RIDDLES! Or, are you smarter than the rest of us?

[XZeroII]

Originally posted by: silverpig

Originally posted by: XZeroII

Originally posted by: Descartes

Originally posted by: XZeroII

Originally posted by: Flyermax2k3
weigh them all to get the overall weight, come up with average
weigh half of them and divide the total by 6, come up wtih average
weigh the other half and divide the total by 6, come up with average
by dividing in half you can isolate the deviant ball from the whole and by weighing two different sets you can see if it's lighter or heavier than the rest

Using the scale 3 times

Put 6 of them in one side, and 6 on the other side. Whichever side is heavier contains the heavier ball. Take those 6 and put 3 on one side and 3 on the other. The heavier side will contain the heavier ball. Take one ball and put it on one side, and put the other on the other side. If one side is heavier, then you have found your heavier ball. If they are the same, then it's the ball you didn't weigh.

Simple

Except that we just explained that you have to not only find the deviant ball, but also whether it's heavier/lighter.

Not so simple now

Divide the 12 balls into 3 groups of 4. Weigh 2 of the groups. If they are the same weight, then the devient ball is in the third group. If one is heavier, weigh that against the 3rd group. If it's heavier than the 3rd group, then that one contains the devient ball and it's heavier. If they are the same weight, then the devient ball was in the lighter group and the devient ball is lighter. You then put one ball on one side and the other on the other. If they are the same, then the devient is the third ball. If not, you have your ball.

Oh yea! no google or help of any kind

WRONG. You have 3 groups of 4. You can only do your last step if you have 3 groups of 2. "One ball on one side and the other on the other"... what about the other two balls in the group?

actually it works with 3 groups of 3. Dang, I forgot about that crappy 4th ball.

 

[silverpig]

Keep trying... and no cheating

 

[Descartes]

I'm not able to get this one. You would have to get down to two balls, because you'd have to take either of the two potentially deviant balls and compare it to a known equal ball to determine whether it was heavier or lighter. I can't see how you could isolate a deviant ball, heavier or lighter, out of twelve balls in just two weighs. It's easy if they are not equal on the first weigh (e.g. if you weight five on each side and they're equal), but if they're not you only have one more weigh to find the deviant, because the last weigh has to be to find whether it's heavier/lighter.

Or maybe not...

[edit]Ok, my only assumption so far is this:

- You have to isolate the deviant by comparing one ball out of a group of no more than two to a known non-deviant

Is that right?[/edit]

 

[RaynorWolfcastle]

ok here's how I do it. let's number the balls 1-12 for simplicity
put 1-4 on side A, 5-8 on side B, leave the others out.

If A = B, then place 9 & 10 on A, 11 & 1 on B, if equal weight use 12 vs other ball to know. If not, then put 9-11 vs 1-2 you can determine which ball and weight

if A != B, then place 1,2, 5, 9 vs 3,6,7,10 if A = B -> weigh 9 vs known-weight ball to determine which ball is wrong and what it weighs

The only other case, requires that you permutate everything in such a way that the 3 weighings together give you the answer, it's too long to explain here because there are so many ways in which this scenario can unfold. Basically you're mapping each position of the ball as a vector whose positions can be 1, 0, -1 (it doesn't really matter what you call them). Make sure no two balls have the same vector, seeing the answer there is only one ball/weight that could have caused the string of 3 outcomes that you got. At least this is what I came up with, half from memory.

edit: IIRC, there was an elegant way to write this in matrix form.

 

[TGregg]

The answer is Black, and I can show you why. There are seven different combinations:

BBB
BBR
BRB
BRR
RBB
RBR
RRB

If the last kid (knowing there are only two red hats) saw two red hats in front of him, then he would easily know he has a black hat. Since he did not speak up, we presume that he sees at least one black hat. So here are the new possible combos:

BBB
BBR
BRB
BRR
RBB
RBR

If the second kid sees a red hat, he knows that he must be wearing a black one (cuz he knows from the first deduction that the first two hats are not red). Since he doesn't speak up, the first kid knows he must be wearing a black hat, and we have one of the following combos:

BBB
BBR
BRB
BRR

 

[silverpig]

Originally posted by: Descartes
I'm not able to get this one. You would have to get down to two balls, because you'd have to take either of the two potentially deviant balls and compare it to a known equal ball to determine whether it was heavier or lighter. I can't see how you could isolate a deviant ball, heavier or lighter, out of twelve balls in just two weighs. It's easy if they are not equal on the first weigh (e.g. if you weight five on each side and they're equal), but if they're not you only have one more weigh to find the deviant, because the last weigh has to be to find whether it's heavier/lighter.

