• JBP (unregistered) in reply to v.

    Two points:

    First, I've seen a lot of the same mistake on the hat problem. The whole "X chances in 8" method of calculating only works if you have not seen the other hats. Think of it in terms of coin flips: the odds of flipping one head in a row is 1/2. It doesn't matter many heads you flipped before, the next one is still a 1/2 chance. Likewise, once you know what the other two hats are, your hat color odds collapse to a 1/2 chance of either color.

    Second, I think I know why the brainteaser persists in the interview. It may not test programming skills, but it's an excellent conversation topic for programmers (135+ posts can't be wrong :).

  • (cs) in reply to akatherder
    akatherder:
    The interviewer wants to find out if you require a 20 page document explaining every functional requirement for each method. Or can they say "Hey, go write BlahBlahMethod()." and you can basically figure out what they need.
    Hm. Which of the two alternatives do you consider preferable?

    I sure as hell wouldn't want someone who'll "figure it out" on his own when faced with incomplete requirements.

  • Carl (unregistered)

    The answer to the Boeing 747 question, obviously, is look it up on Google. Uh, I mean Microsoft Live Search. Wait, I meant to say Live Search! Damn!

  • Carl (unregistered) in reply to KattMan

    How can each guy have sex with each girl safely?

    By not telling the other girl?

  • klagermkii (unregistered) in reply to Chaz

    Just want to add my own thought on the hat thing because I was getting stuck with the probability of it.

    I was pretty adamant that it was a 50% chance of being right with the hat problem because I kept on thinking "We have 3 independent variables A, B, C. Each has a probability of 50% of being Red or Blue. We randomly pick a spokesman and they guess."

    But there's a different probability altogether that changes the odds. "What are the odds that two people have one colour hat, and that a third has the same colour hat?" and NOT "What are the odds that A has one colour hat, and B & C have the same opposite coloured hat". The only reason this works is because ANYONE can call out the answer, not just a pre-picked spokesman.

    So I kept on thinking, "Well if A has red, B has blue, and C has red then looking from A's perspective, I wouldn't say anything and just pass.", but if you turn it around and look at it from B's perspective he would call it correctly.

    Going back to the tables and looking at the different occurances from A, B, and C's perspective should then show where the 75% comes from.

  • (cs) in reply to KattMan
    KattMan:
    KM:
    unknown:
    Okay,

    what is instead of 3 people/hats, we have 1000000.

    You stand there and see that EVERYBODY elses hat is blue.

    Would you risk 10000 to win 100 that your hat is red?

    That's not the bloody point. There is no fixed "you". Look at the scenarios table. Point out a flaw, please.

    That exactly is the point. If you were the one seeing that all other hats are the same color, you should be the one to answer and say the opposite. It matters not the number of hats, whether 3 or 3000000.

    Of course if matters. If you have more than 3 hats you might as well choose 1 person to guess. With 4 hats, following the same strategy, you'd get 16 possibilities; 2 wrongs (all hats the same color), 8 right answers (1 hat different from the other) and 6 'all pass' (the rest, eg. two hats of each color). That's a 50% chance of winning. With five hats, this goes down to 10/32, with six hats it's 12/64. With 3000000 hats, that's (3000000*2)/(2^3000000) which is just about 0% chance of winning.

  • (cs) in reply to tchize
    tchize:
    Interview for a human ressource:

    You ask a teaser to a candidate, he give you a better answer that what is in you list of accepted answer. ... but you still have 2 candidates to interview ... However, the candidate says it's now or never. ...he won't wait for more than 10 minutes,

    I've seen people like that. He's probably very clever, but he's arrogant. He wants to get his own way, and damn the cost to others. It's better not to hire blackmailing pricks like that. Tell him absolutely no deal.

  • JBP (unregistered) in reply to JBP

    "Do I contradict myself? Very well then, I contradict myself" -- Walt Whitman

    I just saw the flaw in my own logic. Your own hat color is independent of the other hats' colors. What is not independent is the total count of red hats. Now I vote for the "if two hats are the same color, vote the other, else pass" strategy.

    The embarrassing part is that it's not the two semesters of probability I just took that convinced me, it's the guy who actually coded the thing up, ran his code and came up with 75.2%.

  • barf indeedy (unregistered)

    You know, usually I don't care for these long stories... But I gotta say, I really really liked this one for some reason.. :D

    A friend of mine was asked all these ridiculous questions, ended up grilling me, I suck at puzzles, and yet after seeing the answers, remembered them. Later I had a Boss who fired off all these questions to a new potential employee. And he was as dumbfounded as I had been. But since I had seen these very same ridiculous puzzles before, as well as the answers that went along with them, I solved all the damn puzzles on the board in a few minutes. People were baffled. So I had to tell them.. "I've seen 'em before. Already knew the answers."

