Atlantic South Logic Problems Thread!

[DJRome]

So, last night, my friends and I went to dinner and we had some logic problems that were really fun to think about. Some of these are easy and some are quite hard. The only rules for this are that you don't search the answer only (really, where is the fun in that) and that you put your guesses/answers in spoilers.

I will add other people's logic problems to the OP and give credit to the first person to solve and cross it out once we have solved it.

[And don't chime in if you know the answer already obviously]

Here's the ones we did to start off.

1. Quarters

3 vending machines. 1 vends only Coke, 1 vends only Pepsi, and one vends both. One night, someone switches the labels so that all 3 machines have the wrong labels on them. It costs 1 quarter to vend a soda. What is the least number of quarters you need to figure out which machine is which, and how would you figure it out?

Solved by EWB

Spoiler

One quarter.
Go to the machine Labeled "Both." Since this is wrong, it is either only Pepsi or only Coke. Buy one soda from there to determine which it is.

Suppose it's Pepsi. The two remaining machines labeled "Pepsi" and "Coke" are actually a Coke machine and a dual drink machine. Since we're given that all 3 machines are labeled wrongly, "Coke" is not the Coke machine. So "Coke" has both, and "Pepsi" has Coke.

2. Light bulb

A house has a light bulb in it. 3 switches on the outside can be switched on and off and only 1 turns the light bulb on. The other 2 are dummies. You can mess with the switches however you want before you enter the house, but once you enter the house, you cannot touch the switches. How do you figure out which switch turns on the light bulb? You cannot see into the house from the outside.

Solved by EWB

Spoiler

Flip switch A. Wait 30 minutes. Turn off switch A, flip switch B, and go in the house quickly.

If the light is on, switch B turns on the light.
If the light is off, feel the bulb. If it is hot, switch A turns on the light. If not, switch C turns on the light.

3. Cue balls

You have 8 cue balls that are all the same except that 1 is lighter. You can't tell from holding it that it is lighter, but you have a balance that you can use. A man bets you $50000 that he can balance twice the balls in any combination and tell you which ball it is. Do you take the bet? Explain.

Solved by Moses

Spoiler

Don't take the bet.

How to find which is lighter:

Take the eight balls, split into three groups: 2 with 3 balls, 1 with 2 balls.

Weigh each group of 3 against each other (1st weigh). If they don't balance, the light ball is in that group of three. Take that group of three, and weigh any two balls against one another (2nd weigh). If they balance, the odd ball out is lighter, if not, the scale will tell which is the light ball.

If the two groups of three balls do balance, then the light ball is one of the two left. Balance them against each other and you will find the lighter ball (2nd weigh).

4. Hats

10 people are stuck on an island. They are found by a cannibalistic tribe. The tribe decides to play a game with them for their survival. They give them a night to decide on a strategy for the game. The next day, a bag of infinite black and red hats with a 50/50 chance to pull either is used. Each person is lined up so that they can see only the people in front of them. Each person is to say the color of their hat and if they are right, they live and if not they die. The people in front of them do not know if the person saying the color of their hat dies or not, but they can hear what color they give.

Each person must say either black or red only. Another restriction is that you can have them say it in a way to mean something or whatever, just straight up red or black.

> > > > > > > > > >

What is the strategy to save the most people guaranteed and how many people can you save this way?

Solved by ZR

Spoiler

So, we have guy 10 who sees 9 hats. Therefore, one color will have an odd number of representations and another an even. So guy 10 codes one color to mean odd and the other to mean even (everyone else knows this code as well). So by letting guy 9 know that say there are an odd number of black hats, guy 9 will know whether he has a black hat or red hat (if he sees an even number of red hats, that means he has to have a black hat). Guy 8 will then know that there are an even number of both hats. So he has the color of the hat which he only sees an odd number amount. And so on.

So 9 are guaranteed to survive. And the 10th has a 50/50 draw.

 

[Moses.]

3. Cue Balls

Spoiler

Don't take the bet.

How to find which is lighter:

Take the eight balls, split into three groups: 2 with 3 balls, 1 with 2 balls.

Weigh each group of 3 against each other (1st weigh). If they don't balance, the light ball is in that group of three. Take that group of three, and weigh any two balls against one another (2nd weigh). If they balance, the odd ball out is lighter, if not, the scale will tell which is the light ball.