Or maybe not...

[edit]Ok, my only assumption so far is this:

- You have to isolate the deviant by comparing one ball out of a group of no more than two to a known non-deviant

Is that right?[/edit]

Not really no. There are other ways of determining if the deviant ball is lighter or heavier based on your previous weighs. In some cases you may have to do a final weigh, in others not.

 

[Metalloid]

Well here is a new one....

What is more powerful than God
more evil than the devil
the rich need it
the poor have it
and if you eat it, you will die?

 

[RaynorWolfcastle]

Originally posted by: Metalloid
Well here is a new one....

What is more powerful than God
more evil than the devil
the rich need it
the poor have it
and if you eat it, you will die?

The answer is nothing, here's mine now since Silverpig won't so much as acknowledge my existence.

It's an easy one but what the hell:
Let's say you're playing russian roulette. I show you the gun and there are no bullets in any of the six chambers, I then carefully drop 2 bullets in adjacent chambers. I point the gun at you, close the barrel and spin it. I pull the trigger, luckily for you nothing happens. Before the next shot I ask you whether I should spin again or not. Should you accept or not? Explain why.

 

[silverpig]

Originally posted by: RaynorWolfcastle

Originally posted by: Metalloid
Well here is a new one....

What is more powerful than God
more evil than the devil
the rich need it
the poor have it
and if you eat it, you will die?

The answer is nothing, here's mine now since Silverpig won't so much as acknowledge my existence.

It's an easy one but what the hell:
Let's say you're playing russian roulette. I show you the gun and there are no bullets in any of the six chambers, I then carefully drop 2 bullets in adjacent chambers. I point the gun at you, close the barrel and spin it. I pull the trigger, luckily for you nothing happens. Before the next shot I ask you whether I should spin again or not. Should you accept or not? Explain why.

You didn't exactly post a complete solution... You're on the right track though.

And the answer to your riddle is yes.

Label the chambers 1-6, with 5 and 6 having bullets in them. On the first spin, you shot nothing, so you were on any of chambers 1-4. As a result of the chambers rotating around once, the gun will now fire from any of chambers 2-5. If you are on chambers 2, 3, or 4, you will be safe, and on chamber 5 you will get shot, leading to a 1/4 probability of you being shot on the next pull of the trigger. A random spin will give a 1/3 chance of you getting shot. Since I want you to die on the next shot, I want you to have the 1/3 chance as opposed to the 1/4 chance.

 

[RaynorWolfcastle]

Originally posted by: silverpig

Originally posted by: RaynorWolfcastle

Originally posted by: Metalloid
Well here is a new one....

What is more powerful than God
more evil than the devil
the rich need it
the poor have it
and if you eat it, you will die?

The answer is nothing, here's mine now since Silverpig won't so much as acknowledge my existence.

It's an easy one but what the hell:
Let's say you're playing russian roulette. I show you the gun and there are no bullets in any of the six chambers, I then carefully drop 2 bullets in adjacent chambers. I point the gun at you, close the barrel and spin it. I pull the trigger, luckily for you nothing happens. Before the next shot I ask you whether I should spin again or not. Should you accept or not? Explain why.

You didn't exactly post a complete solution... You're on the right track though.

And the answer to your riddle is yes.

Label the chambers 1-6, with 5 and 6 having bullets in them. On the first spin, you shot nothing, so you were on any of chambers 1-4. As a result of the chambers rotating around once, the gun will now fire from any of chambers 2-5. If you are on chambers 2, 3, or 4, you will be safe, and on chamber 5 you will get shot, leading to a 1/4 probability of you being shot on the next pull of the trigger. A random spin will give a 1/3 chance of you getting shot. Since I want you to die on the next shot, I want you to have the 1/3 chance as opposed to the 1/4 chance.

