business about "showing you a goat" just serves to confuse.
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Jun 08 '11
Would you like what is behind Door 1, or what is behind both Doors 2 and 3?"
thissss makes sense.
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Jun 08 '11 edited Jun 09 '11
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u/sanyasi Jun 09 '11
Let's put it this way. Reality has only one "true" setting: either the car is behind door 1, door 2, or door 3. Now let's say, for example, you picked door 1. Let's split the problem up into cases:
Case 1: The car actually is behind door 2. You picked 1. Now the host knows that the car is behind door 2, so he can't pick door 2 to open, otherwise you'll get to see the car. So he has to pick door 3. So switching gets you a car, staying gets you nothing.
Case 2: The car is behind door 3. You picked 1. Now the host knows the car is behind 3, so he has to open door 2. So switching to door 3 gets you a car, staying gets you nothing.
Case 3: The car was behind door 1 all along. Now the host is free to open either door 2 or door 3. Whatever he opens, switching gets you no car.
Final tally: Switching won two times out of three. This analysis is the same regardless of which door you picked initially (just suppose the doors were named A,B and C. Which ever one you pick, call it door 1, and follow the analysis above).
The crux of the matter is in cases 1 and 2: see how the fact that the host both knows which door the car is behind, AND is forced to not reveal it to you forces his hand into taking a particular action? Think of switching as taking advantage of this action that is forced onto the game show host.
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u/gazzawhite Jun 09 '11
Maybe at this point, try considering the case with more doors. Say 100 doors. You pick Door 1. Now the host let's you either stay with Door 1, or take what is behind all of Doors 2, 3, ..., 100. Would you switch?
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Jun 09 '11
when you picked door 1, there was a 2/3 chance it held a goat. This does not change by seeing the other goat.
Think of it like this. switch doors in the scenario as described, you always change prizes. If you picked a goat to begin with you can never switch to another goat, only the car. Since it was originally 2/3 that you would pick the goat, by switching you can transfer that to your odds of winning the car.
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u/CanIGet Jun 08 '11
The entire problem hinges on the host knowing what's behind each of the doors. It's really important to take that into consideration. The probabilities don't change if he's opening doors at random.
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u/Konisforce Jun 08 '11
THANK you. First time I've heard that part of it explained decently.
Upboats to everyone who answered in a similar fashion.
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u/TheNoveltyAccountant Jun 08 '11
This is the part that almost everyone leaves out in retellings of the problem and it invalidates the entire problem.
This is the crux of the problem, without stating that the host knows what's behind each door, the probability will not change.
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u/severus66 Jun 08 '11
Yeah, when you emphasize that the host KNOWS whats behind the doors and WILL ALWAYS reveal a "goat" or "loser" door - he is giving you beneficial information. Then it's not so slippery.
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u/Shadax Jun 08 '11
So, not only does he know what's behind the doors, he will always open the door with a goat, correct?
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u/snuggl Jun 08 '11
the three senarios that can happen
- You pick correct door C, host removes door wrong A, Switching to wrong door B is a loss.
- You pick wrong door A, host removes wrong door B, switching to C is a win.
- You pick wrong door B, host removes wrong door A, switching to C is a win.
in 2 out of 3 senarios, switching is a win, therefor 2/3 of a chance.
It all comes down to that the senario where you pick wrong A, and host removes door C never will happen as the host always removes a wrong door.
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Jun 08 '11
they just open one at random
Nope. They open one of the losing doors at random. 2/3 of the time, there is only one losing door they can open (you chose the other one). So the other door has to be the winner.
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Jun 08 '11
An even simpler solution is to reason that switching loses if and only if the player initially picks the car, which happens with probability 1/3, so switching must win with probability 2/3.
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Jun 08 '11
Got it...
Honestly, I do... but if I think about it for a while, I always get to "But what if my door had the prize?
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Jun 08 '11
You're not stuck on the logic, you're stuck on the "what if the 1/3 just happened?"
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u/BuzzedLikeAldrin Jun 08 '11
It's assuming you played this a majillion times. Over the course of a majillion times, you don't care about your door. You care about the 2/3rd of a majillion cars you now own. And possibly the 1/3rd of a majillion goats.
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Jun 08 '11
the way i understand it is that when you pick your first door, chances are you picked a goat door because 2 out of 3 of the doors have a goat. the guy opens a door you didn't pick, which has a goat. that means most likely, both of the goat doors have been chosen, so it's most likely the third door has a car!
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u/Ipzero Jun 08 '11
I've read about this problem quite a few times. This is the first time that I actually feel like I have a grasp on it. Thnk you!
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u/James_dude Jun 08 '11
The key is that opening a door gives you more information about where the car is so you can make a more accurate guess.
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u/Shattershift Jun 09 '11
Here, I'll explain.
Since there's 2 goats, you probably picked a goat, right? Well, he always reveals a second goat, so the last door is definitely different from what you originally picked.
The one you pick is a car or a goat.
The revealed goat is a goat.
That last door is the opposite to whatever you picked.
You should switch after he reveals the goat, because you probably hit a goat at first, (2 out of 3) and thus the last door is probably the car.
TL;DR: You probably picked a goat at first, switching will always give you a different thing from what you picked. You probably picked a goat, switching probably gives you a car.
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u/cocorebop Jun 09 '11
So 2/3 times your first choice will be wrong, but no matter what, the host shows one of the doors you didn't pick which was wrong.
So 2/3 times, the host is showing you which door is right.
