On September 20 2006 03:18 One Page Memory wrote:
Plexa, exact this riddle was solved and explained by Day in other thread about puzzles few months back. But good one for those who never heard it.

lol i never saw it!
i think i was studying for exams then.. i cant remember but i remember not being at tl for ages =( hard times... hard times....

CubEdIn

Romania5359 Posts

September 19 2006 20:19 GMT

#142

I understood what hotbid's logic was. If the two coins are identical, then their flaw would be identical too, so if either coin is pre-determined to land on one specific side 51% of the time, then the other one will too.

Glider

United States1353 Posts

September 19 2006 21:09 GMT

#143

Did anyone got #4 Boat/Brick correct yet? Because i don't think anyone really did since it is a trick question.

It comes down to volume vs mass density or weight. and since there are many kind of bricks and we don't know the volume and mass density of the brick in question. We cannot answer the question. It could be those new light ass clay brick or a gold brick. So really the best thing is to find out what is the brick's dimension and how much does it weigh. Until then, all answer is based on assumptions.

Resonate

United Kingdom8402 Posts

September 19 2006 21:30 GMT

#144

For the boat one:

i couldn't quite see what the answer was from the thread but i heard Archimede's name mentioned and that immediately rang alarm bells in my head cos this is the wrong application for this principle...

"Archimede's principle states that the weight of a floating object is equal to the weight of the water displaced by the object. For an object to be in equilibrium, the upward force of buoyancy must be equal the downward force of gravity (weight). Also, the center of gravity and the center of the underwater volume (center of buoyancy) must be vertically aligned."

however, archimede's law does not apply when something (e.g. the brick) is submerged fully.

who cares if the brick is equal to the WEIGHT of water it displaces? When the brick is submerged it's VOLUME is the important point (so long as it's average density is greater than the waters - thus keeping it submergered).

-----------------------
Consider this example: submerge one 1-cubic cm marble weighing 10 tonnes into some water, the displacement of water is not 10 cubic metres (i.e. 10 tones of water), it is 1 cubic cm (i.e. ~1 gram).

Now consider thes two extensions:
A) putting the 10-tonne marble into a large empty vessel (weighing v tonnes), it will cause the apparent weight of the vessel to increase to 10+v tonnes, thus causing a huge increase in displacement.
B) putting a normal glass marble (weighing a coupe of grams) into the boat. It will not change the boat's apparent weight by a noticable amount and so will not change the displacement by a noticable amount.
-------------------

To answer the question correctly you need to know the following single detail:
What is the brick's average density?

if the answer is greater than the density of water... then the boat displaces more than the brick (so the level of the water falls).

this whole question is works on the assumption that this last 'if' is correct (bricks are generally quite heavy after all).

archimede's principle only applies if you sink both the boat AND the brick.

I hope this makes sense when read, Please point out if this is wrong.

GrandInquisitor

New York City13113 Posts

September 19 2006 21:32 GMT

#145

Resonate: note what it says. The brick sinks quickly - therefore its average density >> that of water.

Hot_Bid

Braavos36374 Posts

September 19 2006 22:07 GMT

#146

ZOMG MAGIC BRICK

Resonate

United Kingdom8402 Posts

September 19 2006 22:43 GMT

#147

On September 20 2006 06:32 GrandInquisitor wrote:
Resonate: note what it says. The brick sinks quickly - therefore its average density >> that of water.

ok fair cop.

But it doesn't change the point of my post. I won't bother editing it or it'll get confusing

On September 20 2006 07:07 Hot_Bid wrote:
ZOMG MAGIC BRICK

LeoTheLion

China958 Posts

December 04 2006 03:35 GMT

#148

I read Hot Bid's two quarter's post, and the responses, and just had to post this.

Hot Bid is right.

It's to your advantage if you pick "same" rather than "different." Why? Examine:

Say that the probability that the coin lands heads is 'p'. Therefore, the probability that the coin lands tails is (1-p). We see that

Probability of same = p^2 + (1 - p)^2
Probability of different = 2*p*(1-p)

where p is between 0 and 1, inclusive.

Doing some algebra (completing the square), we see that

probaiblity of same = 2(p-1/2)^2 + 1/2
probability of different = -2(p-1/2)^2 + 1/2

which implies that

4(p-1/2)^2 >=0 (trivial inequaility, x^2 is always >= 0)
2(p-1/2)^2 >= -2(p-1/2)^2
2(p-1/2)^2 + 1/2 >= -2(p-1/2)^2 + 1/2

probability of same >= probability of different

with equality occuring when p = 1/2.

