Atlantic South Logic Problems Thread!

[ZeldaRox!]

***** it does matter, given a 50/50 chance of each hat theres only a small chance that there will be both 5 black and 5 red, which is vital for the solution you gave

actually you are wrong on 2 accounts;

if it is 50/50 and the sample size is large enough so as to avoid the problem of changing proabilities, the most likely distribution would be 50/50. Have you ever had a statistics class? This is 1st week normal curve stuff.

The second reason you are wrong is because the proportion doesn't matter.

Guy 10 sees only 9 hats....... can you give me a way in which 9 hats can be divided so that there are both even or odd numbers of hats given? NO. Therefore, we know that one color will have odd representation. The other, even. That's all the question rests on. The distribution could be 8 to 1. It wouldn't matter lol. Try actually working the problem out on paper and you will see why.

oh....and sorry if I came off as kind of an azzhole. I only asked if you had ever had a statistics class because this is very simple normal curve stuff.

 

[DJRome]

lol phoot, i left it like that just to show my work

and bbq, lol, i guess, but it's pure math computation.

 

[Diatenshi]

actually this guy was right

not i

 

[Dark Sonic]

@Diatenshi-Zelda Rox's method works perfectly, it's just a little complicated. Let's walk through a full example so that everyone gets it. (and no, the distribution of hats does not matter, you can still guarantee that at least 9 people survive.).

The hats in order are.

Red, Black, Red, Red, Red, Black, Red, Black, Red, Red. I've purposefully put more red hats than black hats to show that it doesn't have to be an even distribution.


Okay, so this will be our code. If the first person says Red, it will mean that he sees an even number of red hats. If he says black, it will mean he sees an odd number of red hats (it's very important to remember that black corresponds to an amount of Red hats, not black hats). Also none of the people's responses after the first are coded. They are simply stating the color of the hat on their head (which can be derived as I will show)

First person says red because he sees 6 red hats (he got lucky that his hat is also red. Yay 50/50 chances!)

Second person, knowing that the previous person saw an even number of red hats, knows that his hat can't be red (as he also sees an even number of red hats, and his hat being red would've made the previous person see an odd number of red hats). He says black.

Third person. The second person saying black means that the number of red hats hasn't changed (second person still saw an even number of red hats). As he only sees 5 red hats (odd), he knows that his hat must be red for the previous person to see an even number. He says red.

Fourth person. The third person saw an odd number of red hats. How do I know this you ask? Because if you were keeping track the second person saw an even number of red hats, and the third person said his hat was red (implying that his view disagreed with the person behind him). The fourth person sees 4 hats (even) which is different from the person behind him, so he says red.

By now you can obviously see the pattern (unless I made my explanation too convoluted to understand.)
Basically, with my particular code you would say Red every time there's a discrepancy between the set of red hats the person behind you saw and the number of red hats you currently see, as it means that there is one red hat missing from the set (the one you can't see that the other person did). If there's no discrepancy, you say black (as the set of red is still in tact, but there's still a hat on your head).

This method works as long as you make sure that the first person says his color based on the amount of one particular color (red and black both correspond to an amount of red, or they both correspond to an amount of black, but never some combination). Also, past the first person, the responses are no longer in "code."


Hope that helped everyone put this riddle to rest.

 

[DJRome]

basically, matt, the answer provides a solution no matter if all 10 hats are red or if all 10 are black. or if there are 9/1, 1/9, 2/8, 8/2, etc.

 

[Dark Sonic]

When I first saw the solution, my mind was blown lol

 

[DJRome]

i used playing cards to figure out the solution and you have no idea my relief when i got it

 

[Todd Bonney]

I want moar

 

[DJRome]

i wish i had more

i'll post more soon. i ordered some logic problem books

 

[Moses.]

I've got a cool question. It's not entirely logic, although it is involved. There are multiple answers and it just depends on how you would justify them.

You are driving your car on a wild, stormy night. You pass by a bus station, and you see three people waiting for the bus: an old lady who looks as if she is about to die, a doctor who had once saved your life and the man/woman of your dreams. There seems to be no one else on the road, including the bus they are waiting for. The doctor has no tools or instruments to work with. You decide to help; however, you can only take one passenger in your car. Explain which one you would choose.

 

[ZeldaRox!]

I've got a cool question. It's not entirely logic, although it is involved. There are multiple answers and it just depends on how you would justify them.