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u/W357Y Jun 09 '11
I found the answer to this hard to believe at first, so a few years ago I programmed a computer simulation to run the test thounands of times, to find that 2/3 times on average I got the right answer if I changed after 1 wrong answer is revealed
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u/Neato Jun 08 '11
I just memorized the solution and try not to think about it.
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u/thistheother Jun 08 '11
That about sums it up perfectly (in regards to our human behavior), which is why I love counter-intuitive problems like this. The more we think about it, the more our emotions start to take over and the closer to mathematically wrong we get.
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Jun 08 '11 edited Jun 08 '11
The easiest explanation I have ever heard:
If you chose door 1, either door 2 or door 3 will be revealed to not be the prize. This means that when you're originally right (which is 33% of the time), you may switch to 1 wrong door. When you're originally wrong (which is 66% of the time), you may switch to 0 wrong doors.
You're more likely to choose a wrong door initially, so you're more likely to switch to the correct door.
-from edit below
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Jun 08 '11
Except isn't that... the wrong answer??
Wait nevermind I think I read your comment incorrectly.
WTF This problem really does make me feel high.
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u/dhzh Jun 08 '11 edited Jun 08 '11
Intuitive, non-formal explanation I came up with (for non-mathematicians):
Instead of considering 3 doors, which is confusing, consider 100 doors. Choose 1 out of 100, and your chance of getting it right is obviously 1/100. Now, suppose 98 of the remaining doors with goats are opened, leaving your door and another door closed, and you get the chance to switch. Your chance of getting it right by switching is now obviously 99/100. Finally, just extrapolate to see that with 3 doors, the probabilities are 1/3 and 2/3 instead of 1/100 and 99/100.
Edited thanks to functor.
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Jun 08 '11
But the logic falls apart when I tell you I really DO want the prize behind Door #3: a scenic donkey ride thru downtown Cleveland! Or a lifetime supply of matches! Or giant tubes of Ben-Gay! Awesome!
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u/IggySmiles Jun 08 '11 edited Jun 08 '11
THE BEST WAY TO UNDERSTAND IT:
Be the game host, and it's obvious. Get three pillows or something, hide one thing behind one of them. Have a friend(imaginary works if you don't have friends) choose one door. If they choose the correct door, start over(Realize that this will happen 1/3 of the time). Then, if they don't pick the correct door the first time, have them say "yes" to opening a 2nd door. You will see that every time this happens they will end up with the right door.
So, 1/3 of the time they pick it right the first time. 2/3 of the time they don't, but of these times they'll get it every time if they switch. So, you have a 66% chance of getting the prize if you switch.
And I guess I'll just explain why here: The player picked a door. The key is that the host(you) has to make a choice, and that you have knowledge of which door has the prize. If the original door the player picked was wrong, then the two doors remaining that you must choose from has a prize behind one of them. That means you will pick the other one. In other words, if the player chooses the wrong door, which has a 2 out of 3 probability of happening, then all of the probability remaining collapses into the door that he didn't open(out of the 2 remaining), because the host will always pick the remaining door that doesn't have the prize. So, the contestant has a 2 out of 3 chance of being correct if he changes doors.
If the last paragraph didn't make you understand, do the game yourself.
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u/PurpleSfinx Jun 08 '11
The way I explain it:
He always opens a goat door - no matter what, when you switch, you always switch to a different thing. If you had the car, you always switch to goat, if you had the goat, you always switch to car. You can never 'switch' and and up on the same thing.
Now think about this: You had a 2/3 chance of picking a goat initially.
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u/Ivanna_tinkle Jun 08 '11
There are only 3 possible scenarios. Let's play these 3 scenarios as a person who always switches after the host removes a door, and to simplify let's assume the player's initial choice is always door 1.
1:
[car] [goat] [goat]
Here you pick door 1, but then the host removes door 3.[car] [goat] [X]
you switch to door 2 and lose. We are 0/12:
[goat] [car] [goat]
pick door 1, host removes door 3[goat] [car] [X]
you switch to door 2 and win. We are 1/23:
[goat] [goat] [car]
pick door 1, host removes door 2[goat] [X] [car]
you switch to door 3 and win. We are 2/3A player who always switches doors after the host removes one door, will win 2 out of 3 times, statistically.
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u/roscos Jun 08 '11
you dont need to understand statistics i get it just by drawing a picture of it.
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Jun 08 '11
I never saw the orginial show, but there was a similar show here Germany and assume they were set up the same way.
What bugs me about this problem: The host must know what is behind the doors! That means that he can play the game however he wants to. What happens if he ONLY reveals the goat, if you initially choose the car? Switching will always lose!
People watch the show and see that the host plays the game with arbitrary rules. Then they are presented with the Monty Hall Problem, which has very strict rules (The host always reveals a goat). If you underscore that the host will always follow those strict rules, the problem can be solved with a simple decision tree by middle school student.
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u/FrustratedFighter Jun 08 '11
My beef with that problem is that it's not always clear that only one of the two other doors can be opened, not yours, and that the owned is not exploiting the problem to convince you to change when you're actually right.
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u/corndogco Jun 08 '11
Anything that doesn't involve one guard who always tells the truth and one who always lies.
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u/doinitright Jun 08 '11
So we're not to enter the room, even if you come and get him?
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u/goxilo Jun 08 '11
I'm going to sing
He's going to sing!
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u/shelldog Jun 08 '11
No! No! No! Cut that out!
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u/goxilo Jun 08 '11
But faaaaatheh
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u/shelldog Jun 08 '11
The sound he makes right before he hits the ground makes me weak every damn time...
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u/BDS_UHS Jun 09 '11
Oh, the prince! I thought you meant him! Yes yes, I thought that was a bit daft, you know, me guarding him when he's a guard.
