Thread: How Good Are You At Reasoning?

Here's some questions, some of which are quite cool, which you have to be quite mathematically minded to answer. I wanted to see how good people are at thinking rationally but creatively on DV. I threw in a couple of standard psychological experiments too. Just do the ones you can. Feel free to add more.


1. You have a row of 1,000 coins, all of which are heads up. You flip the second, fourth, sixth, and all other even coins over so that they're tails. Then you flip over the third, sixth, and all other coins which are multiples of three. Now you do this for every fourth coin, every fifth coin, etcetera, all the way up to every thousandth coin (which means just flipping the last one).

The question is this: which coins are heads up, and why? There is a concise and surprising answer.


2. What's the maximum number of pieces you can cut a pizza into using a cheese wire (i.e. straight cuts) if you can only cut six times? The pizza is to thin to cut horizontally, and too hot to pick up and move around.


3. There's a 1% chance that the average person has cancer. Somebody with cancer has an 80% chance of testing positive when they go for a scan. Somebody without cancer has a 9.6% chance of testing positive (i.e. getting a false positive). You go for a scan and receive a positive result. What is the probability that you have cancer?


4. 100 people out of a group of 10,000 are tree-huggers. You ask all 10,000 people if they hug trees or not. 80 out of the 100 people who hug trees are honest about it and say yes. 950 out of 9,900 people who don't hug trees also pretend that they do. What fraction of people who claim to hug trees are genuine tree-huggers?


5. I have four cards with numbers on one side and colours on the other. I claim that cards with even numbers are red on the other side. I put the cards in front of you. They are

1, 2, red, blue.

Which cards do you need to turn over to check my claim?


6. In the UK you can drink at 18. There are four people drinking at the bar. One is 21, one is 16, one is drinking coke, one is drinking beer. What do I have to check to make sure nobody is breaking the law?


7. In any group of (two or more) people, there are always at least two people with the same number of friends in that group. True or false?

N.B. you can only be friends with somebody who is friends with you!

"How good are you at reasoning"?

I think I'm OK, how about you?

Okay I hope. But only 15% of professional doctors get the one about cancer correct, so we're in the minority, aren't we? Assuming of course that you can answer it, which you showed no evidence of.

These are tricky. I'm guessing 1 instinctively for the first question and 14 slices for the second. For the 3rd, I'm a little puzzled. 72.32?

1, 5, and 6 are the ones that caught my eye.

1) If we let D(n) denote the number of divisors of n that are not 1, then the coins at positions n such that D(n) is even will be in their initial state.

[edit]
Better:

The coins in positions with odd numbers of divisors will be heads up.

[/edit]

5) To test the claim, one would turn over the card showing two and the blue card.
When turning over the blue card, confirm that there's an odd number on it. (or there would be a blue and even card)
When turning over the card showing two, confirm that it's red.

6) That nobody under 18 is drinking.

1, 2 and 7 are probably the most interesting to think about as puzzles. The first two have especially pretty answers. The middle four are more about human intuition and psychology.

You are right about the odd number of divisors thing, but there is a more concise way of saying it.

Ok ok, squares will be in their original state. For 7, is friendship reflexive?

No, but it is symmetric.

And of course, the point of the coins question is to provide a reason. Anybody can assert an answer after checking a few cases.

Originally Posted by Xei

No, but it is symmetric.

Right. Stupid me. Thanks for seeing through my problem with word selection.

And of course, the point of the coins question is to provide a reason. Anybody can assert an answer after checking a few cases.

The reason is inherent in the answer...?

A coin n will be flipped one time for each divisor of n that's greater than 1?

That's fairly obvious once you look at a few cases if not before?

I'll think about the friends problem but might get bored.

Originally Posted by PhilosopherStoned

Right. Stupid me. Thanks for seeing through my problem with word selection.

I wasn't trying to be sarcastic if that's what you mean... friendship could plausibly be reflexive.

The reason is inherent in the answer...?

A coin n will be flipped one time for each divisor of n that's greater than 1?

That's fairly obvious once you look at a few cases if not before?