There's no way in hell I'm posting all possible outcomes of your question, I think it's pretty obvious how to finish up the problem from where I left it.
The reasoning to your answer to my question is right, except I'm shooting at you if you read the question properly (I guess the russian roulette part was a little confusing, I admit); I'm not stupid enough to take your advice on shooting at myself .

 

[silverpig]

You don't have to list ever possible permutation... It's not that long of an explanation.

 

[QueHuong]

Ok, next riddle! No stupid impaled baby "riddle"...need something intelligent.

 

[Flyermax2k3]

Originally posted by: MindStorm
Ok, next riddle! No stupid impaled baby "riddle"...need something intelligent.

Hypocrite.

If you're that desperate for an "intelligent" riddle and can't see my *joke* for what it is, then perhaps you should post your own rather than relying on others to do it for you?

Hypocrite.

 

[NikPreviousAcct]

Originally posted by: MindStorm
Ok, next riddle! No stupid impaled baby "riddle"...need something intelligent.

What am I wearing?

 

[KLin]

Oldie. There's a bomb ready to go off. You have 2 water jugs, a 5 gallon and a 3 gallon. There's a scale on the bomb that requires the weight of 4 gallons of water to deactivate it. 1 weight only and it needs to exact or the bomb goes boom. How do you get 4 gallons of water exactly?

 

[QueHuong]

Originally posted by: Flyermax2k3

Originally posted by: MindStorm
Ok, next riddle! No stupid impaled baby "riddle"...need something intelligent.

Hypocrite.

If you're that desperate for an "intelligent" riddle and can't see my *joke* for what it is, then perhaps you should post your own rather than relying on others to do it for you?

Hypocrite.

LOL...I've never seen anyone so offended or take it so seriously when when their joke is not liked. BTW, notice I put the double quotation marks around riddle - I knew it was a joke. Stop crying.

 

[TechnoKid]

Whats fast, requires skill, precision, risk, asphault to run, but is the same backwards as it is fowards?

 

[silverpig]

Well, seeing as how there's little interest in my riddle now I'll post the solution I came up with:

Divide them up into 3 groups of 4. Call them Group A, B, and C.

Weigh Group A vs Group B.



Case 1: Group A = Group B.

In this case, you know the deviant ball is in group C. Weigh c1, c2 and c3 against a1, a2, and a3.



case 1.1: They are the same.

You know the deviant ball is c4. Weigh it against a1 to find out if it is light or heavy.

case 1.2: c1, c2 and c3 are heavier.

Weigh c1 vs c2. If they are the same, you know the ball is c3 and it is heavier. If they differ, you know that the heavier ball is the one.

case 1.3: c1, c2, and c3 are lighter.

Same as above. Weigh c1 vs c2. If they are the same, you know c3 is the light ball. If they differ, the ball that is lighter is the one you're after.




Case 2: A is heavier than B.

Take a1, a2, and b1, and weigh them against a3, b2, and c1.



case 2.1: a1, a2, and b1 are heavier.

Since you know the heavy ball must be in A or the light ball is in B, you can tell that the ball you want is either a1, a2, or b2. Weigh a1 vs a2.


case 2.1.1: They differ.

You know that the heavier ball is your ball.

case 2.1.2: They are the same.

You know that b2 is your ball and it is lighter.



Case 2.2: a1, a2, and b1 are lighter.

You know either b1 is your ball and it is lighter, or a3 is your ball and it is heavier. Weigh a3 vs c1.


case 2.2.1: a3 is heavier than c1.

You know a3 is your ball and it is heavy.

case 2.2.2: a3 is the same as c1.

You know b1 is your ball and it is lighter.



Case 2.3: a1, a2, and b1 are the same weight as a3, b2, and c1.