    Funny to think that someone's potential new job hangs in the balance based off of "how do you connect all the points using only four lines". Yeah, what a way to determine the quality of your dev team... (Although I'd like to think we turned out alright)

  • (cs) in reply to Mark
    Mark:
    Well I'm not sure if there is much point in posting a comment given the number, but I seriously had an interview (for a job I most definitely didn't get). I was asked "Using nothing but your own reasoning estimate the number of barber shops in the united states." I asked what the point of the question was. They responded with "To see how your reasoning is". So I got up to the white board made some guesses and calculations against known statistics in the US (all taking about 20 seconds) and guessed there where about 35,000 barber shops in the US.

    So anybody know what the point of that stupid question was?

    I believe it was "to see how your reasoning is"

    This isn't a stupid question. The question is to see how you attack a hard problem with no trivial solution.

    The interviwer doesn't care what the actual answer is. He wants to see how you come up with the answer, and does it make sense.

    Let me mention some of my stories. I used to interview college students on campus. One question I asked was "How many blades of grass are on campus." One person, who went to school there for 4 years, said there was no grass on campus! No hire, not very observant. Another person said he would pick a sqaure yard and count the number of blades in that square. Even after I pointed out the flaw, he never changed to picking several squares. He as adament... Until I pointed to the nice clean floor and said, what about that square.

    What was I looking for? I didn't really care what number you came up with. But anyone who'd been going to school there for even a semester should have an idea of how many lawns there are, and give a wild guess how much ground was covered by grass. Then maybe estimate a blade of grass was .01mm by 3mm (or whatever) and do some math and come up with a large number. The math didn't have to be perfect, but it had to be reasonable, and make sense.

    In school you are generally taught to solve easy problems (well the problem might be difficult, but its still straightforward) A train leaves point A and 50 mph, and another train leaves point B, blah blah blah... But how well do you deal with oped ended problems that don't have a real solution.

  • Patrick McCormick (unregistered) in reply to JBP
    JBP:
    "Do I contradict myself? Very well then, I contradict myself" -- Walt Whitman

    I just saw the flaw in my own logic. Your own hat color is independent of the other hats' colors. What is not independent is the total count of red hats. Now I vote for the "if two hats are the same color, vote the other, else pass" strategy.

    The embarrassing part is that it's not the two semesters of probability I just took that convinced me, it's the guy who actually coded the thing up, ran his code and came up with 75.2%.

    It's hard to argue with results.

  • (cs) in reply to Anonymous Coward
    Anonymous Coward:
    Via Google:

    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.

    In the "canonical" question and answer, half constitutes a majority, so #2 can keep all 1000 coins. But yeah, the approach is right.

  • Tom (unregistered) in reply to Matthew

    For the condom problem, if you're not concerned for the safety of the women, you only need one condom. If you are concerned, then it's impossible with less than 3 condoms, unless you can wash the condoms. Assuming disease can transfer from condom to condom as well.

  • (cs) in reply to Anonymous
    Anonymous:
    I interviewed for MS once and was asked a few of these questions. I probably failed miserably, but not because I am incapable of answering the questions. Answering these is just not in my frame of reference when I am interviewing. I prepare carefully for the subject of the job, and I guess I'm blindsided when concentrating on having a good interview and then suddenly thrown a question out of context. I think this would be the equivalent of walking over to a software engineer's cube and asking them the appropriate temperature at which meat can be assumed to be fully cooked. Why the hell would most engineers know that?

    I did, however, ask why the person was asking me the question. The response was, "I want to see how you think."

    Well....I have since come up with the exact answer I will give the next time (if ever...I doubt I'd ever apply at MS again) I am asked this question:

    If you are asking this question to see how I think, then there can be no right or wrong answer. Unless you have specific criteria in which you will give me some sort of "grade" by which to compare me to other candidates that are asked the same question, what you are really asking for is my opinion. In my opinion, a question that involves solving a puzzle is hardly relevant to providing a software solution to a non-arbitrary, real world business problem. But since you asked, here is what I think:

    1. Did the people that architected, approved the release of, and implemented that disaster called ActiveX prove that they were exceptional people by answering this type of question? Would you hire them now if you could see the future? If I let you watch me write an ActiveX control right now that pwnz0r3d your box, would you hire me?

    2. How about the people that came up with COM/COM+ as the future of cross platform development? I guess that little performance problem called marshaling that is done WITH EVERY SINGLE METHOD CALL sort of killed it, eh? Not to mention that it never actually went mainstream with those bogus cross platform problems.

    3. Microsoft Bob. Enough said.

    4. If this interview method lets you only choose brilliant people, why is it that so many security holes are found in your software that involve simple buffer overruns?