Of course, if your answer is 'coins with an odd number of divisors', then it follows immediately. But that's not far off from trivially answering 'coins which are flipped to heads by the given process'. The meat of the question is in explaining why odd number of divisors is equivalent to the more concise condition you stated of being a square.

I'll think about the friends problem but might get bored.

It's quite a quick and linear question (it took me thirty seconds or so, but I'm used to these kinds of things).

Originally Posted by IndieAnthias

1. all of the squares (I did it with 10 coins)
2. 24 (?)
3. 85.2% (?)
4. 8/103
5. 2 & red
6. the age of the beer drinker and the drink of the 16 yearold.
7. true (?)

(?) = guess

1. Yep, why though? (Nobody's nailed this one yet).
2. Close, try to think of how you'd answer it if it were 100 cuts.
3. Nope, but try calculating the probabilities for various things in question 4.
4. Ya.
7. Maybe.

7. True.

Here's my answer to why.

Let n be group size
Let x be the number of friends a person has
Let z be the number of people in the group with 0 friends
Let d be the number of discreet values x may take

For this to be false, group size n must have a d value of n.

A person can only have a maximum of n-1 friends, x cannot be negative, and a person with 0 friends cannot increase the value of x for someone else therefore:
0<= x <= (n-(1+z))

If z=0, d=n-1, therefore for the statement to be false, z must =1.

Since the number of people with zero friends must be positive
0<= z <= n

However, a person with no friends cannot increment the friend count of someone else so for z=1, d = n-z

for z=0 and z=1 d>n, therefore statement is true.

5. I have four cards with numbers on one side and colours on the other. I claim that cards with even numbers are red on the other side. I put the cards in front of you. They are

1, 2, red, blue.

Which cards do you need to turn over to check my claim?

The 2 card and the blue card.

Originally Posted by Xei

A 9.6% chance of the results being false means 9.6% of the people with positive results have no cancer. But this is different from 9.6% of people with no cancer having positive results.

I recommend you draw a rectangle representing a population of 10,000 typical people, then subdivide it into people with cancer with positive tests, people without cancer with positive tests, etcetera.

Since a person without cancer can only have a 9.6% chance of testing positive, no person without cancer can have a >9.6%> chance of testing positive. If a person is tested positive, which assumes a person is 100% positive, then it follows that with exclusion to this assumption, I may have a < 90.4% chance of having cancer. (This was actually the answer before the "original" one I tried to force. Albeit a bit broad, it's still possible. Who needs such silly complex math when you can make fenced guesses!) I win


I'm an engineer undergrad. How sad

Originally Posted by Xei

Of course, if your answer is 'coins with an odd number of divisors', then it follows immediately. But that's not far off from trivially answering 'coins which are flipped to heads by the given process'. The meat of the question is in explaining why odd number of divisors is equivalent to the more concise condition you stated of being a square.

I thought the meat was in figuring out the odd factor thing. The squares thing is (imo)just a cute way to use a well known fact from number theory to cover it up and make it look cool. I hate it when people do that.

At any rate, an integer has an odd number of factors if and only if it's a square.

proof:
(if)
Suppose it's a square, N=n². We need to show that |S| is odd where S is the set of factors and |S| is the number of elements in S. Then for every divisor, d, N/d is also a factor. Hence we get a map f from S to itself given by f(d) = N/d. Note that f(f(d)) = f(N/d) = N/(N/d) = d. Hence we can partition S into distinct sets, S_d = {d, f(d)}. if d =/= n then d =/= f(d) and so each set formed from such d has two distinct factors. But n lays in S and so S_n = {n}. Hence S is partitioned into some numer k of sets with 2 elements and 1 set of 1 elements, and we can count it as |S| = 2k + 1. This is trivially an odd number as it is one greater than the even number 2k.

(only if)
Now suppose it has an odd number of factors. We need to show that it must be square. (This is the part that everybody seems to have forgotten so far.) Using the same labels as above, partition S into sets S_d = {d, f(d)} using the same f. Again, f(f(d))=d and so there is a maximum of two elements in a given partition. Let k be the count of partitions with 2 elements and p be the count of partions with 1 element. Then we have |S| = 2k + p. For |S| to be odd, p must be odd and hence not zero. Hence there is at least one d with |S_d| = 1, and f(d) = N/d = d and so N = d².


