• Konstantin Surkov (unregistered) in reply to Anonymous Coward
Anonymous Coward:
Grandpa's puzzle sounds like a trick. The color of your hat is independent from the color of the other hats, so seeing the other hats gives you no information. It's possible that order of guessing / passing could be used to offer information, but that's been ruled out as well by the fact that everyone has to go simultaneously with no communication. So, basically, you've been given useless information and are stuck with no better than a guess. Pick one person and have them guess wildly while the others pass, you have a 50% shot.
The puzzle has very simple solution. It will actually work not only with two colors, but with any number that is less or equal to the number of people in the room. No guessing involved.
• Anonymous Coward (unregistered) in reply to quamaretto

People continuing to attack brainteasers: You are at most attacking dumb interviewers who apply them wrong. As others have noted, it's the thought process that's interesting. Bad interviewers will continue to be bad interviewers, counting off TLA's and buzzwords.

Also, I can't believe that people are still ignoring the "no communication" bit. That means "no communication," including order of passes and everything. For what it's worth, my comment about your hat being independent is correct, though the 50/50 solution is wrong as noted. As someone else hinted with the reference to Marilyn, this is the Monty Hall Problem in disguise; while your personal chance of your hat is still 50/50, "the guy who sees two hats the same color" is in fact dependent on some variables.

quamaretto:
I confess that I have "How Would You Move Mt. Fuji" at home somewhere, and worked on a large number of the riddles.

The "five pirates" riddle (google it) is still one of my favorite riddles ever, and the inmates and two switches is probably the second.

A pirate ship captures a treasure of 1000 golden coins. The treasure has to be split among the 5 pirates: 1, 2, 3, 4, and 5 in order of rank. The pirates have the following important characteristics: infinitely smart, bloodthirsty, greedy. Starting with pirate 5 they can make a proposal how to split up the treasure. This proposal can either be accepted or the pirate is thrown overboard. A proposal is accepted if and only if a majority of the pirates agrees on it. What proposal should pirate 5 make?

This is one of those backwards induction game theory problems, right? Here's a gander at it.

The 1 pirate case is easy. 1000 doubloons for me. The 2 pirate case is also easy. 1 gets all 1K for free if he vetoes the proposal, so 2 will offer 1 all 1000 gold coins in hopes that 1 spares his life.

3 pirates case: 1 will get all the treasure otherwise, so he's impossible to payoff and will always vote against. Luckily, 2 is really easy to bribe, since he gets nothing in the above case. 2:1 3:999

4 pirates case: Now 1 is on the outs, expecting nothing if it comes to 3 pirates. 2 isn't expecting much either, so they can be bribed. 1:1 2:2 4:997

5 pirates case: Assuming the above solution will pass for sure, getting three votes for 5's offer would involve 1:2 3:1 5:997 This is a better deal for 1 and 3 than the 4 pirate case, so it should pass.

• Anonymous Coward (unregistered) in reply to jwray
jwray:
Grandpa:
Three players enter a room and a red or blue hat is placed on each person's head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players' hats but not his own.

It's pretty obvious that all strategies are equally effective here, because the only information a player gets before guessing is independent of his own hat color.

What you are saying is absolutely true, but does not prove the 75% strategy to be wrong. A single person has no information to determine his own hat color, but the strategy relies on the fact that the information being provided (the other hats) is used to determine who should be speaking!

• Franz Kafka (unregistered) in reply to TimS
TimS:
Ideally, these aren't riddles, per se, but rather problems that require solving. For instance, asking where survivors are buried isn't going to teach you anything about the applicant, but something that involves doing some math to work out various possibilities is going to give you an idea about how the person approaches a problem.

Really, it's the approach that you're trying to see, not the answer. If someone shouts out the answer immediately, and recites the answer, as memorized, from a book of problems, that's an immediate check in the "no" column for me.

Great, so if you ask a riddle I've already heard, then I'm DQed. Why don't you, instead of a stupid riddle, ask something more open ended, like how to build an exchange clone, what's important, and what tradeoffs you'd make in the design?

• jwray (unregistered) in reply to vt_mruhlin
vt_mruhlin:
I actually saw a good discussion of this one, and how that particular example can be used to guage programming skills. Correct solution was to do a binary sort on on. Put 4 coins on each side of the scale, discard the lighter half, weigh the remaining 4, then weigh the remaining two...