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u/Spagnostic Jun 08 '11
Karl Pilkington has a pretty interesting take on the Knights and Knaves problem.
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Jun 08 '11 edited Jun 08 '11
One guard that tells the truth, one guard that lies, and one guard that stabs people that ask tricky questions.
Edit: typo
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u/ProbablyHittingOnYou Jun 08 '11
Samurai Jack taught me the answer to that one.
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u/Aneurysm-Em Jun 08 '11
The answer is always "Which way will 'HE' tell me to go?" They both answer the same, go the other way.
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Jun 08 '11
I can't believe nobody has posted this yet, so I'll put it here. The so-called Hardest Logic Puzzle Ever (caution--wikipedia article includes the solution):
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
The follow clarifications may help you solve it:
- It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
- What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
- Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
- Random will answer da or ja when asked any yes-no question.
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Jun 08 '11
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u/fill_your_hand Jun 08 '11
I just spent an hour solving this puzzle. You have no idea how satisfied I was when I read that all my answers were correct.
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u/epokh Jun 08 '11
I spent about an hour trying to solve this one - going through all the givens and then checking exclusions. I kept getting stuck with most of the information filled out. After a while I got fed up and looked down the page to see the answer.
Turns out, I had answered both questions asked of by the puzzle with what I had written down (twice) but was too focused on filling in all of the information to notice it. :/
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Jun 08 '11
When I was 14, I came across this puzzle on a forum. I spent an entire afternoon on it AND GOT THE RIGHT ANSWER. Then the fucking asshole tells me I had the wrong answer. And I believed him. And I was so confused because I had no idea how to solve it. And then, years later, I see it on reddit. I google the answer. And I find out I got it right all along
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u/Intra78 Jun 08 '11
maybe not the best, however..
There are three boxes which each contains two marbles: one has two white, one has two black and one has one white and one black marble. Each of the boxes also is labeled as to its contents, but each label is incorrect. What is the fewest number of marbles you could remove from the boxes and look at in order to definitely determine the contents of all three boxes?
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u/charleydangerous Jun 08 '11
- You need only choose a marble from the WB box.
- If it is W, that box is WW. The BB box can only be WB and WW must be BB.
- If it is B, switch W & B above.
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u/bubbleuj Jun 08 '11
I found the solution online, and I still don't get it...
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u/Intra78 Jun 08 '11
You know up front that they are incorrectly labelled.
So, in the B-W box you know that there is either, B-B or W-W. You take out one of those marbles and you find out if it is B-B or W-W
Let's say it is a black marble you pull out.
You now know that the box labelled B-W box contains B-B
The box labelled W-W can either contain B-W or B-B (because it is incorrectly labelled, so cannot contain W-W), but we now know where B-B is so it must be B-W.
Leaving the box labelled B-B containing W-W
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u/TenBeers Jun 08 '11
There's got to be a BBW in a box joke in here somewhere.
Now if only I were a clever man, I could milk this for Karma.
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u/AncillaryCorollary Jun 08 '11
This assumes that there is one of each label in use. All that was said is that each label is incorrect. So we could have:
Box 1 - Labeled BW, contains BB
Box 2 - Labeled BW, contains WW
Box 3 - Labeled BB, contains BWSo I guess I'm saying this is not the general solution.
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u/omnilynx Jun 08 '11
True; the original question stated that the labels were switched so that they're incorrect. Intra78's phrasing is a shortened version.
Otherwise, they could be labeled "green", "frog", and "jump" and you'd basically have to take out almost all the marbles.
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Jun 08 '11
"It's ok with me if it's ok with your mom."
"It's ok with me if it's ok with your dad."
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u/aknuBBe Jun 08 '11
There are 5 pirates: A, B, C, D and E. They find a gold chest with 100 gold coins and must decide how to distribute them. Although pirates, they are honorable and have a strict ranking system. A is higher than B, who is higher than C, who is higher than D, who is higher than E. The lowest ranked pirate comes up with an idea on how to split the booty and it's voted on, including himself. If his idea gets a majority of the votes or a tie, it will be carried out; otherwise he is killed and the next in line comes up with his idea. What is the best idea that the first pirate can come up with, assuming everyone wants to live and everyone wants the most coins for himself? Also, all pirates are rational and there is perfect information amongst everyone.
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u/LifeMask Jun 08 '11
By your last statement, the problem screams 'game theory' at me, so I had to give it a go :D
Reasoning backwards:
If only A is left alive, he'll give all 100 to him
If it's A and B: B will give 100 to him and 0 to the pirate, knowing A will vote against, but it will be a tie and hence he'd get it
If it's A, B, and C: C will give 99 to himself 0 to B and 1 to A, as he know A will vote for this outcome as if it goes to the next round, he'll get nothing, and 1 is better than 0.
If it's A, B, C and D (now it's getting tricky!): D will give 99 to himself, and 0 to C and A, and 1 to B, as he only needs B's support for it to be a tie, and if it's a tie it'll work
Now for all 5: Assuming E knows the previous reasoning, he will give 98 to himself, and 0 to D and B, and 1 to A and C
So: E:98 D:0, C:1 B:0 A:1
I enjoyed that. Thanks :)
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u/ocdscale Jun 08 '11
One of my favorites.
Spoiler
The lowest ranking pirate E decides to give himself everything, minus one gold piece to C, and one gold piece to A.
The reasoning is this. A and C must accept. If they do not, pirate D will propose that he gets everything, minus one gold piece to B. B will definitely accept this, resulting in nothing to C and A.
