Logic Puzzles

[Rabid Monk]

Here's the rules:
I will post an initial logic puzzle.
The person who solves it correctly, and provides justification for their solution, gets to post the next logic puzzle.

Let us begin:

There are 4 cards before you on a table. On the top side, you see that one card has an "O", on card has a "L", one card has a "5", and on card has a "2".
You are told that every card with a vowel on one side has an even number on the other side. Which 2 cards must you flid to test the validity of that statement?
(This one is easy, but 90% of people get it wrong.)

[noodle]

Here's the rules:
I will post an initial logic puzzle.
The person who solves it correctly, and provides justification for their solution, gets to post the next logic puzzle.

Let us begin:

There are 4 cards before you on a table. On the top side, you see that one card has an "O", on card has a "L", one card has a "5", and on card has a "2".
You are told that every card with a vowel on one side has an even number on the other side. Which 2 cards must you flid to test the validity of that statement?
(This one is easy, but 90% of people get it wrong.)

"O" and "5".

if "O" has no even number on the other side, the statement is false.
if "5" has a vowel on the other side, the statement is false.
"L" is not a vowel, so the other side is irrelevant to the statement.
"2" can have a vowel or consonant on the other side, as the statement says nothing about consonants and even numbers, so it is also irrelevant.


my turn, no cheating.

FACTS OF THE PUZZLE

1. There are five houses in five different colours.
2. In each house lives a person of a different nationality.
3. These five owners drink a certain beverage, smoke a certain brand of
cigarette and keep a certain pet.
4. No owners have the same pet, smoke the same brand of smoke or drink
the same drink.

HINTS

1. The Brit lives in the red house.
2. The Swede keeps dogs as pets.
3. The Dane drinks tea.
4. The green house is on the left of the white house.
5. The green house owner drinks coffee.
6. The person who smokes Pall Mall rears birds.
7. The owner of the yellow house smokes Dunhill.
8. The man living in the house right in the center drinks milk.
9. The Norwegian lives in the first house.
10. The man who smokes Blend lives next to the one who keeps cats.
11. The man who keeps horses lives next to the man who smokes Dunhill.
12. The owner who smokes Blue Master drinks beer.
13. The German smokes Prince.
14. The Norwegian lives next to the blue house.
15. The man who smokes Blend has a neighbour who drinks water.

The question is: WHO OWNS FISH?

[Rabid Monk]

It took awhile. I checked my answers.
(Does that count as cheating? I hope not.)

Anyway:

Short answer:
The German owns the fish.

Long answer:
And the Logic said: Draw for thyself a grid, 5 rows by 5 columns, and let each colum represent Nationality, Colour, Drink, Smoke and Pet. And it was done, and there was drawn a grid of 5 rows and 5 columns, and each row represented, in turn, Nationality, Colour, Drink, Smoke and Pet. And the Logic saw it, and it was good. And this was the begining and end of the first step. (sorry, I couldn't resist.)
To be honest, I stumbled across that particular problem last year, and solved aboiut a third of it before getting bogged down and giving up. I resolved to come back to it at some point in time, so I never looked up the answer. So when i saw you post it, I went and dug up my paperwork from the last attempt, and went through it all again. This time, I also kept pushing. Yay?
Actually, instead of detailing out my rather random process, replete with much backtracking and misreading before coming to a solution, I'll just link to the site I used to compare my answer. It still confused me a little, because I did it by columns, not rows, but I get the same final answer. Hopefully, as I aid, this does not constitue cheating. I did solve it on my own, first.
Here


NEXT PUZZLE:
There are three prisoners in a certain jail, two have normal vision and the third is totally blind. The jailor is a bit of a sadist and likes to play games with his charges. He tellls the prisoners that, from 3 white hats and 2 red hats, he will select 3 and put them on the prisoners' heads. No prisoner will be able to see what colour hat he is wearing. The jailor says he will offer freedom to any prisoner if he can tell the jailer what colour hat he is wearing.

The jailer asks the individual prisoners' if they knw what colour hat they are wearing.
The first prisoner can not tell what colour hat he wears.
The second color prisoner also can not tell what colour hat he wears.
The jailer turns to the blind prisoner and says that since he is blind, he obviously cannot answer the challenge. So the jailer begins to lead them back to their cells.
The blind prisoner, however, stops the jailer and says that he has no need of sight, and that he knows what colour his hat is. He tells the jailer, and is set free.

What colour is the blind man's hat?

[remistress]

White. The only way he could have gone free is if the first guy guessed white and the hat was red. The second guy guess that the hat was white and it was also red. This would only leave white hats.

Unfortunately I don't have any logic puzzles.