That is suboptimal. You can eliminate two thirds of the quarters in each weighing, so you only need to use the scale twice.

• bramster (unregistered) in reply to Jim
Jim:
vt_mruhlin:
This only works in the case where you know if the odd coin is lighter or heavier. If you check the post above, it could be either, in which case splitting into three groups does not help you: if the one of the weighed groups is heavier, it does not tell you which of the two has the odd coin, so you may need to do more weighing (2-4 weighings total). So in this instance, binary search is the optimal solution.

If you don't know if the odd coin is heavier or lighter, the binary search will have exactly the same problem as the ternary solution. I haven't worked through the worst case scenarios, but in both cases it should be significantly more than 4 weighings.

Minimum two weighings, Maximum 5

• (cs)

This reminds me of one very painful interview. The interviewer asked me a brainteaser. It was one I already knew, but he messed up a couple of the details, resulting in a puzzle with no possible solution. I thought it was a trick and tried to be clever by mathematically proving there was no solution. Needless to say, I wasn't offered the position.

• Franz Kafka (unregistered) in reply to fist-poster
fist-poster:
Concerning the 8 coins.

If you don't know if one is heavier or lighter, how is binary search going to work? You weigh 4+4. One side is heavier. Is it because the heavier side has the heavy coin or is it because the lighter side has the light coin?

split the coins into 2 groups. 1: weigh one group (2+2). if the scales balance the odd duck is in the other group 2: repeat for the group with the odd coin you now have 2 coins, one of which is odd 3: weigh one of those coins against the known good coins. if the scales balance, it's the other one.

• Ken (unregistered) in reply to fist-poster
fist-poster:
Concerning the 8 coins.

If you don't know if one is heavier or lighter, how is binary search going to work? You weigh 4+4. One side is heavier. Is it because the heavier side has the heavy coin or is it because the lighter side has the light coin?

Take the heavier group and split it 2/2. If they weigh the same, then it's a lighter coin in the other four. (Two more weighings -- total 4.) If they are different, then it's a heavy coin in the heavier half currently on the scale. (One more weighing -- total 3.)

• Nonymous (unregistered) in reply to Milkshake
Milkshake:
```(fox)(grain) -----(chicken)----->
(fox)(grain) <------------------- (chicken)
(grain) -------(fox)-------> (chicken)
(grain) <----(chicken)------ (fox)
(chicken) ------(grain)------> (fox)
(chicken) <------------------- (fox)(grain)
-----(chicken)-----> (fox)(grain)
After all that...
(interview) -----(me)----->
```

Shooting the fox is definitely more efficient.

why call boeing when you have google?

empty weight for a 747-100 is 358100 lbs, and the 747-400 weighs in empty at 399000 lbs

That doesn't tell you the weight of a specific plane (who knows what they have in there). The answer is obvious (and it doesn't involve building a barge, Alex). You measure the air pressure in the tires and the contact area with the tarmac. Problem solved with essentially no cost. Do I get the job?

• Hmmmm (unregistered) in reply to Anonymous Coward
Anonymous Coward:
Three women want to [redacted] with you, but you only have two condoms. How do have safe [redacted] with all three?

This was an actual interview question!

Oral (you), oral (her), anal, and if the condoms are edible, you each have a snack when you're done

• jbrecken (unregistered)

There's a solution to the coin problem that requires exactly three weighings, always. (No best case or worst case options) It works because the number of coins here was a power of two.

Set half (4 coins) aside. Weigh the others 2 against 2. If they balance, the odd coin is is the 4 you set aside. If they don't the odd coin is in the 4 you weighed.

Repeat the process with the 4 coins you know to contain a bad one. Set half (2 coins) aside. Weigh the others 1 against 1. If they balance, the odd coin is one of the 2 you set aside. If they don't, the odd coin is in the 2 you weighed.

Now you have 2 coins and you know one of them is bad. Set one aside and weigh the other against one of the six good coins. If they balance, the odd coin is the one you set aside. If they don't balance, the odd coin is the one you weighed.

• Darien H (unregistered) in reply to kimos
Kimos:
Turn on switch A for a couple minutes. Turn A off then turn on switch B. Quickly open the box. The bulb that's on is switch B, the bulb that is off but is still warm to the touch from being lit is switch A, and the remaining bulb is switch C.