Now that's quite excessively nailed

It's quite a quick and linear question (it took me thirty seconds or so, but I'm used to these kinds of things).

Yeah, you're right. Took me longer than thirty seconds and Photo made it to the comp first but I'll post my argument anyways as it's cleaner and actually correct rather than just being very close. ("it cannot be proven false, therefore it's correct"? Flirting with Goedel there...)

For a group of N people, there will be two with the same number of friends.

Proof:
For a person P, let f(P) be the number of friends.

First suppose that there are n people with zero friends. If there are two people with F(P_i) = F(P_j) among the N-n people with friends, then F(P_i) = F(P_j) among all N people as somebody with zero friends won't change the value of F. Hence we can assume that n=0, that is everyobdy as at least one friend.

Hence we have 1 <= F(P) < N for all P.(we're not counting friendship with oneself).

But then there are N values of F(P) to be chosen from among N-1 values. Hence by the pigeon hole principle, at least one value will occur at least twice.

1. oxxoxxxxoxxxxxxoxxxxxxxx etc etc
2. 64 assuming I can reorientate the pizza between slices.
3. I'm a little confused on this one. I'm thinkin 1%. I'll ponder it a little longer though.
4. 80/1030
5. You didn't limit my card flips. I will flip them all.
6. The IDs of those drinking alcohol.....
7. False. You never claimed friendship had to be mutual.

EDIT: looks like you changed the rules on the last question so my answer no longer applies. Don't have time to rethink it at the moment.

Okay, I'll bite: I'll probably fall into all the traps..

1) I'm not atall mathematical, but I notice that the first and second coins remain untouched.. and seeing as you said that the answer was concise and surprising, I'm gonna guess 2; [BTW.. a "stack" of 1000 coins? How does one flip without dismantling the stack?]

2) Using 2D for a few minutes, I managed 19; but some sort of nagging voice is saying something about dual parallel perspective in 3D;

3) 4 in 5;

4) 80/1030ths = 40/515ths;

5) a) All of them: any card could have been tampered with;

b) None of them: make sure that the cards are placed on a glass table and look from underneath;

6) First you need to check if the bar needs a license: it could be a bar built in one's living room, i.e. private- the UK 18-year-old thing only applies in certain premises;

7) False: could be any group of people who don't like each other.

Originally Posted by Oneiro

7) False: could be any group of people who don't like each other.

If they don't like each other. They all have 0 friends in the group so it would be true.

Originally Posted by LikesToTrip

1. oxxoxxxxoxxxxxxoxxxxxxxx etc etc
2. 64 assuming I can reorientate the pizza between slices.
3. I'm a little confused on this one. I'm thinkin 1%. I'll ponder it a little longer though.
4. 80/1030
5. You didn't limit my card flips. I will flip them all.
6. The IDs of those drinking alcohol.....
7. False. You never claimed friendship had to be mutual.

EDIT: looks like you changed the rules on the last question so my answer no longer applies. Don't have time to rethink it at the moment.

1. What do you mean by etcetera?
2. How come?
4. Yep.
5. But I did ask which ones you need to flip, and you don't need to flip them all.
6. Why would you need their IDs? The question tells you their ages.
7. Are you sure it'd be false? But okay, I clarified it.

Originally Posted by Oneiro

Okay, I'll bite: I'll probably fall into all the traps..

1) I'm not atall mathematical, but I notice that the first and second coins remain untouched.. and seeing as you said that the answer was concise and surprising, I'm gonna guess 2; [BTW.. a "stack" of 1000 coins? How does one flip without dismantling the stack?]

2) Using 2D for a few minutes, I managed 19; but some sort of nagging voice is saying something about dual parallel perspective in 3D;

3) 4 in 5;

4) 80/1030ths = 40/515ths;

5) a) All of them: any card could have been tampered with;

b) None of them: make sure that the cards are placed on a glass table and look from underneath;

6) First you need to check if the bar needs a license: it could be a bar built in one's living room, i.e. private- the UK 18-year-old thing only applies in certain premises;

7) False: could be any group of people who don't like each other.