You are driving your car on a wild, stormy night. You pass by a bus station, and you see three people waiting for the bus: an old lady who looks as if she is about to die, a doctor who had once saved your life and the man/woman of your dreams. There seems to be no one else on the road, including the bus they are waiting for. The doctor has no tools or instruments to work with. You decide to help; however, you can only take one passenger in your car. Explain which one you would choose.

Spoiler

well its really more of an ethical/moral dilemma. I've heard this one before and apparently its been used on job interviews. Its often been said that the best answer is to give the doctor, who is usually substituted for 'friend', the keys and let him drive the old woman while you sit and wait for the bus with the woman of your dreams.

 

[exarch]

Well actually, I doubt you were any good at math then. Cause if 10 isn't the answer, and you know it has to be more than 5, then you have a 1/4 chance of getting it right. And picking something surprisingly high tends to work well with these sorts of problems. But its clear that you don't know what you're doing when you answer logic problems. These problems ask you to give a runthrough of your problem solving strategy. Which you of course did not. You had no system. You just said its not this or this so it must be this. Too sad. Too sad. People like you are developing the world. I see bad things coming in our future. So, I'm guessing if you were put in this situation, you would have no clue how to do it, am I right?

Condescending douche
I assume you were a math major too then?

so if you really were a math major...and just do the ****ing problem like normal people

I assume you weren't a math major then?

A customer at a hardware store is told that the items sell for $1 each. The customer says, "I'll take 36." The total cost is $2. What did the customer buy?

 

[~Twitch~]

A customer at a hardware store is told that the items sell for $1 each. The customer says, "I'll take 36." The total cost is $2. What did the customer buy?

well he obviously didn't buy the items...but hmmmm...

 

[X1-12]

Condescending douche
I assume you were a math major too then?

I assume you weren't a math major then?

A customer at a hardware store is told that the items sell for $1 each. The customer says, "I'll take 36." The total cost is $2. What did the customer buy?

Spoiler

numbers for his front door or something, like a number 3 sign and a number 6 sign. Can't really word it any better

Its okay that I'm not actually from atlantic south right?

 

[_Keno_]

I have one, but I havent look through the thread to see if anyone else has posted it.

So 100 geniuses are stuck on an island. Each one has either a black or red dot on his forehead (and he does not know the color if his own dot, and nor is he allowed to see it in anyway). There is at least one dot of each color. The geniuses are told that each day, a boat arrives and leaves five minutes later. For the boat to make a successful voyage, every single person aboard must have either a red dot (only red) or black (only black) on his forehead. If a person with the wrong color dot boards the ship, or if a person with the correct color dot is not on the ship, it will sink on its voyage. They are also told that the red dots MUST make their voyage first. Communication of any kind is not allowed between the geniuses, but they can still see everything else going on. How do all the geniuses survive, and how many days does it take them to leave the island (if half are red dots)?

 

[Todd Bonney]

Spoiler

It takes X+2 days.

If there is only one red, then after the first day, the guy with the red dot sees that no one got on the boat, so he knows he must be the only one with the red dot. So on the 2nd day he gets on the boat and goes by himself. On the 3rd day, the boat comes back, meaning the voyage was successful and all the red dots are gone. So all the people with black dots get on and go.

And so on. Induction and ****. QED.

 

[Saki-]

Ahhh there was one that stumped me about two islands, a bear and some people......O: is there an answer to the "What came first, the chicken or the egg?" question?

 

[Todd Bonney]

Let's not turn this into that.

 

[Toneh]

I can never guess but i like readin the answers.

More questions gogogo!

 

[ZeldaRox!]

Spoiler

It takes X+2 days.

If there is only one red, then after the first day, the guy with the red dot sees that no one got on the boat, so he knows he must be the only one with the red dot. So on the 2nd day he gets on the boat and goes by himself. On the 3rd day, the boat comes back, meaning the voyage was successful and all the red dots are gone. So all the people with black dots get on and go.

And so on. Induction and ****. QED.

This works only if there is 1 red dot. What if there are 2 or 3? No one will ever board. If I have a red dot and I see 2 others do then I will never board until I see the other 2 do. And so is the case with the other 2.

 

[ZeldaRox!]

Condescending douche
I assume you were a math major too then?

I assume you weren't a math major then?

lol

edit: I once was

 

[_Keno_]

Can someone try to give a better explanation that EWB? He doesnt go into any details.
Also, his answer is off (but it is the right idea)

If anyone wants, I can give a full explanation, but I'd rather people get the full answer first.