Ah, but do you even know the initial settings of the switches? That is, are they labeled "on/off"? Could some of them be mounted backwards? They could all be switched in the same direction while the bulbs have different states. (Let's assume there is a one-to-one relationship of switch to bulb, although the original phrasing leaves some weasel room.)

As a corollary to your solution, if all the bulbs are initially on, you find the cool bulb and the unlit-but-warm bulb and it still holds.

Here's the "sneaky solutions"...

1. When you open the box, trace the wiring!
2. Figure out which switches control the side bulbs by measuring the temperature of the outside of the opaque box. (If anyone calls you on this "It's a black box!" just tell them that you deal with the real world and that you're thinking outside the box.)
3. If the box isn't battery-operated, Get your electrical tools and check the energy usage of the box from it's power cord. You can at least establish which direction is on or off for each switch.
• jwray (unregistered) in reply to Ken
Ken:
Select one person, who will give a predetermined answer. The rest remain silent. You have a 50-50 chance of being correct.

Bingo. You can't do better than 50%.

• (cs)

Here's another... I don't remember where I heard it:

Given a random 32-bit integer, write some code that will count the number of set bits (i.e., 1's) in O(n) time, where n is the number of set bits. Use your favorite language that has bit-wise operators.

• Anonymous Coward (unregistered)

• anon (unregistered) in reply to vt_mruhlin

How about, put 3 and 3 on the balance, keeping 2 aside. If it balances then just compare the 2. Otherwise take the heavy side (3 coins) and compare two, same thing, if they balance it's the odd one out, else you know it's the heavier one. Just 2 weighings, not 3.

• Ken (unregistered) in reply to jwray
jwray:
Ken:
Select one person, who will give a predetermined answer. The rest remain silent. You have a 50-50 chance of being correct.
Bingo. You can't do better than 50%.
Yes, you can. I saw the 75% solutions, and agreed that my instincts were wrong here.
• Ken (unregistered) in reply to anon
anon:
How about, put 3 and 3 on the balance, keeping 2 aside. If it balances then just compare the 2. Otherwise take the heavy side (3 coins) and compare two, same thing, if they balance it's the odd one out, else you know it's the heavier one. Just 2 weighings, not 3.
Again, re-read the specs. You do not know whether the coin is heavier or lighter.
• (cs) in reply to anon
anon:
How about, put 3 and 3 on the balance, keeping 2 aside. If it balances then just compare the 2. Otherwise take the heavy side (3 coins) and compare two, same thing, if they balance it's the odd one out, else you know it's the heavier one. Just 2 weighings, not 3.

They're arguing semantics because no one said whether the "odd coin out" was heavier or lighter than the others.

• (cs) in reply to jwray
jwray:
Ken:
Select one person, who will give a predetermined answer. The rest remain silent. You have a 50-50 chance of being correct.

Bingo. You can't do better than 50%.

There are three options though. Blue, red, and pass. If you were forced to say blue or red, then you would be correct. As it stands, you are not.

• jwray (unregistered) in reply to v.
v.:
Agree beforehand to shout out the answer only if both the other hats are the same color. Should only fail when the three hats are the same color, which should only be 2 out of 8 possible combinations. Thus, my odds at winning are 75%.

That works, strangely.

• (cs) in reply to Patrick McCormick
Patrick McCormick:
You people, who think that one person should make a 50% guess at the hat color, fail at probability (or learned it from Marilyn vos Savant). The correct strategy is for each person who sees the others wearing the same color hat to guess the opposite color, and for those that see one of each hat on the other heads to pass. This has more than a 50% chance of getting a winning condition.

I love all these people who think this is an obvious answer. This thing has been floating around the Internet for a little while now, most recently showing up on Digg. Nice try with your smug little "oh I'm so smart" attitude. Better luck next time.

• Ken (unregistered) in reply to Anonymous Coward
Anonymous Coward:
If Monty were to truly pick a door/box at random, then 1/3 of the time, the "wrong" one would be exposed. Then, out of the 2/3 of the times that remained, the statistics do agree that you have a 50% chance of having the good prize.

However, Monty has knowledge about the locations of the prizes, and will never reveal the "wrong" box accidentally, leaving you with the original 1/3 chance of having the good prize, and you should switch.

• unknown (unregistered) in reply to akatherder

Hat problem:

People are saying that you can get a 75% chance by guessing the opposite colour IFF the hats you see are the same colour.