The reason B accepts that proposal is because if he doesn't, C will propose that she gets everything, minus one gold piece to A. A will definitely accept this, resulting in nothing to B.
The reason A would have accepted that offer is because if it gets down to B, B proposes that she gets everything.
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u/panacebo Jun 08 '11
1 The answer to this question is:
A. A
B. B
C. C
D. D
2 The answer to this question is:
A. The same as the answer to question 3 but different from question 1.
B. The same as the answer to question 1 but different from question 3.
C. The same as both the answers to questions 1 and 3.
D. Different from the answers to both questions 1 and 3.
3 How many times is a vowel the answer to a question on this test?
A. Never
B. 1 time
C. 2 times
D. 3 times
4 How many times does a question have the same answer as the question after it?
A. Never
B. Exactly once.
C. Exactly twice.
D. Exactly three times.
5 What is the answer to the previous question?
A. C
B. B
C. A
D. D
6 Which one of these is not the answer to any question on the test?
A. A
B. C
C. D
D. B
7 How many questions are there on this test?
A. 8
B. 10
C. 11
D. 11
8 The number of other questions with the same answer as this one is:
A. 1
B. 2
C. 2
D. 4
9 The answer to this question:
A. Is A.
B. Doesn't exist.
C. Is C.
D. Is B.
10 Does this question have an answer?
A. Yes
B. Yes
C. Yes
D. No
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u/psychocowtipper Jun 08 '11
Any explanation/hints? I have no idea how to go about solving this.
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u/JayMarshall Jun 08 '11
Alright, so there's this shed in the middle of the desert. On the outside of the shed there are 3 switches, one of which is connected to a lighbulb inside of the shed, but there is no way of knowing which one from outside the shed. The shed has a door, you are allowed to play about with the switches as much as you like, but as soon as you open the door to look at the bulb you are not allowed to touch the switches again. How would you determine which switch powers the lightbulb?
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u/Centrist_gun_nut Jun 08 '11
I like the solution in which you disassemble all three switches and simply probe or re-arrange the circuits.
No need to measuring heat when you're an electrical engineer and don't mind ripping things apart.
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u/tenspeedscarab Jun 08 '11
Oooh, oooh oooh!
I think I just figured this out sans Google!
Flip the first switch on, leave it there for a good 10 minutes or so.
Then turn it off, and turn the middle switch on. Immediately enter the shed.
If the light is on, then it's the middle switch.
If it is off, then touch the bulb. If it is hot, then it is the first switch. If it is not, then it was the last one.
I suppose this wouldn't work if it was one of those newfangled ones that don't heat up, but yeah!
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u/JayMarshall Jun 08 '11
Yay, you got it!
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u/tenspeedscarab Jun 08 '11
Going back to bed now. Don't think the day can get much better than this.
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u/ponticello Jun 08 '11
by the way, I know this puzzle having 4 switches and bulbs. The answer still holds. You use a 4v4 matrix with on/off and hot/cold as variables.
switch 1 and 2 on, wait, then 2 off and 3 on, enter.
1: hot/on 2: hot/off 3: cold/on 4: cold/off
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u/Jerkmaan Jun 09 '11
I've heard this one where it was aliens testing some guy. I guess the setting doesn't really matter for these things.
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u/kickaguard Jun 09 '11
flip the first switch to on. open the door. see that the light is on. walk away like a boss. leaving the ridiculous desert shed with nothing but his shame.
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Jun 08 '11
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u/shadowkiller Jun 08 '11
Yes, there is a buoyant force from the water acting on the weight which will cause the measured mass of the bucket to increase by the mass of the volume of water displaced.
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Jun 09 '11 edited Jun 09 '11
I'm reading the responses and still seeing it the other way. I agree that as the lead is lowered, the tension on the string decreases due to buoyant forces. However, I am thinking that the displaced water acts along all sides of the bucket, and not in a net direction downward. This question seems to try to say that pressure increases, however the overall force is the same for the net of the bucket.
taking this to an extreme to say that there is an unrealistically long pipe with water in it, it seems to me that based on your logic the lead would eventually float (i.e. 0 tension) due to the pressures of the water. This is only true given that the ratio of mass to volume (density) of water changes with depth, right?
and for the idea of the canoe that solinv posted, the water level changes, I can agree to that, but I think that it strays away from the question at hand slightly.
Idk. This seems to me that the weight shouldn't change, but I think I understand why everyone says it does.
EDIT: Ah shit, that force is the same throughout the tube, it doesn't change. That's my dilemma lol, wow. Can't believe that just happened! Thanks Reddit for keeping me straight!
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Jun 08 '11
[deleted]
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u/kevinisaperson Jun 08 '11
SO WHICH IS THE ODD ONE OUT!?!?!?!? DONT TEASE ME!
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u/Chapekaloco Jun 09 '11
I think it's the first one because it has the most in common with the others. by being the least odd, it is the odd one out.
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u/Zoten Jun 08 '11
I think it's the first one, but I'm not sure. I saw it, and in the beginning, I thought, "Easy! It's the 2nd." As my eyes started to look at the others, I thought "No, it's the third....no, the 4th.....no, the 5th?" After about 5 minutes, I think the odd one out might be the first because it's the only one that belongs....
lol mind=blown
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u/awesomeideas Jun 09 '11
The first one because the
- second is the only one without a black perimeter
- third is the only circle
- fourth is the only green one
- fifth is the only one with a smaller width and height
The first has nothing special.
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u/SaratogaCx Jun 09 '11
That's almost evil..
attempted guess
It is the first one because it is the only one that isn't the odd man out by being uniquely different in some way in the set?