[Robert Smith]

Same answer, different reasoning. The only way a prisoner can know his own hat colour is if both the other prisoners are wearing red hats. So, if the first prisoner doesn't know, it's because he doesn't see two red hats. So, if the blind prisoner was wearing red, the second prisoner could safely deduce that he himself must be wearing white, else the first prisoner would have seen two reds. If he can't make this deduction, then it must be that the blind prisoner is wearing white.


I don't have an original problem for you, but I can offer the Monty Hall problem for anyone who hasn't come across it before. Quoting from Wikipedia,

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

The answer may surprise you; it did me.

[noodle]

Same answer, different reasoning. The only way a prisoner can know his own hat colour is if both the other prisoners are wearing red hats. So, if the first prisoner doesn't know, it's because he doesn't see two red hats. So, if the blind prisoner was wearing red, the second prisoner could safely deduce that he himself must be wearing white, else the first prisoner would have seen two reds. If he can't make this deduction, then it must be that the blind prisoner is wearing white.


I don't have an original problem for you, but I can offer the Monty Hall problem for anyone who hasn't come across it before. Quoting from Wikipedia,

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

The answer may surprise you; it did me.

the monty hall problem is a distortion of statistics, imo.

[keith]

the monty hall problem is a distortion of statistics, imo.

truthiness reaches mathematics?

http://math.ucsd.edu/~crypto/Monty/monty.html

click until it gives you the answer you want

[Robert Smith]

the monty hall problem is a distortion of statistics, imo.

In what way? It appears to hold true that changing doors is the choice to make. I couldn't believe it - still not sure why it's so - so I ran up an Excel spreadsheet to test it.

Column 1 (the actual door) =INT(RAND()*3) +1
Column 2 (the guess) =INT(RAND()*3) +1
Column 3 (the door that Monty opens) =IF(INT(RAND()*2)+1=1,MOD(B2,3),MOD(B2+1,3))+1
Column 4 (the result if you stick with your first choice) =IF(B2=A2,1,0)
Column 5 (the result if you switch) =IF(B2=A2,0,1)

Copy these down for a couple of thousand rows or so and sum columns 4 and 5. You'll find the sum of column 4 is consistently around half the sum of column 5. Strange but true...

[noodle]

In what way? It appears to hold true that changing doors is the choice to make. I couldn't believe it - still not sure why it's so - so I ran up an Excel spreadsheet to test it.

Column 1 (the actual door) =INT(RAND()*3) +1
Column 2 (the guess) =INT(RAND()*3) +1
Column 3 (the door that Monty opens) =IF(INT(RAND()*2)+1=1,MOD(B2,3),MOD(B2+1,3))+1
Column 4 (the result if you stick with your first choice) =IF(B2=A2,1,0)
Column 5 (the result if you switch) =IF(B2=A2,0,1)

Copy these down for a couple of thousand rows or so and sum columns 4 and 5. You'll find the sum of column 4 is consistently around half the sum of column 5. Strange but true...

no no, i know how and why it works.
and its one of the reasons why i don't like statistics.

[Rabid Monk]

We discused a problem like this in my Critical Thinking class.
That's the same class i pulled the prisoners' hats problem from. I figured out the prisoner problem, but the doors threw me. Still does, mostly, and it was a large part of the reason I detested that class.
As I recall, it has something to do with the removal of the false choice, and the fact that he opened the door with the goat and not the other door.
I'll leave you there instead of continuing. I've seen the solution to this before (even if I barely remember it), and I can't think of another puzzle at the moment.
Work off what I gave you.

[Rabid Monk]

Okay, nobodies biting, so let's see how well I remember this odd puzzle (it still doesn't make much sense).

When you first choose, each door has a 1/3 chance of being the corect door.
So you've chosen a door that has a 1/3 chance of having the big prize.
The host opens another door, showing a dud, and asks if you want to switch.
The trick here is that the common sense says that the door you chose and the door remaining now both have a 50/50 chance of being correct.
However, as I remember this puzzle (and I think it has something to do with the fact that the host chose a particular door), it does not mean this.
The door you chose still ha s a 1/3 chance of being the correct one.
Of the two remaining doors, there is a cumulative total of 2/3 chance that the prize lies behind one of them.
When one of these doors is opened and does not have the prize behind it, this cumulative 2/3 chance shifts to the remaining door.
So now the door you chose has a 1/3 chance of having the prize. The other door has a 2/3 chance of having the prize.
So 'tis better to switch.
At least, that's how I rememver it.

It still doesn't make much sense, though. I'd be interested to see and experiment that would determing whether the reamining door had the prize 50% of the time of 67% of the time.

Now we wait for Robert Smith to tell me if I remembered correctly, so I can post another puzzle.