Basically, at this point, you have a4, b3, and b4 left. From your first measurement, you also know that a4 cannot be the lighter ball, nor can b3 nor b4 be the heavier ball. Take a4 and b3, and weigh them against c1 and c2.


case 2.3.1: a4 and b3 are heavier.

Since you know that b3 cannot be the heavier ball, a4 must be the ball you want, and it is heavy.

case 2.3.2: a4 and b3 are lighter.

Since you know that a4 cannot be the lighter ball, b3 must be the ball you want and it is lighter.

case 2.3.3: a4 and b3 are the same as c1 and c2.

You know that b4 is the deviant ball, and from your very first measurement, you know that it must be light.


Case 3: B is heavier than A:

Just repeat case 2 over, but switch all the balls labeled a for b.

DONE

 

[merlocka]

Originally posted by: KLin
Oldie. There's a bomb ready to go off. You have 2 water jugs, a 5 gallon and a 3 gallon. There's a scale on the bomb that requires the weight of 4 gallons of water to deactivate it. 1 weight only and it needs to exact or the bomb goes boom. How do you get 4 gallons of water exactly?

Take the jugs to blockbuster, sell them to the checkout dude, and use the $3 bucks to rent Die Hard.

 

[merlocka]

Here is one.

2+11=1

 

[jaydee]

Originally posted by: KLin
Oldie. There's a bomb ready to go off. You have 2 water jugs, a 5 gallon and a 3 gallon. There's a scale on the bomb that requires the weight of 4 gallons of water to deactivate it. 1 weight only and it needs to exact or the bomb goes boom. How do you get 4 gallons of water exactly?

This was in Die Hard 3. Put 3 gallons in the 3 gallon jug and pour that in the 5 gallon jug. The refill the 3 gallon jug and top off the 5 gallon jug. You now have 1 gallon in the 3 gallon jug. Empty the 5 gallon, pour the 1 gallon in the 5 gallon, refill the 3 gallon jug and pour all that into the 5 gallon.

 

[sillymofo]

Originally posted by: KLin
Oldie. There's a bomb ready to go off. You have 2 water jugs, a 5 gallon and a 3 gallon. There's a scale on the bomb that requires the weight of 4 gallons of water to deactivate it. 1 weight only and it needs to exact or the bomb goes boom. How do you get 4 gallons of water exactly?

Easy, you take the 5 gallons, fill up the three gallons, so you're left with two gallons in the 5. Wait, aren't you missing a jug? Dumbfcuk... you forgot a fcuking jug going to defuse? What kinda dumbfcuk it this... man, I'll never go defuse any fcuking bombs with you again.

What he ^ said.

 

[jaydee]

Two people, named S and P, are talking about two numbers x and y.
(note: x and y are both integers greater than or equal to 2). S
knows their sum (x+y), while P knows their product (xy); however,
initially NEITHER knows x and y.

S: Hey P! I don't know what the numbers are.
P: I'm not surprised. I KNEW you didn't know. However, I too don't
know.
S: You don't? Really! Then I know what the numbers are!

What are the two numbers?

 

[sillymofo]

Ok... so you're a tough guy, you roughed up some kids for lunch money and now you're behind bars on an island. This island is surrounded by lazers-on-head shark infested water, so swimming or crossing the water is out of the question, the bridge is only accessible by a 5 miles long bridge and at the center is a guard shack, maned by a big hairy monkey. The monkey's job is to guard against intruders and also tough guys like you that are trying to break out. The monkey's only weakness is epilepsy, it will fall asleep for a 3 mins period, and when it's awake, it will take you back to where you came from when it sees you. Assuming that you can run at the rate of 1 mile/min, how do you escape? (and no, this monkey is not tempted by your cheap bananas).

 

[Zorba]

Originally posted by: bootymac

Originally posted by: Descartes
Try this one:

There are three switches in a hallway. One switch controls a light fixture in a room at the far end of the hall. The door to the room is closed, and you can't see whether the light is on or off. You need to find out which of the three switches controls the light. How can you be certain of finding that out, making just one trip to the room?

[edit]I'll tell everyone what answer I gave if no one gets it.[/edit]

How about walking down there and opening up the damn door so you can see the light .