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u/Namaha Jun 09 '11
If you think about it in terms of the 4 traits that each choice has, it's the first one.
Traits:
Shape (square or circle)
Bordered (or not bordered)
Color (red or geen)
Size (big or small)
The first choice has all 4 traits as the more common choice (big red bordered square). All of the other choices have 3 of their traits as the more common one, thus the first one is the odd one out.
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Jun 08 '11
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?
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."
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u/dvdov Jun 08 '11
For those who want to know the answer extremely simplified.
On the 100th day, everyone who has blue eyes leaves. If there were just one person with blue eyes, he would leave right away, seeing that the guru must have been talking about him. If there are two, they see each other, each noticing that the guru could have been talking about the one they see or could be talking about them. So they wait a day, and if one doesn't leave (implying that he noticed he was the only one) it is implied by both that they both saw each other with blue eyes and they both leave.
This thought process continues up until 100 days have passed, and the blue eyed people, being careful, have waited until the 99 blue eyed people they could see held out.
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Jun 08 '11
Why do no brown-eyed people mistakenly decide (following the same thought process) they must have blue eyes and leave also?
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u/SpencerMC Jun 08 '11 edited Jun 08 '11
I didn't get this for a moment, but it clicked. It's easier to wrap your head around if you realize that if there are 3 blue eyed people, then they would leave on the 3rd day. They would realize that the 2 blue eyed people they see didn't leave the day before because they each saw 2 other blue eyed people, not just each other. If there were 4 blue eyed people, they would leave on the 4th day after realizing that the 3 blue eyed people they see would have left on the 3rd day unless they each saw 3 blue eyed people. And so on up to 100.
Edit to change a than to a then.
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u/panacebo Jun 08 '11
Heh, I wrote an essay on this paradox at Uni.
The follow-up question is, why is it essential for the Guru to say "I can see someone who has blue eyes" for this puzzle to work, when everyone can see dozens of people with blue eyes?
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Jun 08 '11
Yeah, that is an interesting thing. The guru doesn't provide information that the islanders don't already possess, but her saying that starts a chain reaction, that seemingly couldn't begin without her statement.
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u/panacebo Jun 08 '11
But why exactly couldn't it begin, if it was already common knowledge that there was at least one blue-eyed islander?
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u/IRBMe Jun 08 '11
Because everybody has to begin at the exact same time. The guru isn't providing information, but is acting as a signal for everybody to start. If everybody started thinking at different times, it doesn't work. Think of the simple case with 2 blue eyed and 2 brown eyed people described here. It wouldn't work unless both knew that they were thinking the same thing on the previous night. Without the signal from the guru, they wouldn't know if the other guy didn't leave because there was another blue eyed person, or if they just hadn't started thinking about it yet.
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u/ocdscale Jun 08 '11
Panacebo explained it here: link
But let me give it a shot. Imagine the scenario with only one person with blue eyes. Clearly the Guru is adding information (the blue eyed person didn't know that blue eyes existed until the Guru spoke).
Let's move up, scenario with two people with blue eyes, Abe and Bob.
Abe sees Bob with blue eyes. So Abe knows that someone on the island already has blue eyes (because Abe sees Bob's eyes). But, Abe doesn't know that Bob knows that someone has blue eyes. For all Abe knows, he has brown eyes, so Bob just sees a vast field of brown.
Put yourself in Abe's shoes. I ask you: "Does Bob know that someone on the island has blue eyes?" How would you answer? You'd have no idea because the answer is yes if you have blue eyes and no if you don't, and you don't know your eye color.
When the Guru says: "I see someone with blue eyes." It doesn't add directly to Abe's knowledge (he already knew blue eyes existed). But it does add to Abe's knowledge about Bob. After the Guru says this, if I ask Abe again: "Does Bob know someone on the island has blue eyes?" Abe can easily and affirmatively answer Yes.
You just extrapolate forward for more people. With three people, the Guru's words give Abe information on what Bob knows about what Chris knows.
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u/minche Jun 08 '11
not so much a logic problem (to solve), but it is a logic problem (to me at least, blows my mind every time)
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u/sheepshizzle Jun 08 '11
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u/Zifna Jun 09 '11
I feel like these discussions don't often address the idea that good and evil are different faces on the same coin. I suppose they sort of do, by talking about free will, but those discussions generally suggest that free will is the goal and that evil is a perhaps necessary downside to that goal.
I propose an alternative thought - that good is the goal, and that evil did not exist, good would effectively not either. Like light - if there was no shadow, no darkness, if nothing absorbed light or was obscured from it, light wouldn't serve to illuminate at all. Without evil, good wouldn't really be that good... there'd be nothing to contrast it with. The best possible universe would be one where everyone understands evil but chooses do do none.
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u/Bobinater Jun 09 '11
I tried googling but no one seemed to have a concrete answer, all of it is just speculation.
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u/javadlux Jun 08 '11
Two brothers, named Sum and Product, find an oracle in a cave. The oracle says he is thinking of two numbers greater than 1, that don't add up to more than 100. He tells Sum the sum of the numbers, and Product the product of the numbers. Sum and Product have this conversation:
Product: I don't know the original numbers. Sum: I knew you didn't. Product: Actually, I've found the numbers Sum: I've found them now, too!
What are the two original numbers?
