The riddle thread

Since the number of quarters in infinite but the number that start in tails position is finite, you have 0 chance of picking a tails quarter if you just reach out and grab one. So what you do is grab 20 quarters, flip them all over, put them in a separate pile, and declare victory. This assumes the tails start randomly distributed, or that you have some means of selecting at random.

It's simpler than that. You don't mess with chance at all. Just grab 20 and flip them. If you grabbed all 20 tails, each pile will have 0. If you grabbed 19, each pile will have one. Etc, etc.

Just flip them all and make an infinite line. There are infinite tails at any side

Djabenete is right though, technically if you grabbed 20 coins they are all heads anyway.

You could grab any number of coins and as long as you flipped 20 over you would win.

The infinity thing doesn't really add anything to the riddle.

switch camels

Agree to split the money 50/50?
Yeah, he pointed out that the other Brother was not currently on his Camel so they could jump on the wrong camel and race it to the line.

Open the Bronze chest. If it were in the Silver, all answers would be true. If it were in the Gold, all answers would be false.

You open the Bronze chest.
All the possibilities of true and false create contradictions EXCEPT Silver false, gold true, bronze true.
(my table)
Silver bronze gold Chest?
t t f fail
t f t fail
f t t Bronze
f f t fail
f t f fail
t f f fail

throws away one bag, says I chose that one, show me what's in the bag I did not choose


he's too short

both contained white stones, the king thought that the general would outhink him if he put two black stones in them by choosing the one he doesnt open, but the general being smarter knew exactly that the king knows that he is too smart to fall for such a cheap trick

Does he take a bag and place it face down on the floor. Then he rubs the stone in the bag against the floor (assuming he's standing on a stone stage like an amphitheater). The stone will crumble on the outsides creating a layer of white rubble. Then he simply removes the stone from the bag and it appears to be white.

The second one works better when spoken aloud, since you do not immediately reveal that you were just kidding about the concrete. This greatly perplexes all but the most astute of men.

1coin from the 1st machine, 2coins from the 2nd, 3from 3rd, and so on.. just take all the coins you got, 210??? and calculate how much it should weigh, if 1ounce less its the 1st machine, if 2 the 2nd and so on

The guy is a midget and can only reach the buttons for the 28th floor. If someone else is there, he asks them to press the 30th floor button for him.

This one is wrong though =p

yes it does, because 20/infinity = almost almost almost zero
thats why you can pick up most coins and get heads
in math u learn to assume anything/infinity = 0

I was about to say that it was because he was banging some chick that he didn't want his wife to know about.

4 exit wounds = 4 more entry wounds, for a total of 8 holes so far. The man's ass and mouth count for the other 2 holes. The bullets punctured the walls and flew out windows so they are not in the room.
Am i rite

nostrils n ears though?

C.

an anchor

A ring or /ragequit going "Gollum, Gollum!"

goose, fox and bring goose back, corn, goose

He takes a stone in swallows it before anyone can see it. The leftover stone will be black, so the king would have to admit A: he got the white stone or B: he was cheating.

The bullets were gone because they were made of ice. The gun was hidden in the tap dancing shoes.

Oh no, it took me many hours to figure out. You're not getting it that easy.

It can work; anything going at that velocity can kill someone. It doesn't melt instantaneously and if it shatters it'll just look like a shotgun shell...i just said it was a revolver cause of the other question =p

Use 1 weigh to find out the weight of 1 ball. Take a book or a flat surface and angle it slightly. Take all the balls and roll them off from the top off slope. The one that is slightly farther or closer is the ball with the odd weight. Take the weight of that ball and compare it to the first one. Take the weight of another ball and compare it to that. If the odd ball is > than the other two balls then it's heavier. If it is <, vice versa.

Sorry, there are no books or flat surfaces in the world of this riddle. Unless you chose to interpret the weighting scale as one of course. Besides, if you roll the same ball down the slope twice it will most likely roll different lengths.