 

[Todd Bonney]

I think my answer works, though admittedly I got lazy. I'm not in school, I don't want to do proofs.

But I'll start one.

Spoiler

I was wrong about the case where there is only one red dot. If there are n=1 red dots, the red dot knows it immediately. He looks around and sees 99 black dots. Given that there is at least one red dot, he comes to the conclusion that his dot is red. On day 1 he gets on the boat and goes. On day two, the boat returns indicating a successful voyage. Everyone else gets on and goes.

So change my original guess to n+1 instead of n+2

In the cases you mentioned ZRox,

Spoiler

n=2
The two people with red dots see only n=1. If that were the case, the red dot they see would have left on day 1, in accordance with logic above. After day 1 comes and goes, they each realize there are at least 2 . Since they look out and see only 1 red dot, each concludes that his dot is red. On day 2 all red dots board, on day 3 all black dots board.

n=3
These 3 people each think n=2. As day 2 comes and goes, they realize their dot is red. on day 3 the reds go.


Proof by "And so on."





I also have this notion that it could be only 3 or 4 days, or n/2 + 1 or 2. I'm starting to get confused.

 

[ZeldaRox!]

EWB I still think you are wrong. If I have a black dot on my head what's to stop me from making the same conclusion as the red dotted person? This is the point I was getting at. Your logic works for both the people with red dots and the people with black dots.

No one would ever leave the island....I'm confused by how you say in accordance with the logic above. Red dot person is none the wiser than black dot person. If all I see is 1 red dot, how is it that I can get him to leave?

 

[_Keno_]

*Genius problem spoilers*

Spoiler

EWB I still think you are wrong. If I have a black dot on my head what's to stop me from making the same conclusion as the red dotted person? This is the point I was getting at. Your logic works for both the people with red dots and the people with black dots.

No one would ever leave the island....I'm confused by how you say in accordance with the logic above. Red dot person is none the wiser than black dot person. If all I see is 1 red dot, how is it that I can get him to leave?

Spoiler

Actually it does work, the person with a red dot only see x-1 red dotted heads, while the person with the black dot sees all x red dotted heads; therefore they would come to different conclusions depending on the number of days that have passed. EWB is now correct.

Please spoiler discussion of answers

 

[ZeldaRox!]

*Genius problem spoilers*

Spoiler

Actually it does work, the person with a red dot only see x-1 red dotted heads, while the person with the black dot sees all x red dotted heads; therefore they would come to different conclusions depending on the number of days that have passed. EWB is now correct.

Please spoiler discussion of answers

Spoiler

the only way your logic works is if the number of red dots is known and the wording of the question *strongly* implies that this is unknown. The person with the black dot does not know he sees all the red dots no more than does the person with the red dot. If you don't know your color you gain no certainty as to whether you see all of the other colors.

To put it simple: if I have a red dot (and I don't know this) and I see other red dots I can draw no conclusions what to do; I just have to hope the other red dots leave the island so I can get better certainty as to what I have. The other red dots are thinking the same way. And the black dots are thinking the same way too.

I'm resolved to say that the puzzle is either unsolvable or is poorly worded. Without knowing the number of red dots that are on the island, everyone is at a stalemate. No one can make the first move.

 

[ZeldaRox!]

cheap peach I would greatly appreciate it if you would tell me how the first person figures out he has a red dot. Cause as far as I can tell, he can't as long as he doesn't know the number of red and black dots. If there are 5 red dots on the island (95 black dots), and no one knows the ratio, person 1 sees 4 dots. As do persons 2, 3, 4, and 5. Persons 6-100 see 5 dots. If I do not know the number of dots, then I can draw no conclusions.

And if I do know the ratio of dots, this is a stupid question and hardly a logic probelm.

 

[_Keno_]

Spoiler

the only way your logic works is if the number of red dots is known and the wording of the question *strongly* implies that this is unknown. The person with the black dot does not know he sees all the red dots no more than does the person with the red dot. If you don't know your color you gain no certainty as to whether you see all of the other colors.

To put it simple: if I have a red dot (and I don't know this) and I see other red dots I can draw no conclusions what to do; I just have to hope the other red dots leave the island so I can get better certainty as to what I have. The other red dots are thinking the same way. And the black dots are thinking the same way too.