Whilst it is true that there is a 75% chance that any 3 hats will have that kind of distribution, you are missing conditional probability rules;

given that the other 2 hats you see are the same colour, the probability that your hat is the opposite colour is only 50%

• Some Guy (unregistered) in reply to Anonymous Coward
Anonymous Coward:
Three women want to [redacted] with you, but you only have two condoms. How do have safe [redacted] with all three?

Do the employees at this company face this problem a lot? If so, sign me up! :-)

But to answer the question: Technically you only need one condom. That way it's safe for you, anyway.

• Fuz (unregistered) in reply to aardmoose

int f(uint32_t n) { int count = 0; while (n) { n = n & (n-1); ++count; } return count; }

I got the bit count one in an interview, and none of them had seen this solution before!

Of course, the constant-time solution to this problem (for 32 bits) is even better...

• Anonymous (unregistered) in reply to aardmoose

count = 0; for(i = 0; i < 32; i++) { count += (integer >> 1) & 1; }

Constant time, and therefore O(n). Remember, since the size of the integer is constant, so is the running time.

• Undefined Reference (unregistered) in reply to bramster
bramster:
Jim:
vt_mruhlin:
This only works in the case where you know if the odd coin is lighter or heavier. If you check the post above, it could be either, in which case splitting into three groups does not help you: if the one of the weighed groups is heavier, it does not tell you which of the two has the odd coin, so you may need to do more weighing (2-4 weighings total). So in this instance, binary search is the optimal solution.

If you don't know if the odd coin is heavier or lighter, the binary search will have exactly the same problem as the ternary solution. I haven't worked through the worst case scenarios, but in both cases it should be significantly more than 4 weighings.

Minimum two weighings, Maximum 5

Nope, Min 2, Max 3.

{0,1,2} < {3,4,5} => {0,3} < {1,4} => {0} < {3} => (0, lighter) {0} = {3} => (4, heavier) {0,3} > {1,4} => {0} < {3} => (3, heavier) {0} = {3} => (1, lighter) {0,3} = {1,4} => {0} < {5} => (5, heavier) {0} = {5} => (2, lighter) {0,1,2} > {3,4,5} => {0,3} > {1,4} => {0} > {3} => (0, heavier) {0} = {3} => (4, lighter) {0,3} < {1,4} => {0} > {3} => (3, lighter) {0} = {3} => (1, heavier) {0,3} = {1,4} => {0} > {5} => (5, lighter) {0} = {5} => (2, heavier) {0,1,2} = {3,4,5} => {6} < {7} => {6} < {8} => (6, lighter) {6} = {8} => (7, heavier) {6} > {7} => {6} > {8} => (6, heavier) {6} = {8} => (7, lighter) {6} = {7} => (8, UNKNOWN)

• jwray (unregistered) in reply to jwray

In the problem of 8 quarters, when you don't know whether the odd one out is heavier or lighter, starting with a 4v4 weighing doesn't give you any information. You could put half the quarters off of the scale, and half the remaining quarters on each side of the scale. If the scale is balanced, you eliminate the ones on the scale. Otherwise, you eliminate the ones off the scale. 3 weighings.

• Fuz (unregistered) in reply to Anonymous
Anonymous:
count = 0; for(i = 0; i < 32; i++) { count += (integer >> 1) & 1; }

Constant time, and therefore O(n). Remember, since the size of the integer is constant, so is the running time.

"O(n) time, where n is the number of set bits."

Yours may be constant time, but it's not O(n).

• drew (unregistered) in reply to Bezalel

guess which color correctly?

• PW (unregistered)

Fill both jugs.
Tip both jugs so half of the water pours out from each (waterline from corner to corner). Pour the 3 litre jug into the 5 litre jug (1.5 + 2.5 = 4).

This assuming the jugs are not some odd shape, which you can quickly determine visually.

• Anonymous (unregistered) in reply to Fuz

Big O notation expresses an upper bound to the running time. So yes, any program that is O(1) is also O(n). What you're talking about is capital theta notation, which specifies a tight bound rather than an upper bound.

• (cs) in reply to unknown
unknown:
Hat problem:

People are saying that you can get a 75% chance by guessing the opposite colour IFF the hats you see are the same colour.

Whilst it is true that there is a 75% chance that any 3 hats will have that kind of distribution, you are missing conditional probability rules;

given that the other 2 hats you see are the same colour, the probability that your hat is the opposite colour is only 50%

You're actually calculating the probability that a coin flipped three times will be heads every time (or tails every time).