Out of the box solutions are always nice, and some riddles are only solvable if you find loopholes in the premises. But this not one of these.

juggle, idk xD

If that's the answer, the riddle is pretty lame. Why would they be hidden in tap dancing shoes? Why not say the murderer was a box maker and hid the gun in a box. Second, nothing hints at ice bullets besides no bullets being found, which could mean almost anything. Besides, I'm pretty sure Mythbusters did a test on ice bullets, and yes they would melt pretty much instantly from the heat. You know that an explosion propels bullets out, right?

all the coins must be weighed once, because otherwise you won't figure out which one is fake if the it stays on the outside. Two coins cannot appear together on the same weighing because if the fake one is one of them we wont know which it is. So numbering the coins from 1 to 12 we have the following weighings:

1) 1,4,7,10 x 3,6,9,12
2) 3,6,9,10 x 2,5,8,12
3) 3,4,8,12 x 2,6,7,11

Damnit how do u use 3 measures? I keep having to use 4.

put sand in the box til its half full and keep resetting the buttons and pressing the 4 buttons to see which one turns it on. You don't need the fan; you just want the oasis =p

But yes, it is correct. Though my answer was a one page long algorithm.

I base this solution on the premise that when you put some sand in the box it will lie in a cone formed shape right under the hole. If you then turn the fan on it will distribute evenly.

First set all buttons to off, then put some sand in the box. Then turn buttons 'red' and 'blue' on for a while before turning them off again. Now put some more sand in the box and let button 'red' be on for a while before turning it off again. Finally, turn button 'green' to on.

When opening the box his will result in different outcomes:
Sand cone on top of a flat layer of sand=>Blue
Sand cone=>Yellow
Flat sand and fan turned off=>Red
Flat sand and fan turned on=>Green

Yes

chicken came first
eggs cant cum

A towel?

I know one method of solving this, but I need to know whether the 1 ball is either heavier OR lighter. T_T

his insulin shot

Parachute

The antidote

Imagine that the monk has an identical twin who starts climbing the mountain exactly like he did on the way up exactly one week after. At 9am when the monk starts making his way down, his twin is somewhere along the route going up. Obviously, they're going to meet up briefly at one point - and that's the time/place we're looking for.

Let f(x) = height on the way up, defined for x in [6am to 6pm]
Let g(x) = height on the way down, defined for x in [9am to 6pm]
Assume f(9am) < height of the mountain (i.e. he didn't get to the top in 3 hours)

Let h(x) = f(x) - g(x) defined for x in [9am to 6pm]. h(9am) < 0, h(6pm) > 0... so by the IVT (f, g, and h are clearly continuous functions) there must be some point t where h(t) = 0.

Saint stole it, as he saw Shark and Light having very manly gay sex at 3AM in Light's room last night

he declares that he will choose the black stone, since it's 50/50, the other stone "must be white". he takes out a black stone, so the other stone is white by process of elimination, unless the king rigged it (which he did, but the public doesnt know that). so the general lives

Guys, the key here is 'dessert'.
He was unknowingly about to devour a strawberry cheesecake, but he has a deadly allergy to strawberry.
If he had opened his knapsack, he would have taken the chocolate pudding in there which just so happens to be his favorite food, and would have eaten that instead thus saving his life.

Open the refrigerator, put in the giraffe, and close the door.
This question tests whether you tend to do simple things in an overly complicated way.

Did you say, "Open the refrigerator, put in the elephant, and close the refrigerator?" (Wrong Answer)

Correct Answer: Open the refrigerator, take out the giraffe, put in the elephant and close the door.

This tests your ability to think through the repercussions of your previous actions.

Correct Answer: The Elephant. The elephant is in the refrigerator. You just put him in there. This tests your memory.

OK, even if you did not answer the first three questions correctly, you still have one more chance to show your true abilities.

Correct Answer: You swim across. All the crocodiles are attending the animal conference.

This tests whether you learn quickly from your mistakes.

This probably isn't the right solution but I would just stick a stick or needle or string into the hole and press the buttons one at a time. Eventually the fan will turn on and hit the stick.

I don't see how this is possible. At 6 in the morning on the first day he is at the foot of the mountain whereas at 6 am the 7th day he is on top of the mountain. I've tried reasoning with him going up down up again and with his starting point being the top also but I don't think it's possible that way either.

I've got another solution based on another doubtful premise; am I allowed to press down some buttons, open the box and then let go of the buttons as the box is opened and observe what happens? If I am, then this is a solution:

First press red and blue, then press blue and green and hold them down. Open the box and let go of blue and green.

If the fan was on when I opened the box and it stayed on when I let go of the buttons it is the red one that is connected to the fan.