If the two groups of three balls do balance, then the light ball is one of the two left. Balance them against each other and you will find the lighter ball (2nd weigh).

 

[DJRome]

yep

 

[~Twitch~]

1. quarters

edit: oh wait, now i know.

Spoiler

it only takes one quarter

Spoiler

you do that little trick where you tie a string around it and you yank it out when the vending machine thinks you're paying for it.

Spoiler

that way, you can try the vends as much as you want.

 

[exarch]

1. Are all of the labels definitely wrong? Or were they rearranged randomly? (So the Pepsi machine could still have a Pepsi label.)
By "vends both" does that mean that it has one button which randomly dispenses either Pepsi or Coke? Or does it have 2 separate buttons which dispense different drinks?

Spoiler

If randomly assigned with 1 button, 3 with some luck
Select two and get a soda from each. Hopefully you get the same soda from both machines. With the third quarter pick one of the two and get another. Hopefully it gives you a different soda that it originally did. That one dispenses both, the others you know from the first round.

2. How many trials do you get?

4. Not sure how this one works either. You see the person's hat in front of you, are they chopping people down from the front or the back? (Do you guess your color before the person in front of you does?)
Couple of ways: Have everyone guess 1 color... or whatever color they want. This CAN save all 10 people (which is all the question asks.) On average though, only 5 people make it.

Spoiler

As far as guarenteed, depending on how it works, have person N*2-1 guess person N*2's hat color (assuming they can see it.) This on average saves 7.5 people, and with some nice luck can save all 10.

 

[~Twitch~]

2. How many trials do you get?

i've heard that riddle before and i know the answer. you only get to open the door one time. and no, you can't prop the door open and flip all the switches as much as you want.

 

[DJRome]

1. the labels are definitely wrong.

3 quarters is wrong.

2. you can do whatever you want as many times as you want outside. once you're in you can't leave and touch the switches.

4. to clarify, the person in the back (#1) starts talking first. it must be how many guaranteed

1 > 2 > 3 etc

edit: lol no twitch.

 

[Masky]

..........

 

[DJRome]

no, it is random

 

[Sneak8288]

the answer is

Spoiler

2 quaters

 

[Masky]

..........

 

[DJRome]

you don't win anything cause u didn't explain your answer

 

[Masky]

..........

 

[Todd Bonney]

Quarters

Spoiler

One quarter.
Go to the machine Labeled "Both." Since this is wrong, it is either only Pepsi or only Coke. Buy one soda from there to determine which it is.

Suppose it's Pepsi. The two remaining machines labeled "Pepsi" and "Coke" are actually a Coke machine and a dual drink machine. Since we're given that all 3 machines are labeled wrongly, "Coke" is not the Coke machine. So "Coke" has both, and "Pepsi" has Coke.

Light Bulb

Spoiler

Flip switch A. Wait 30 minutes. Turn off switch A, flip switch B, and go in the house quickly.

If the light is on, switch B turns on the light.
If the light is off, feel the bulb. If it is hot, switch A turns on the light. If not, switch C turns on the light.

Cue Ball
I've heard this one before.

Hats
I've heard this one or one similar, but can't quite remember. Who starts? The person in back or in front?
Edit: K, just saw your followup.

Spoiler

All right, here's what I'm thinking. I'm the person in the back, #10, and there's 9 people in front of me. I don't have anything else to go on. There is not going to be an even split. At best it'll be 5-4 in favor of one color. I'll choose whichever color is in the minority. Suppose it's red.

K so now I'm #9. #10 has just said "Red." I got 8 people in front of me. There are 2 scenarios to consider.

If it's a 4-4 split in front of me, then given that #10 said "Red" there must be more black hats in front of him. So I say "Black"

Otherwise, I choose whichever hat is in the minority, which will be the same as #10. So I say "Red."

So if you're an even number, choose whichever hat in front of you is the minority.
If you're an odd number, do the same. If there's a tie (including if there's no one in front of you), say the opposite of the previous guy.

Not interested in doing the math and shiz, but I'm guessing it's better than if everyone said the same thing. At least I hope so. So I'm going to say you can expect to save 7 people this way.

 

[ZeldaRox!]