In short, this means, that picking same will always give you better or equal odds than picking different. (The only time when the odds are equal is when p = 1/2.)

QED

Xodiac

Sweden55 Posts

December 04 2006 06:59 GMT

#149

Hey, do anyone have the link to where [9]day solves and explains that question PLexa posted? I'm very curious about it, I won't sleep before I find it out. . .

Last Romantic

United States20661 Posts

December 04 2006 07:05 GMT

#150

+ Show Spoiler +

Ask either man "what would the other man say the correct path is?"

Liar knows other man would tell you the right path, so he would tell you the wrong path.
Honest man knows other man would tell you the wrong path, so he will tell you the wrong path.

Either way, you know not to go down whatever path they tell you to, and pick the other one.

^response to Plexa's question. I'm not sure how Day did it, but it should be similar.

Xodiac

Sweden55 Posts

December 04 2006 07:22 GMT

#151

I just about figured it out myself, or well, yeah similar to it aswell :D This one was better explained, easier to understand/explain though. Thanks!

Daveed

United States236 Posts

December 04 2006 07:32 GMT

#152

-__- don't write QED after that kind of proof

Ilintar

Poland794 Posts

December 04 2006 08:05 GMT

#153

#5 is the most trivial case for a much more general problem.

See http://www.cs.huji.ac.il/~dolev/pubs/cheating.pdf for a serious paper which describes that case and many more complicated (great to catch any friends of yours who might have heard the original).

snowbird

Germany2044 Posts

January 25 2007 23:57 GMT

#154

On September 19 2006 20:16 IIICodeIIIIIII wrote:
simple answer for #2 + Show Spoiler +

question 1 "is the car behind door 1 red?" question 2 "is the card behind door 2 red?" if the watch flashes a color it means that the car is behind that door. if the car isn't behind that door, the question doesn't make sense and therefore, the watch wont flash anything. if the watch doesnt flash something on both doors, it means the car is behind door 3.

sounds nice, but doesn't work
if the premise is wrong, the answer will always be yes (thus a light will flash)

if i ask you if birds are blue when the sun is green, the answer is yes. it's a rule of propositional logic. (correct me if i'm wrong)

CCHS

United States614 Posts

January 25 2007 23:58 GMT

#155

wow snowbird why did u revive this

snowbird

Germany2044 Posts

January 25 2007 23:59 GMT

#156

there was a link in the other riddle thread, and i didn't know this thread before, and i think it's a good one

also i wanted to clarify that mistake (correctly i hope)

Haemonculus

United States6980 Posts

January 26 2007 00:59 GMT

#157

All the answers for #6 are crazy complicated. It's pretty simple.

Put 6 marbles on each side. One side will have the heavier ball. Keep those 6 marbles, discard the rest.

Put 3 marbles on each side. One side will have the heavier ball. Keep those 3, discard the rest

Of the 3 you have, one is heavier. Randomly put 1 on each side, and hold on to the 3rd. If one side goes down, that's the heavier one. If they are even, you're holding the heavier one.

snowbird

Germany2044 Posts

January 26 2007 01:28 GMT

#158

On January 26 2007 09:59 Haemonculus wrote:
All the answers for #6 are crazy complicated. It's pretty simple.

Put 6 marbles on each side. One side will have the heavier ball. Keep those 6 marbles, discard the rest.

Put 3 marbles on each side. One side will have the heavier ball. Keep those 3, discard the rest

Of the 3 you have, one is heavier. Randomly put 1 on each side, and hold on to the 3rd. If one side goes down, that's the heavier one. If they are even, you're holding the heavier one.

your answer does not include: "Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light."

Haemonculus

United States6980 Posts

January 26 2007 01:36 GMT

#159

On January 26 2007 10:28 snowbird wrote:

Show nested quote +

your answer does not include: "Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light."

LOL. My bad. Didn't even read the whole question. gg me.

CharlieMurphy

United States22895 Posts

January 26 2007 01:38 GMT

#160

As for number 4 the brick in the lake one. The water level stays the same because it was boyant in the boat which is holding its weight in the first place. So throwing it in the lake wont change the water level at all.

Don't know if this has been said. old thread.