If the fan was on but turned off it means it is the blue one.

If it was off and turned on it is the green one.

If it was off and stayed off it is the yellow one.

Though this would mean that the hole in the box is a red herring.

Yeah but you need the intermediate value theorem for that so you need to be in classical logic. In intuitive logic this fail, even if it is not obvious. ;-) I think you can find counter-examples in the topos of sheaves over [0,1], but i suck at intuitive logic. (but "topos of sheaves" looks good so i can say it to be a smart-ass :p)

thought of a ephemera... either it didnt hear anything about it yet, or its alrdy dead.

footprints in the butter

1. Divide the balls into two groups of 6 and put each group on the balance. The ball that is lighter must be in the group of 6 that is lighter.

2. Divide that group of six into two groups of 3 and repeat this process, the ball that is lighter must be in the lighter group of 3.

3. Take any two of the the group of three and weigh them on the balance. If one is lighter, there you have it. If they are in fact, balanced, then the one you didn't weigh from that group of three must be the lighter ball.

It is impossible to go by the elimination method I outlined earlier because you do not know if the ball is heavier or lighter, so you would need an extra weighing to know which group to pick each time. So instead I thought that you could assign a 3 letter code for each ball that would be the result of the three weighings. If the first weighing makes the left side go down, then the first letter in the code is "L", an "R" if the right side goes down, or an "e" if it is even. There are then 14 unique pairs of codes that correspond to each ball being an incorrect ball, as follows:

LLL - RRR
LLe - RRe
LLR - RRL

LRL - RLR
LRe - RLe
LRR - RLL

LeL - ReR
LeR - ReL
Lee - Ree

eLL - eRR
eLR - eRL
eLe - eRe
eeL - eeR
eee - eee

So if ball one's code is eLR and it's heavier, then that corresponds to ball one not being on the scale for the first weighing, then being on the left side for the second then on the right side for the third. If the ball is lighter, then it's code would be eRL, and hence the pairing system.

Our problem now is to assign each ball a unique code pair, and then assign the weighings that will result in that code pair if that ball is the odd ball. Since there are 14 paris and 12 balls, I throw out LLL and eee as uninteresting, though it might be possible to solve it with any combination of the 14... that would require more work to prove, which I won't do here.

Taking my first weighing as

I. 1 2 3 4 vs. 5 6 7 8

I'll assign codes just in order (pretty arbitrary, I think you can make multiple ways work). (Note: 5-8 will use reciprocal codes because they start on the right)

1 - LLe 2 - LLR 3 - LRL 4 - LRe
5 - RLL 6 - ReR 7 - ReL 8 - Ree

So now we have some information about our second two weighings. We must have

II. 1 2 5 _ vs. 3 4 _ _
III. 3 5 7 _ vs. 2 6 _ _

for the codes to work. We also have 4 pairs left to assign, which again, I'll just do in order

9 -> eLL/eRR 10->eLR/eRL 11->eLe/eRe 12->eeL/eeR

again, there are a couple ways to pick which pair you want to use, but the bolded ones are the ones I picked.

Now our three weighings are:

I. 1 2 3 4 vs. 5 6 7 8
II. 1 2 5 11 vs. 3 4 9 10
III. 3 5 7 10 vs. 2 6 9 12

Do these weighings, and match what code the experiment gives you with the correct ball for that code, and viola, you are done, because each ball has a unique code. (If it's a reciprocal code, that simply means the ball is lighter, not heavier, as the code corresponds to which side goes down.)

phew, you were right, that was a bit of work!

The letter codes leads straight into my follow-up question. Would you mind going for that one?

There are actually elimination methods that can be used too. Here's one:
+ Show Spoiler +

If it's the codes I picked, it's heavier, or if it's the "reciprocal" code then it's lighter.

<3 The Office

24 mins

17 min: 1 and 2 cross, 1 comes back, 5 and 10 cross, 2 comes back, 1 and 2 cross: 2+1+10+2+2 = 17

I'd go with 20 minutes unless im misreading the problem or just being stupid

17 minutes.
1minute and 2 minutes go = 2 minutes
2 minute comes back = 4 minutes
10 minutes and 5 minutes go = 14 minutes
1 minute comes back = 15 minutes
1 minute and 2 minutes go = 17 minutes

if the prisoner is going there for the first time, leave the light off. if he is there for the 2nd time, then turn it on. if a prisoner goes in there and sees it on and it's his first time, he turns it off. he knows that the person before him has been there. if a prisoner goes in there and sees it on and he's also been there before, leave it on. eventually, everyone will go in and see it on.

this is where im not sure what to do next. supposedly over time it will always be on, but i dont know how they can be 100% sure everyone has been in the room

maybe some one can take it from here

He's walking, not driving.