Um Emma Watson's Boyfriend......you may want to read the scenario over again. It's an infinite bag of hats. The point of it being infinite is that there is no change in ratio as hats are removed.

but I will venture to give an answer:

by the conditions of the question, it is apparent that the captives will give there answer in reverse order (the person last in line first; the front person last).

So every even numbered person gives the hat color of the person in front of him. The odd number people say the color that the person behind them gave.

This guarantees 5 people are saved. And probability would say that between 2-3 more people would be saved.

So we are guaranteed 5 people are saved with a possibility of 10 being saved but most probably between 7 and 8 being saved.

 

[DJRome]

5 is not the answer. there are about 10 ways to guarantee saving 5 but that is not the answer

the hat one is pretty hard. it took me like 30 minutes of hardcore thinking to get it

 

[exarch]

9 is the answer, my explanation:
5 is not the answer, therefore there must be a way to communicate to the person in front of you while still guessing your correct hat color. Even though the rules of the game say you're not allowed to guess in a way to signal to the next person--If this was truly enforced, I'm relatively confident 5 is the best you could do. In that case you can only communicate 1 piece of information at a time. To do better than 5 there must be a way to communicate more than 1 piece of information at a time, which means everyone can communicate 2 pieces at once.

Since the first person to go has no one before him to tell him what is hat color is, he will never be guarenteed to survive. Therefore only 9 are guarenteed.

Perhaps pausing before guessing black or red signals to the person in front that their hat is black. That way they are not "saying it" in a way which means something. This may not be the actual solution that you're looking for, but I would be incredibly surprised if the answer isn't 9.

--
Another potential solution to this problem is they look in the ocean and see their reflections, since they're on an island. 10 win guarentee.

 

[DJRome]

no, you cannot look at reflections. the only information you get is that you can hear the people behind you saying their colors (but not whether they die) and the strategy you come up with beforehand.

and no, each person only receives 1 piece of information from the people behind them, that is the color they say.

and no, the "pausing" is not it. your rationale for why 10 cannot be guaranteed is true, but still nothing about how to guarantee 9 if that is your answer on the max guaranteed

 

[ZeldaRox!]

Oh............!!!! By golly I got this. This is somewhat similar to another riddle I heard. No wonder I was thinking odd and even.

Spoiler

Spoiler

So, we have guy 10 who sees 9 hats. Therefore, one color will have an odd number of representations and another an even. So guy 10 codes one color to mean odd and the other to mean even (everyone else knows this code as well). So by letting guy 9 know that say there are an odd number of black hats, guy 9 will know whether he has a black hat or red hat (if he sees an even number of red hats, that means he has to have a black hat). Guy 8 will then know that there are an even number of both hats. So he has the color of the hat which he only sees an odd number amount. And so on.

So 9 are guaranteed to survive. And the 10th has a 50/50 draw.

 

[exarch]

9 is max because you can never have 10, and 5 isn't correct.
That's my rational. Tell me 9 isn't right.

Edit: Ninja'd and confirmed.

 

[DJRome]

as i stated, u should put it in spoilers

exarch, it's irrelevant what you guess because you have no logic. the point is to identify the strategy, not the number. what good is the number 9 to these people if they don't know how to save 9?

god mods are ********

 

[ZeldaRox!]

it wasn't a spoiler. I said I've heard a similar one. But it was only like 3 or 4 hats. Used the same logic though.

and I heard it when I was in like 8th grade.

 

[DJRome]

no, i said put your answer in spoilers. like everyone else has ^^^^

 

[ZeldaRox!]

9 is max because you can never have 10, and 5 isn't correct.
That's my rational. Tell me 9 isn't right.

Is this guy serious? Dunno if its a troll or not. Either trolling or just being stupid.

"I've picked a number that can possibly be the answer and I'm justified because its not an answer that has been deemed incorrect and its not an answer that cannot belong in the set. Geeze this is so easy."

 

[ZeldaRox!]

no, i said put your answer in spoilers. like everyone else has ^^^^

oh u mean in that black block......how do I do that?

 

[DJRome]

they are spoilers and share the

 

[exarch]

Process of elimination is too good. And it wasn't a guess... well I suppose it was an educated guess.

Also don't tell me that without logic answers don't matter. I was a Math major. I know they don't. I just use different logic a lot of the time to get the correct answer and work backward from there.