I'm resolved to say that the puzzle is either unsolvable or is poorly worded. Without knowing the number of red dots that are on the island, everyone is at a stalemate. No one can make the first move.

There is no first move to be made, they all figure out that they have red dots on the same day. I'll do a few examples.

Spoiler

One person has red dot: He sees nobody has red dot, and figures it must be him, and so he leaves the first day.

Two people have a red dot: I see exactly one person with a red dot, I also see that the person does not leave on this first day, which means that the other person sees a person with a red dot (if he didnt, he would have followed scenario 1.) I see no other red dots, therefore I must have the other red dot.

Three people have a red dot: I see exactly two people with a red dot, I wait two days and see if they follow the scenario two. If they do not, then there must be another person with a red dot (because they must have also waited two days to see if two people follow the second scenario). I can only see two people with reds, so I must be the third. We all three leave the next day.

Do you see where I'm going with this? It continues like this forever.

 

[Todd Bonney]

No one would ever leave the island....I'm confused by how you say in accordance with the logic above. Red dot person is none the wiser than black dot person. If all I see is 1 red dot, how is it that I can get him to leave?

Spoiler

So we've agreed that my solution works for n=1.

If n=2, then 98 people see 2 red dots, 2 people see 1 red dot.

Let's call the 2 people A and B, and I am A. You're right, I still don't know if my dot is red or black. I just know that n=1 or n=2.

If n=1 (the base case, which we've shown works), and B is that one, then B would have left on day 1. So then n=2 and I must be one of the two. And remember, B has the exact same logic, so we both know we are the only ones with red dots, and we board on day 2.

The important thing is that there is at least one of each color.

 

[ZeldaRox!]

I really only see this logic working if I know the number of people with red dots. And what exactly keeps the people with the black dots from making the same conclusions.

And I'm still confused with n=3. At this point, it seems like your not making it clear why waiting 2 days guarantees I must have a red dot.

 

[ZeldaRox!]

okay nvm. I see it now. but I'm going to have to think about it for the extreme cases (n>50)

 

[_Keno_]

Well, here are two more riddles (fairly common ones actually).

1. You are stranded on the top of a mountain with one 150 meter piece of rope, a knife for cutting the rope, and amazing knot tying skills (anything you can think of that is possible you can do). The mountain itself has two levels: from ground to the middle is a 100 meter cliff, from middle to the top is another 100 meter cliff. Explain how you can make it to the ground (there may be more than one answer). You cannot jump or fall.

2. Your wife dies, and the devil sees your worry and comes to make a deal with you. If you beat him in a game of his own creation, then he will give you your wife, but if you lose, he will take your soul too. The game goes as follows: You and the Devil place pennies on the top of a gamecube, the last one to place a penny before the top surface runs out of room wins. The devil flips a coin, and you win the toss. Should you play the game? If so, do you wish to go first or second? What is your strategy for beating him?

 

[DJRome]

lol @ 2. what else do we know about the game? doesn't seem to be very clear

 

[_Keno_]

lol @ 2. what else do we know about the game? doesn't seem to be very clear

I changed the surface, it should be more clear. I figured people thought i meant that the pennies had to be inside the squares or something, which they dont.

 

[DJRome]

i have an idea on how to play this game, but i can't really explain it well

 

[Todd Bonney]

1. So you can't detach the rope after the first level?

2. You can't stack the pennies?

 

[_Keno_]

yeah, I should be more specific, but I am recalling these from memory...

You cannot detach rope connected to the peak of the mountain
Pennies cannot overlap in any way.

 

[exarch]

2. I'm not sure whether this is more of a game theory question or a packing problem. If the former (which seems more likely as you changed the shape we were packing the pennies into.)

Spoiler

Go first; if there is a winning strategy in any game, it is held by the first player. This is due to the fact that if the winning strategy was held by the 2nd player, the first player merely makes a random move first, and then becomes the second player himself. I don't remember the strategy for this particular game (I think I've heard this one before,) but that's 2/3 of the way there.

Also how come the devil has power over my wife? She couldn't have gotten into that much trouble in the kitchen...

 

[_Keno_]

Simply going first will not assure you victory, and neither will going second.

 

[DJRome]

i think it depends on the ratio of the diameter to the length of the square surface and how the even and odd ratios of that (actually im thinking more of the square of that ratio) depend on either placing it in the center or placing it in an "optimized" way to force the last configuration