Nothing.

Which road would your twin tell me to take. Then take the opposite.

It is yes when n < (3^k - 3)/2.

Basically, it is yes when the possible number of outcomes for the weighing is less than the number of balls weighed. Each weighing has 3 possibilities (Left heavier, Right heavier, or equal.). With k weighings, that means that there are 3^k outcomes.

3^k

Because we are constructing our system in such a way that all Left heaver or all Right heavier or all equal are not viable outcomes, those 3 cases are subtracted from our possible outcomes.

3^k - 3

Since we do not know if the odd ball is heavier or lighter, we have half as many outcomes to work with.

(3^k - 3)/2

For 3 weighings, this gives us (3^3 - 3)/2 = (27 - 3)/2 = 24/2 = 12 outcomes, so we can determine the odd ball out of a maximum of 12 balls.

There is a sure way to get this right, but it takes a long time.
The prisoners create a process of selecting a leader. The other prisoners will be ordinary prisoners. The leader has the job of declaring all 100 prisoners have been to the room.

The first time an ordinary prisoner enters the room and finds the light bulb turned off, the prisoner turns the light bulb on. If the light bulb is on, the prisoner does nothing. If the prisoner enters the room and finds the light bulb turned off a second time, the prisoner does nothing.

Whenever the leader enters the room, the leader turns the light bulb off if it is on. If the leader does this, the leader adds 1 to the tally. When the tally reaches 99, the leader declares that all prisoners have been inside the room.

Optimally, the leader is the prisoner that enters the room on the second day. The process takes 20+ years on average.

The alternate solution is to use statistical probability such that the probability of getting all 100 prisoners into the room is so high that it might as well be certainty. Something like a 1000 days is enough time to wait. Each prisoner expects to enter the room 10 times during that long a duration. The probability of each prisoner never going to the room is near zero, but over 100 prisoners, it still adds up. Yet there is still more than 99% probability.

So spend 3 years in jail and take your chances. Easy enough right?

Your solution has one "leader" to tally up counts.
This usually takes around 10,000+ days, or ~28 years. What's interesting is that we can optimize even more. You chose the leader to be the person who enters the room on the second day. If we choose the leader to be the first person to enter the room twice. That is, if the light is still on, then only unique prisoners have entered on everyday. This means that if I enter the room on the 40th day (and it's my second time) and the light is still on, then I know for a fact that 39 unique prisoners have entered the room already. This does a somewhat significant slimming on the solution time (~9000 days instead of ~10,000).

Now with this track of mind, we have been using only one leader to count. But...

What happens if you use more than one? How much does this reduce the average time?
Lastly, what do you need to add or change to the prisoner's planning in order to make it work with the new plan?

For those who think this is too easy, can you think of a better solution if the room now has two lightbulbs instead of 1?

What if there are n prisoners and k light bulbs?
(I really love this riddle, at first glance many find it impossible, and upon hearing one solution people fail to realize how deep the rabbit hole can go).

Nice job. Though LLL/RRR is a usable outcome, so you're one off.

The velocity of the snail is the summ of 2 terms: the velocity of the snail relatively elastic and the velocity of elastic (each point of elastic is moving while it stretching).

where

with new variables

this equation will take simple form

and the solution (after returning to the original variables) will be

Now from the equation

we can find time T required to catch up the bicycle and this time will be

Ask twin A "What would twin B say which way I should pick not to die?" If the answer is left, you go right.

the solution was something retarded with ice bullets, except they're pretty much impossible to make so the solution could be whatever the hell you think it is. i say the bullets were made of fairy dust and disappear upon contact.

Ask one of the twins if the other twin would say that a(choose either) passage is the safe one to go. If its the correct passage the liar would say it isn't. Even if it was the truth teller he'd say it isn't also. If its the unsafe passage the liar would say its safe and the truth teller would say it isnt safe. This way the passage that is always "unsafe" is the correct one.