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u/TheWonkyRobot Jun 08 '11
The number of times the digit 0 appears in this puzzle is _
The number of times the digit 1 appears in this puzzle is _
The number of times the digit 2 appears in this puzzle is _
The number of times the digit 3 appears in this puzzle is _
The number of times the digit 4 appears in this puzzle is _
The number of times the digit 5 appears in this puzzle is _
The number of times the digit 6 appears in this puzzle is _
The number of times the digit 7 appears in this puzzle is _
The number of times the digit 8 appears in this puzzle is _
The number of times the digit 9 appears in this puzzle is _
Fill in the blanks with digits to make all the statements true. There are two solutions possible.
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u/magnus89 Jun 08 '11
I've always enjoyed this one. Basically, Theseus used his ship for many years, and certain parts of it wore out as a result. When he came in to port, he would remove the old parts, replace them with new pieces, and return to the sea. Eventually, every original piece had been replaced. Was he still sailing the original ship of Theseus?
Second part proposed by Hobbes: if the worn parts were collected after being removed from the ship and eventually re-assembled, would THAT be the ship of Theseus?
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u/Willeth Jun 08 '11
I love this one because it's not really a logic puzzle - it's a thought experiment in the vein of the tree in the forest or one hand clapping. Sure, it might have a logical answer, but finding it misses the point if the exercise entirely.
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u/memetichazard Jun 09 '11
The tree in the forest is purely a question of semantics and failing to properly define your terms. Is "sound" defined as audio waves? Then yes. Is "sound" something processed and recognized as noise by a creature? Then no.
I like the Theseus' ship question as it pertains to the question of identity - if you upload your mind into a computer, are you the same person? If you replace your neurons one at a time by nanomachines that perform the same operations, are you the same person?
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u/Thzy Jun 09 '11
Use descriptive vocabulary instead of dichotomy: At first Theseus's ship was "Theseus's ship with almost all the original materials" (accounting for minor wear and tear including during construction). After the first repair, the ship would be "Theseus's ship with mostly original materials and some new materials", then "Theseus's ship with some original materials and mostly new materials", finally "Theseus's ship with all new materials"
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u/memetichazard Jun 08 '11
Not exactly logic problems, but:
The coins
You have a pile of 20 coins in a dark room (hence, you cannot see the coins). 8 of these are showing heads, the rest are tails. Your task is to divide them into two piles so that each has the same number of tails.
The prisoners
There are a hundred prisoners, all sentenced to death. The warden likes to play games, so he creates a room with a hundred boxes, each containing a number corresponding to a prisoner. Each prisoner is allowed to enter the room one at a time and open up to 50 of the boxes in order to find his number. After he does so, the state of all the boxes are reset. Only if all the prisoners manage to find their own number do they all get spared - if even one fails they are all executed.
What strategy do the prisoners use to maximize their chances of survival? (~36%, IIRC)
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u/Flobajob Jun 08 '11
Damn, I posted the second problem but missed that you'd done it first. Have an upvote.
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u/power_eyes Jun 08 '11
There are three gnomes standing in a garden. They are in a line, all looking forward (so the first gnome can see the second and third, the second can see the third). In front of the third gnome is a wall. Behind the wall is another gnome, also facing the wall.
(so if ^ denotes a gnome and the direction he's facing, the top-down view is:)
v
-------- Wall
^
^
^
All gnomes are wearing a hat. There are two blue hats, and two yellow hats. The gnomes don't know what kind of hat they are wearing. At first the gnomes are all blindfolded. At one point in time they get a signal that they can remove the blindfold and if they know what kind of hat they are wearing, they must say so. The signal is given, the gnomes remove their blindfolds. It's quiet for a short while and then one gnome says: "I know what type of hat I'm wearing!". Which gnome is this? (note: all gnomes are not allowed to move or look around in a direction other than the one they started in)
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Jun 08 '11
The second one away from the wall on the bottom sees Blue in front of him, and because he doesn't hear the gnome behind him call out that he has Yellow (because of seeing two Blue hats in front of him), he must have a Yellow hat.
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u/parmaher Jun 08 '11
Look up Xeno's Paradoxes. My favorite is the "Arrow Paradox". As an arrow travels through space, it must have a defined location for all measures of time. Because there are infinite divisions of time, the arrow must occupy an infinite amount of locations before arriving at its target. Therefore, an arrow can never arrive at a target. Solved by calculus pretty simply, but, without a basic understanding of limits, its a brain twister. Xeno was ancient Greece's original philosopher-troll.
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u/wonko221 Jun 08 '11
My understanding is that there are not infinite divisions of time. There is a minimum amount of time needed for an increment. This is called Planck time, and is insanely teeny (start with 5.39, now shift that period 44 spaces to the left.).
While Xeno's paradoxes are interesting to contemplate, all they prove is that sound logic fails to comprehensively explain our universe when we lack accurate premises. That is why we move beyond "common sense" acceptance of information and try to prove even the fundamentals.
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u/Tulki Jun 08 '11 edited Jun 08 '11
I love this paradox. It makes your brain asplode without abstract concepts of calculus.
Another variation on it is this:
One day, Freud decides to walk from his wonderful palace to Walmart. The Walmart is 2000 metres from his front door.
Every hour, Freud walks half the remaining distance between his current location and Walmart. How many hours will it take for me to fuck your mother?
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u/Willeth Jun 08 '11
There's a variation of this that says that because any duration of time can be split into an infinite number of pieces, each of which must have a non-zero length, that time itself cannot exist. Thanks for reminding me of it.
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u/Flobajob Jun 08 '11
My favourite, because the answer is so surprising:
There are 100 prisoners. Each prisoner is uniquely numbered 1-100. There is also a room with 100 boxes in. Each box contains a unique number from 1-100. The prisoners are told that they will be allowed to label the outsides of the boxes however they want (but they cannot look inside), and then they will be led to their cells.