[parrythis]

New problem:

You are in a room where there are no metal objects except for two iron rods. Only one of them is a magnet. How can you identify which one is the magnet?

[noodle]

New problem:

You are in a room where there are no metal objects except for two iron rods. Only one of them is a magnet. How can you identify which one is the magnet?

take one rod and touch the tip to the center of the other. if it moves, magnet is in your hand. if not, its the other.

edit: here's an easy one:
if you had an infinite supply of water and a 5 quart and 3 quart pail, how would you measure exactly 4 quarts?

[noodle]

Okay, nobodies biting, so let's see how well I remember this odd puzzle (it still doesn't make much sense).

When you first choose, each door has a 1/3 chance of being the corect door.
So you've chosen a door that has a 1/3 chance of having the big prize.
The host opens another door, showing a dud, and asks if you want to switch.
The trick here is that the common sense says that the door you chose and the door remaining now both have a 50/50 chance of being correct.
However, as I remember this puzzle (and I think it has something to do with the fact that the host chose a particular door), it does not mean this.
The door you chose still ha s a 1/3 chance of being the correct one.
Of the two remaining doors, there is a cumulative total of 2/3 chance that the prize lies behind one of them.
When one of these doors is opened and does not have the prize behind it, this cumulative 2/3 chance shifts to the remaining door.
So now the door you chose has a 1/3 chance of having the prize. The other door has a 2/3 chance of having the prize.
So 'tis better to switch.
At least, that's how I rememver it.

It still doesn't make much sense, though. I'd be interested to see and experiment that would determing whether the reamining door had the prize 50% of the time of 67% of the time.

Now we wait for Robert Smith to tell me if I remembered correctly, so I can post another puzzle.

i'll explain it easier. and i'm approximating percentages.

each door, including the one you just chose, has a 33% chance of correctness. so, in picking a door, you have 33% chance of being right, 66% chance of being wrong.

so, despite the fact that they reveal the one door, that 66% chance of you being wrong still sticks. there's a 66% chance that the one other door left is correct.

[larkmaj]

noodle's I think is the easiest to understand and basically the way I think of it. The trick is to think of your probability of being wrong. The probability of being wrong (or right) on the door you first select DOES NOT CHANGE when the wrong door is opened. Considering that that this probablity does not change and that a wrong door is removed, the probability that the leftover door is correct goes up.

Another explanation

Before you choose:
Door, right, wrong
A, 33, 66
B, 33, 66
C, 33, 66

One door is right, 2 doors are wrong, if you add up the probabilies I listed, the right adds to 100, the wrong adds to 200. This is not a coincidence.

Now, I choose door A, that locks the probability of that door. Either B or C is now opened revealing a wrong choice. This effectively combines B and C and removes 100% probability of a wrong door, so you end up with:
A, 33, 66
B/C, 66, 33

Instead of 50/50 odds you have 66/33 on the door you did not choose. The TOTAL odds are still even that one is right and one is wrong (they both add to 100).


I apologize if these explanations make little sense... I have trouble explaining my logic to others at times.

[Rabid Monk]

New puzzle. The burning ropes:

There are two lengths of rope.
Each one can burn in exactly one hour.
They are not necessarily of the same length or width as each other.
They also are not of uniform width (may be wider in middle than on the end), thus burning half of the rope is not necessarily 1/2 hour.

By burning the ropes, how do you measure exactly 45 minutes worth of time?

[Robert Smith]

It still doesn't make much sense, though. I'd be interested to see and experiment that would determing whether the reamining door had the prize 50% of the time of 67% of the time.

If you have MS Excel handy, try the formulae I posted earlier. They reproduce the situation accurately, I think. It still stinks, though, even if it is true.

By burning the ropes, how do you measure exactly 45 minutes worth of time?

You light one at both ends. Once it's burnt out, light the other. Doesn't matter how unevenly they burn, the one lit at both ends will take only 15 minutes to burn out.

Sorry, no new problem comes to mind right now.

[Rabid Monk]

You light one at both ends. Once it's burnt out, light the other. Doesn't matter how unevenly they burn, the one lit at both ends will take only 15 minutes to burn out.

Sorry, no new problem comes to mind right now.

You've got the right idea, but the timing is off.
(Each rope burns in one hour, not half-an-hour.)

[larkmaj]

I don't have a new one, but:

For the pails:

fill 5, then pour 5->3, dump 3, pour 5->3, fill 5, pour 5->3. There are now 4 quarts in the 5.

The ropes:

Light one rope at one end and the second rope at both ends, when the second rope burns out light the second end of the first rope.

[Rabid Monk]

Yes and yes.

You get to post the next puzzle.

I recommend using Google as a source.