It looks like a mutli-tiered counting approach would work faster. The fastest is to tier by counting to as close to e as possible. That is 3, and so that gives 4 tiers, and all 100 prisoners have to be involved in one way or another, but the goal is to minimize their involvement.

(strange 4 tiers of 3 = 12, but 6 tiers of 2 = 12 - except with more tiers - don't know what this means - perhaps I want to use 2 instead?)

So that looks like that will be
66->22->8->3->1
which gives a total of exactly 100. nice.

But there's no way to coordinate when the first tier is done - or when any tier is done. Only the last 1 can know when it's down. Hmmm, that means that the prisoners have to guess by calendar date, and there will need to be a way to return to any or all of the tiers if necessary if they are unlucky.

Ahh yes, they have to set calendar dates ahead of time in order to coordinate all the prisoners, and they'll also have to work out calendar dates for when they redo the tiers.

On the first pass, prisoners will want to be lightweight and progress through each cycle with a reasonable chance of 100% completion. If they are too conservative, then they are just wasting time, since if they miss 100% completion, they can just go back through a redo cycle.

On the redos they need at least 100 day per tiers to make meaningful progress. A redo cycle has to be effective at matching up 1 countee and 1 counter. The first redo cycle maybe might have to be more conservative. And there needs to be some kind of tier transition mechanism. I'll have to work that out first. A programming exercise is probably in order.

Conservatively, I would have to think that it would take 4 tier * 3 counts * 3 probability estimate * 100 men = 3600 days or 10 years. Ha!

Cut the rope into 2 pieces, one 500m and one 250m.
Make a noose at the end of the 250m piece
Slip the 500m piece through. You now have a rope that is 500m.
Slide down to the rock and stop. Pull the rope through the noose
Slide down the the bottom

Cut the rope into 2 pieces, one 500m and one 250m.
Make a noose at the end of the 250m piece
Slip neck through the noose.
Slide down to the rock and stop. Pull the rope tight
Jump down to the bottom

The first time an ordinary prisoner enters the room and finds the light bulb turned off, the prisoner turns the light bulb on. If the light bulb is on, the prisoner does nothing. If the prisoner enters the room and finds the light bulb turned off a second time, the prisoner does nothing.

Whenever the Leader enters the room, the Leader turns the light bulb off if it is on. If the Leader does this, the Leader adds 1 to the tally. When the tally reaches 99, the leader declares that all prisoners have been inside the room.

If it is your first time in the room, leave the light on! If it is your second time in the room and the light is still on, turn it off and count yourself as the new leader, and tally the amount of days that have passed.

This is commonly referred to as a "snowball round." Here is an in-depth explanation (this is the first introduction to having stages in data collection):

A common atypical stage, called a snowball round can be applied to several schemes. The snowball round occurs at the beginning of the scheme and (for schemes with leaders) passes out a single badge. The prisoner on day 1 leaves his simple token by turning on the light. Later visitors leave their token by leaving the light on. The first prisoner to visit who is unable to leave a token (either because he has been in the room before and no longer has a token to leave, or because he is the last visitor of the stage, or because special rules of the scheme prevent him from doing so) turns off the light and picks up all the tokens left in the room. This number is determined by the day number. If he visits on day k, then he picks up k-1 tokens. In addition, he is also given a badge (usually, the crown). No one is allowed to turn on the light during the snowball round except the prisoner on day 1, so once the light is turned off, the rest of the round is inactive. The effect of the snowball round is transfer a large number of tokens to a single prisoner in just a few days. Unfortunately, it is only worthwhile when a large majority of the prisoners have tokens to leave, which is why it is only done at the start of the scheme.

In general doing things this way - using stages to create a leader rather than relying on a pre-defined leader can slim down solutions significantly.

For example if we were two use four counters that each collect 25 tokens (light switch counts) each. During the token collection stage they all have the ability to turn off the light switch.

During the counter collection stage, the first person in the counter room leaves his counter (higher level token worth 25 single tokens), and removes his count. The very next person in the room who finds the light on during this sequence turns off the light, becomes the master token counter, and adds the other players tokens to his total. This is more optimal than relying on one specific person to tally the tokens (since any master counter entering the room can do it).

a ring

an echo

coals

snail

None!
The elephant ate them all

reminds me of chuck and larry