Then people get all pissy because I abused the system and got the right answer in a way they didn't want me to.

Also @ Zelda, if I had really done that, I only have a 1/8 chance to get it right. Ya think I maybe put a bit more thought into it than that?

 

[Diatenshi]

Spoiler

there is no way to guarantee saving anyone given infinite hats, there is no way to eliminate any possibility of you having a differently colored hat.
if everyone gets to say only a single color the possibilty of everyone being wrong is .5^10 which is 0.00976, and while its a statistical improbability its very possible that nobody is saved given any method of guessing seeing as there is an infinite sample space of hats.

now if they get to say only the words black or red, but may say either twice they can save the person infront of them by saying first their color then the color of the hat infront of them, though the questions vague, if this was allowed it would save a guaranteed 9 people, plus a 50/50 chance of saving the man in the back.

 

[Todd Bonney]

Yeah I read the problem. You can still expect half of them to be red and half to be black.

But I missed the part where they have a night to decide on a strategy. That's useful.

I can think of a way to save an expected 7.5, but it's not guaranteed.

Darnit, not gonna look at that answer.

K, I can guarantee 6 now... but that's not enough is it?

Come on, I can induct.

 

[DJRome]

no, you only get to say red or black once. obv lol

and exarch, that's not using a different method. i asked for a strategy and how many. you want me to confirm half the answer so that you can reduce the thought process. pretty much avoiding the thinking part altogether

so if you really were a math major, you would stop thinking you're in the right and just do the ****ing problem like normal people

 

[ZeldaRox!]

Process of elimination is too good. And it wasn't a guess... well I suppose it was an educated guess.

Also don't tell me that without logic answers don't matter. I was a Math major. I know they don't. I just use different logic a lot of the time to get the correct answer and work backward from there.

Then people get all pissy because I abused the system and got the right answer in a way they didn't want me to.

Also @ Zelda, if I had really done that, I only have a 1/8 chance to get it right. Ya think I maybe put a bit more thought into it than that?

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?

 

[ZeldaRox!]

no, you only get to say red or black once. obv lol

and exarch, that's not using a different method. i asked for a strategy and how many. you want me to confirm half the answer so that you can reduce the thought process. pretty much avoiding the thinking part altogether

so if you really were a math major, you would stop thinking you're in the right and just do the ****ing problem like normal people

LMAO. agreed.

I'm thinking Exarch doesn't like me and is gonna hit me big time for this double post.... lol

 

[Diatenshi]

the hat answer is wrong

you cannot expect half and half hats if theres an infinite number of hats, if there were only 10 hats it would be obvious theres 5 black 5 red but with an infinite number there is no concievable method to accurately show the amount of the hat color

 

[DJRome]

hats are pulled from the bag @ a 50/50 chance

you can look at it like a hypergeometric distribution with n = infinity such that it becomes a binomial distribution in which the number of hats is irrelevant.

furthermore, the probability of hats pulled is irrelevant. it could be 99/1 from a bag of 1000000 hats. doesn't matter in regards to the answer

 

[X1-12]

4. hats:

Spoiler

5



the farthest back person calls out whichever colour there is most of infront of him (there are 9 people infront of him), after that everyone says what he says.

 

[BBQ°]

math team question lol-

Sarah pours four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?

 

[Diatenshi]

hats are pulled from the bag @ a 50/50 chance

you can look at it like a hypergeometric distribution with n = infinity such that it becomes a binomial distribution in which the number of hats is irrelevant.

furthermore, the probability of hats pulled is irrelevant. it could be 99/1 from a bag of 1000000 hats. doesn't matter in regards to the answer

***** 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

 

[DJRome]

sorry x1,that is not correct

lol bbq, i don't think that's the kind of problem we're looking for

but since 2/3 of the 2nd cup is cream, 2/3* 3 ounce/ (3+2 ounces of liquid) is your final answer

matt, it does not matter. the "chance" of anything is irrelevant since we're looking for guaranteed. it should work for any permutation of 10 people given 2 hat color choices

another way to look at it would be to posit any permutation of black and red hats and then apply the said strategy. it will work in that case the same as for any other permutation.

 

[ph00tbag]

2/5ths of the first cup is cream, because DJRome apparently can't read word problems.

 

[BBQ°]

my bad. I thought it would be okay since it requires logic, and stuff.