1. I changed the question to a row of coins if that makes it clearer for you.
2. That is close to my answer. Don't worry about 3D, the pizza is 2D.
3. Way off.
4. Yep.
5. No tampering, no glass tables. None of these are trick questions.
6. See above.
7. What LikesToTrip said.

1. All primes will be tails. Since all non primes (save 1 which is not flipped) must have an even number of factors they will be flipped an even number of times and thus be heads.
2. I managed 21 playing around trying to get the maximum number of intersections with each cut.
3. 7.76% chance of actually having cancer with a positive result (or 800/10304 more precisely). For bonus points you have a 0.22% (or 200/89696 more accurately) chance of having cancer when you get a negative result.
4. 80/1030 obviously.

Will look at the rest later.

Originally Posted by Photolysis

1. All primes will be tails. Since all non primes (save 1 which is not flipped) must have an even number of factors they will be flipped an even number of times and thus be heads.
2. I managed 21 playing around trying to get the maximum number of intersections with each cut.
3. 7.76% chance of actually having cancer with a positive result (or 800/10304 more precisely)
4. 80/1030 obviously.

Will look at the rest later.

1. Along the right lines, but doesn't check out with the actual numbers (2 is non-prime but tails for example).
2. Pretty close. If you formalise your technique you should get it right. What about n cuts?
3. Yep; via what general method did you do this? Why was 4 a more obvious question?

Originally Posted by LikesToTrip

1. There will always be 2 more tails in between each heads all the way up to 1000. Why it does this, I have no clue. I'm sure there is some mathematical or logical equation you could write that sums it up, but that is beyond me.
2. Cut in in half stack, cut in half stack, cut in half stack. You end up with 64 pieces.
5. Herp a derp. I'm retarded. Flip the 2.
6. You said four people. 21, 16, beer, coke. Aren't those the 4 people? We don't know the drinks of the 21/16 and we don't know the ages of the beer/coke. So we need to check the drink of the 16 and the age of the coke. Did I misinterpret the question?
7. I think the answer is now true. I'm not completely sure though. I'll ponder it more.

1. Try numbering the coins for a more concise description of which ones are flipped.

The point of this question is in giving a good reason. I recommend you mull it over. None of these questions require mathematical knowledge or skill. They just require you to think like a mathematician, which is the important bit.
2. God damn you lateral thinkers. Okay no, no moving the pizza around.
5. Nope. You fell into the psychological trap that the vast majority of people do.
6. Yes that's completely right, and yes it is an easy question. The funny thing is this is logically identical to question 5.

Originally Posted by Xei

1. Along the right lines, but doesn't check out with the actual numbers (2 is non-prime but tails for example).

This brought me joy to an otherwise dull day.

Originally Posted by Xei

1. What do you mean by etcetera?
2. How come?
4. Yep.
5. But I did ask which ones you need to flip, and you don't need to flip them all.
6. Why would you need their IDs? The question tells you their ages.
7. Are you sure it'd be false? But okay, I clarified it.

1. There will always be 2 more tails in between each heads all the way up to 1000. Why it does this, I have no clue. I'm sure there is some mathematical or logical equation you could write that sums it up, but that is beyond me.
2. Cut in in half and stack, cut in and half stack, cut in half and stack. You end up with 64 pieces. 2^6=64
5. Herp a derp. I'm retarded. Flip the 2.
6. You said four people. 21, 16, beer, coke. Aren't those the 4 people? We don't know the drinks of the 21/16 and we don't know the ages of the beer/coke. So we need to check the drink of the 16 and the age of the coke. Did I misinterpret the question?
7. I think the answer is now true. I'm not completely sure though. I'll ponder it more.

Oooh! Clever.. I said I'd fall into all the traps..

(Is zero a number?)

3) Oh yeah! I just read the question properly.. 1%;

It's definitely not 1% either.

I was referring to my brain power..

2) I've managed 20 in 2D;

A photon checks into a hotel and the concierge says: "Any luggage, Sir?" The photon replies: "No.. I'm travelling light.."