First flip doesn't matter. Then you have a 50% chance twice.

1.00 * .5 * .5 = .25

So you have a 25% chance of three consecutive heads (or three consecutive tails).

You have a 75% chance of success if the three people follow the rules:

1. Pass if you see two different colored hats
2. Say the opposite if you see two of the same colored hats
• Some Guy (unregistered) in reply to unknown
unknown:
Hat problem:

People are saying that you can get a 75% chance by guessing the opposite colour IFF the hats you see are the same colour.

Whilst it is true that there is a 75% chance that any 3 hats will have that kind of distribution, you are missing conditional probability rules;

given that the other 2 hats you see are the same colour, the probability that your hat is the opposite colour is only 50%

Sigh

Look, no one's saying that if the two hats you see are the same color, yours has a 75% chance of having the other color.

But if everyone who sees two hats of the same color guesses that they have the opposite color, and everyone who sees two different color hats abstains from guessing, then this strategy will work for 75% of the 8 possible hat combinations. If you don't believe it, write down the 8 combinations, figure out who guesses what color for each combo, and you'll notice that the only time when this strategy fails is when all three hats are the same color, which is 2 of the 8 cases.

• (cs) in reply to jwray
jwray:
Grandpa:
Three players enter a room and a red or blue hat is placed on each person's head. The color of each hat is determined by a coin toss, with the outcome of one coin toss having no effect on the others. Each person can see the other players' hats but not his own.

It's pretty obvious that all strategies are equally effective here, because the only information a player gets before guessing is independent of his own hat color.

And it's pretty obvious you don't understand what you're saying. Proof by example: strategy of "no one guess." Fails 100% of the time. Run some sims if you don't understand probabilities. Pre-agreeing that one person should guess randomly gets you 50% wins.

Having everyone stand silent if they see two different colors, and guess the opposite if they see two identical colors gets 75%. The eight possibilities are each equally probable. You do only get information about the other hats, but that's enough to eliminate many of the possible answers. Part of the key is that the information about the other hats determines whether or not you make a guess at all, like dynamically choosing who is making the guess.

• Dan (unregistered) in reply to Anonymous Coward

Yeah, I have been asked that one before. It would be OK if they have come up with that one themselves. If you are going to ask these questions. At the very least put some efforts into preparing the questions and come up with an original one. Not one you copied off from the web or from a riddle book! So now each time before I go to an interview, I have to read up on the riddle books as well?!

• SteveD (unregistered) in reply to Rodyland

Manhole covers aren't round (at least not in the UK). Over 90% are either square or a rectangle made up out of two triangles. Why? To stop people stealing them by rolling them away. Where in the world has round manholes?

• (cs) in reply to SteveD
SteveD:
Manhole covers aren't round (at least not in the UK). Over 90% are either square or a rectangle made up out of two triangles. Why? To stop people stealing them by rolling them away. Where in the world has round manholes?

It's the opposite in the USA. They make them round so sewer workers can move them easier (and so they can't fall in the hole as mentioned earlier). The covers weigh a ton so i couldn't imagine it would be a very popular theft item.

• Lazgen (unregistered) in reply to Rodyland
Rodyland:
- You've got a chicken, a fox, and a bag of grain on one side of the river, and a boat that holds you and one of the chicken/fox/grain. If the fox is left alone with the chicken, he'll eat it. If the chicken is left alone with the grain, he'll eat it. How do you get all across the river so that nobody/nothing gets eaten.

Take the Chicken Over, return and take the fox over. Upon returning for the Grain, take the chicken back over. Take the grain. Return for the chicken.

It's not that hard.

• unknown (unregistered) in reply to poochner

Poochner:

So you are standing there and you see 2 red hats on the other guys. You know that - it is a priori information now.

You also know that your hat was chosen randomly. Are you telling me that there is a 75% chance that your hat is blue?

P(Your Hat == Blue | Other Hats == Red) is 0.5

This is because the two events are independent.

• unknown (unregistered) in reply to Some Guy

Sigh (I can do that too)

there are 2 possible groups of people.

A) Those that can see the others have hats of the same colour B) those that can see that the others have hats of different colours.

You are telling me that, even though the colour of your hat is randomly chosen, that these groups have different properties and that A can make a more informed decision?

You are wrong. The events are inependent. It does not matter what you see.