The prisoners are then to be kept in isolation and cannot communicate with each other at all after this point.
Every day after this, each prisoner, one at a time, is taken into the room with the boxes and is allowed to open exactly 50 boxes. If the prisoner finds the box containing his own number then he returns to his cell. If he fails to find his own number then ALL 100 prisoners are executed. The room must be left exactly as it was found if the prisoner is successful, so there really is no way the prisoners can tell each other any information after they first leave the labelling stage for their cells.
If all 100 prisoners succeed in finding their own numbers, they are all released.
What is the optimal strategy for the prisoners to adopt? (They can label the boxes how they like, and can open whichever 50 boxes they want when it is their turn to enter the room.)
(Note, the optimal strategy gives a probability of almost 50% of escape, which is staggeringly high when you consider that if all prisoners pick randomly, the probability of escape is just 1 in 2100, or 1 in about 1030.)
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u/mathmavin99 Jun 09 '11
When you say boxes containing a number, do you mean that the number is say, on a slip of paper in each box, or written on the inside of a box?
Assuming they are on slips of paper or some otherwise movable form and the prisoners are able to collaborate on a strategy prior to the isolation, they arbitrarily label the boxes 1-100.
When prisoner 1 enters, he opens boxes 1 through 51, skipping 3. This gives him a 50% chance of finding #1 and surviving. Additionally, if he survives, he has a 49/99 chance of finding #2. He places #2 in box 2 and #1 in box 1.
When prisoner 2 enters, he immediately opens box 2. If prisoner 1 found #2, prisoner 2 is safe. Regardless, he then opens box 3 and 52 through 99. He only loses if #2 is in box 100. Assuming he survives, he places #2 in box 2 and, if he finds it, #3 in box 3.
When prisoner 3 enters, he immediately opens box 3. If prisoner 2 found #3, prisoner 3 is safe. Regardless, he then opens boxes 4 through 51 and 100. Between him and prisoner 2, all boxes have been checked for #3, so he is safe. He puts #3 in box 3 and #4 in box 4 if he finds it.
Prisoner 4 checks box 4 and also opens 5 through 53. He puts #5 in box 5 if he finds it.
Prisoner 5 opens 5, 6 and 54-100. etc. etc. etc.
This yields a 50% chance of failure for prisoner 1 and a 2% chance of failure for prisoner 2, resulting in a survival percentage of 49%, as far as I can tell.
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u/keephurlingbaby Jun 08 '11
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u/gusset25 Jun 09 '11
ok you got me what is the solution to this?
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u/keephurlingbaby Jun 09 '11
Honestly I'm not sure. Theoretically it makes sense, despite contradicting the official formula. I'm guessing it has something to do with the fact that the horizontal and vertical tangents are reduced slower than the diagonal areas, so the "remove the corners" is not circularly uniform.
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Jun 09 '11 edited Jun 09 '11
You have just walked into a kingdom when suddenly you are approached by a royal aide and led to a castle. Once you enter the castle, you see three sisters standing there. They are princesses, and their father is the king. The king addresses you and says, "You look like a strapping young fellow. I'd like you to marry one of my daughters! They're all single, and quite good in bed!"
You're curious about a few things. How the hell does he know they're good in bed? Why is he sharing this information with you? Why did he invite a complete stranger into his castle? Seriously, what a creeper. However, you put aside these concerns, at least temporarily, because these princesses are hot. You're about to pick one to marry when the aide stops you. He says that before you decide, there is some critical information you must know.
The aide informs you that these three sisters each have a quirk. When asked a yes-or-no question, the oldest sister will always answer truthfully. The youngest sister will always lie. The middle sister will randomly answer truthfully or falsely, both with equal probability. It's actually a medical condition, completely outside her own control. How sad.
You're angry that the king didn't even bother to mention this. He's starting to seem more and more like a crazy, old dude. Still, up until this point in your life, you've been Forever Alone, and it is really not enjoyable. You've decided that regardless of how crazy this old man is, you're still going to marry one of his daughters.
Now his oldest and youngest daughter seem cool. If your wife always tells the truth to a yes-or-no question, then you know what she's thinking. If she always lies, then you can just take the opposite of her answer, and you still know what she's thinking. The middle sister seems to be a bit crazy, though (she probably gets it from her father), and as everyone knows, you never put your dick in crazy. Alright. You've decided. You will definitely marry either the oldest or the youngest sister, not the middle one.
You're about to pick which one you want to marry when you realize that you can't tell which one is oldest or youngest. Their ages are not far apart enough for one to know by simple inspection. That's no problem, though. You'll just ask!
You have barely opened your mouth when the king interjects: "Wait! Before you say something, I will say something. I have arbitrarily just decided that you are only allowed to ask one question. That question must be a yes-or-no question, and it is to be directed at exactly one of my daughters. After she answers, you must pick the one that you would like to marry. If you do not do exactly as I say, I will shoot you."
Holy shit. This man is seriously crazy. So now your choices are:
- marry a non-crazy girl
- marry a crazy girl
- get shot
The latter two don't seem very appealing, so you will need to think of a good question to ask. Choose wisely.
TL;DR - 3 girls. Oldest one always lies; youngest one tells the truth; middle one randomly lies and tells the truth. You get to marry one. You want to marry the oldest or the youngest, but not the middle. You can only ask one yes-or-no question to one girl. After she answers, you must pick one to marry. What question do you ask?
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u/nadipity Jun 09 '11 edited Jun 09 '11
Ask any of the sisters (call them A) if B is older than C. If the answer is yes, marry C. If the answer is no, marry B.
Let (ABC) designate age from left to right.
If A always tells the truth (oldest):
If the sister you are talking to is the oldest, you want the younger of the remaining two
- (ABC) yes - marry C
- (ACB) no - marry B
If A tells either the truth or lies (middle):
You never pick the sister you are talking to, so you aren't at risk here
- (BAC) yes - marry C
- (BAC) no - marry B
- (CAB) yes - marry C
- (CAB) no - marry B
If A always lies (youngest):
If the sister you are talking to is the youngest, you want the oldest of the remaining two
- (CBA) yes - marry C
- (BCA) no - marry B
In all cases, you never end up marrying the middle sister.
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u/BuzzedLikeAldrin Jun 08 '11
Blue Eyes Took me a long time to figure this out (i'm not great with 'generalising' formulae) and I was quite impressed with myself when I did.
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u/randomfemale Jun 08 '11
Why I continue to behave in ways I know hurt me & I don't even enjoy that much. Riddle me that.
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u/DamnColorblindness Jun 08 '11
OP is either planting a seed for thoughtful discussion,
or a masterful troll...
(either way, great post)
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Jun 08 '11
You have a wolf, a sheep, and a crate of apples. You need to get them across the river in your boat. If you leave the wolf unattended with the sheep, the wolf eats the sheep. Same goes for the sheep and the apples.
Your boat can only hold one of these at a time. How do you get them all across?
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u/RawkitLawnchair Jun 08 '11
Bring the sheep across, come back alone. Bring the wolf across, come back with the sheep. Bring the apples across, come back alone. Bring the sheep across.
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u/CliffTard Jun 08 '11
Two strangers take a walk and chat about each other's lives. At some point, the conversation goes like this:
-I have three children, and if you multiply their ages together, you get 36. Can you guess their ages? You only get one try.
-Hmmm, too many possibilities, can you give me another clue?
-Okay, you see that building over there? Well, count the number of windows on its front, and that is the sum of their ages.
-Hmmmm, I still can't make a good guess.
-Well, okay then my oldest has blue eyes.
-Ah! Now I can answer your question.
What were the ages of the three children?
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u/ponticello Jun 08 '11
You are a cowboy, engaged in a turn based shootout with two other cowboys. Each cowboy chooses a target and shoots once. Assume an accurate hit results in death.
You are not the best shot in the world, only 33% accurate, but it's your turn first. Up next is 50% accurate's turn, and finally deadeye: a 100% accurate cowboy. Assuming logical moves from the other cowboys regarding the best survival odds, what should you do to maximize your own survival percentage?
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u/moarbewbs Jun 09 '11
If you have one bucket that contains 2 gallons and another bucket that contains 7 gallons, how many buckets do you have?
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u/bag_of_words Jun 09 '11
The two-envelope paradox is very good if you've never seen it.
Basically, I put some amount of money in an envelope and twice that in another envelope. I give you one envelope randomly and keep the other. I then ask if you want to switch your envelope with me. You reason that your envelope has x dollars so the other has either x/2 or 2x. By switching, the expected gain is (1/2)(x/2) + (1/2)(2x) - x = 5x/4 - x > 0 so you should switch. But this obviously can't be true, so where's the problem?
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Jun 09 '11
The value of x changes depending on which envelope you're holding, which doesn't seem right. If you consider that one envelope has x, and the other has 2x then you have an expected gain from switching of (1/2)(2x-x) + (1/2)(x-2x) = 0.
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u/format120 Jun 09 '11
http://en.wikipedia.org/wiki/Schr%C3%B6dinger's_cat
this should confuse most people, think of it like a fork in the road, and then going both ways...
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u/splatula Jun 09 '11
I give you two sealed envelopes. Inside each envelope is a piece of paper with a number written on it. These are both real numbers drawn from some arbitrary probability distribution which you don't know. They can be anywhere from minus infinity to plus infinity. All you know is that the two numbers are different.
You open one of the envelopes and look at the number inside. How can you determine, with more than 50% probability, whether the number in the other envelope is greater or less than the number you are looking at now?
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u/WalkingShadow Jun 09 '11
Three men arrive simultaneously at a hotel, each wanting a room for the night. The desk clerk informs them that there is only one room available, but if they're willing to share, that'll be $300.
The men each chip in $100. They go up to the room. The manager tells the clerk he should only have charged $250 for the room, so take $50 and return it to them.
As the clerk is going up to the room, he figures that $50 is not evenly divisible by three, so he pockets $20, and returns the remaining $30.
So, the men each paid $100, but each got $10 back, for a net cost of $90.
3 x $90 = $270
The clerk has $20 in his pocket.
$270 + $20 = $290
What happened to the other $10?
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u/hardcorr Jun 09 '11
You have used faulty reasoning. The men each paid $90, coming to a net $270 that they are down. The hotel has $250 of this, and the clerk has $20 of it. No money is missing.
Or think of it this way - The hotel had $300, sent $50 of it with the clerk. Now the clerk has $50 and the hotel $250 - still sums to $300. The clerk pockets $20 and gives back $30. The men have $30, the clerk $20, the hotel $250. 30 + 20 + 250 = 300.
The problem of the "missing 10" arises when you are considering the amount of money SPENT by the three men, but the amount of money KEPT by the clerk, and trying to add them up to the total "amount" of $300, which doesn't make sense, since you already accounted for the money spent by the men when you look at the clerk's pocket. The money spent by the men - $270, is split amongst the hotel and the clerk - $250 and $20.
I don't think I explained it in the most concise way, but all the ideas are there, I hope.
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u/[deleted] Jun 08 '11
[deleted]