• Undefined Reference (unregistered) in reply to unknown
unknown:
Poochner:

So you are standing there and you see 2 red hats on the other guys. You know that - it is a priori information now.

You also know that your hat was chosen randomly. Are you telling me that there is a 75% chance that your hat is blue?

P(Your Hat == Blue | Other Hats == Red) is 0.5

This is because the two events are independent.

Yes. You have a 50% chance of answering correctly any time you choose to answer.

Now what is the probability that you choose to answer.

Further, when you are going to answer incorrectly, what is the probability that your group following this strategy would have otherwise gotten it correct.

Basically, you arrange so that all people answer incorrectly simultaneously. But when someone would answer correctly, he is the only person to attempt an answer.

• (cs) in reply to unknown
unknown:
Poochner:

So you are standing there and you see 2 red hats on the other guys. You know that - it is a priori information now.

You also know that your hat was chosen randomly. Are you telling me that there is a 75% chance that your hat is blue?

P(Your Hat == Blue | Other Hats == Red) is 0.5

This is because the two events are independent.

You're looking at it in the perspective of one event in time and therefore you are calculating the probability of the wrong thing.

There are 8 combinations of hats that the three of you can be wearing. Assume red=0 and blue=1 (since R's and B's look a lot alike).

000 Everyone guesses 1 and gets it wrong 001 The two 0's say nothing. 1 says 1 and is correct. 010 The two 0's say nothing. 1 says 1 and is correct. 100 The two 0's say nothing. 1 says 1 and is correct. 110 The two 1's say nothing. 0 says 0 and is correct. 101 The two 1's say nothing. 0 says 0 and is correct. 011 The two 1's say nothing. 0 says 0 and is correct. 111 Everyone guesses 0 and gets it wrong

Now follow the previously agreed upon rules:

1. pass when you see two different colored hats on the other two guys
2. say the opposite color when you see two of the same colored hats on the other two guys.

This only fails when you have 000 or 111 (in which case it fails spectacularly because you all guess AND you all guess wrong).

• unknown (unregistered) in reply to Undefined Reference

Yes,

whenever you choose to answer, the prob = 0.5 that you are correct.

If all 3 answer then the prob is 0.875 that someone guesses wrong and you all lose.

If 2 answer (doesn't matter which) prob is 0.75 that you lose.

If 1 answers (dones't matter which one) prob is 0.5

If 0 answer , prob is 1.0 that you lose

• Robot (unregistered)

We use brainteasers in our interviews as a way to see how the interviewee thinks and reacts to not being able to figure it out. The wrong answer is "brainteasers suck, i give up."

• KM (unregistered) in reply to unknown

That's not what matters. Everybody's stuck on what one particular person does. That's not what matters. What matters is what all three people do together.

Look at all the scenarios. Persons A, B, C, given hats 1 or 0. Strategy of anybody who sees two identical hats calling out the opposite, and everybody else stays silent. Here are the possible hat assignments,

A B C 0 0 0 Everybody says 1, loss. 0 0 1 C says 1, win. 0 1 0 B says 1, win. 0 1 1 A says 0, win. 1 0 0 A says 1, win. 1 0 1 B says 0, win. 1 1 0 C says 0, win. 1 1 1 Everybody says 0, loss.

• YourMoFoFriend (unregistered)

It is frightening how many people can't read or comprehend the problem definition. What's worse is that many of those brush off "puzzle on an interview" as useless while it clearly shows that at the very least those questions test your comprehension abilities. Now, there will always be stupid interviewers who will expect you to know or figure out the right answer completely missing the point of the exercise. Oh well, perhaps you don't want to work with those people anyway. I've been asked those questions and I asked them myself and I can tell you, you can find out a lot about someone by the way he handles the challenge. A typical question I used was "How many trips a tooth fairy makes per night". Here are some answers I got:

• without a pause: "3000"
• "there is no tooth fairy"
• a refusal to answer a "stupid question unrelated to the position"
• an attempt to estimate number of teeth lost per night by kids based on population of countries that do believe in tooth fairy

Guess which answer is more likely to land a job for a candidate... I'm not saying that one is "righter" than the others, it just I wouldn't want to work alongside someone who when presented with a nontrivial problem dismisses it outright, just pulls a guess out of his a\$\$ or goes on the offensive because the question is "not related to the position he's interviewing for".

CAPTCHA: craaazy - how appropriate :)