    5. Did the developers and QA people responsible for letting BizTalk 2004 out the door when it would crash when receiving a file larger than 2GB (or was it 3GB?) prove they were k-k00l, aw3s0m3, and l33t people by answering these questions?

    I could seriously find a lot more points I think. If your interviewer, imho, can't figure out if you're skilled from just asking you questions about relevant past experience or even questions relevant to the job position, they are the wrong interviewer.

    No Hire. Candidated can't tackle ambiquious problem, instead points out other peoples failure.

    But seriously, the company is hiring you to solve problems. Not on how well you can lie on a resume. What? You didn't lie on the resume, how do we know without asking you questions?

    I start my interviews off by asking someone to tell me about a favorite project they worked on. This guy told me about an XML parser he just wrote. Then I asked him to write code for strcmp. Not only could he not write strcmp, he didn't understand the concept of COMPARING TWO STRINGS! How do you write an XML parser, if you can't understand how to compare strings?

    By the way... The subject of the job is to solve problems. Perhaps you should study up on it next time.

  • Robot (unregistered) in reply to Eggtastic

    Anyone memorizing scripts of how to pretend to think through brainteasers is obviously going to fail some other aspect of the interview.

  • Ahnfelt (unregistered) in reply to nini

    Or fill the 3L jug. Pour all from the 3L jug to the 5L jug. Refill the 3L jug. Pour from the 3L jug to the 5L jug until it's full. Empty the 5L jug. Pour the remaining 1L from the 3L jug to the 5L jug. Fill the 3L jug. 3L + 1L = 4L.

    captcha: muhahaha, now the captcha wasted your time too!

  • (cs) in reply to YourMoFoFriend
    YourMoFoFriend:
    A typical question I used was "How many trips a tooth fairy makes per night".
    That's easy. A tooth fairy makes one trip per night. A more interesting question would be: "How many tooth fairies would you need to cover <geographical region>?".
  • dirkpitt! (unregistered)

    For those of you making it this far....Let's Make a Deal! This is an oldie but goodie, which has an answer that some people refuse to believe, even if you prove it with a handy program (like rbowes did with the hat riddle).

    There are three doors (label them A, B, and C) that you cannot see through. Behind exactly one door there is a prize. Behind the other two are nothing.

    You get to choose one door. The host of the show, Monty Hall, after you make your choice, does not show you if you chose correctly. Instead, he opens one of the other two doors. That door always has nothing behind it.

    Now there are two doors unopened, one of which has a prize. He offers to let you stick with your original choice or switch to the other closed door.

    What should you do (assuming you want the prize)? Does it matter?

  • (cs) in reply to tchize
    tchize:
    Interview for a human ressource:

    You ask a teaser to a candidate, he give you a better answer that what is in you list of accepted answer. He did great at rest of interview. Probably the best candidate of today, but you still have 2 candidates to interview. You don't comment on his answer, the interview is nearly over. However, the candidate says it's now or never. When he quit the room, if there is no contract for him, he will never come back here. He also says he won't wait for more than 10 minutes, it's enough to make an offer. What do you do?

    A) You start to negociate term end ensure him he will be accepted with a very good salary

    B) You ask him to wait while you get to the boss for a contract. During this time you try to quickly test 2 remaining candidates

    C) You let him go, there are still 2 other candidates

    D) You report to you boss about it and ask his opinion about it

    E) You inform him he's not as self-important as he thinks and ask that he only comes back after learning humility.

  • anon (unregistered) 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.
    You missed the part where all three hats are assigned by independent events. The probability that Alice got a red hat is unrelated to whether or not Bob/Carol got a red hat. P(A=red, B=red, C=red) is 1/8, but P(A=red, B=red, C=red | B=red, C=red) is 1/2.

    Rephrased... Initial possibilities: {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}. All are equally likely. Let's take just the situations where Alice sees two red hats in front of her. The two ways this can happen are BRR and RRR, which (as stated before) are equally likely.

    I like Tom Woolf's answer. It's (captcha) yummy.

    Actually... by reading the comments here, I think I have discovered the true value of brain teasers for these interviews: So many people don't read the ****ing requirements before proposing a solution.

  • (cs) in reply to AnonCoder
    AnonCoder:
    Someone who say there silently for 60 seconds and then gave the correct answer would get less credit from me than someone who talked coherently and briefly through several possible solutions and made a justifiable choice out of them.

    Think of it like an exam question with the little bit at the end that says "show your working".

    See, I always got into trouble in maths for not showing enough working. I wrote down every step that I took, but I just took fewer steps than everyone else - which taken to a pathological extreme, would result in marking me down for being more naturally able at maths because I didn't have to put in as much effort. (For myself, I couldn't understand why I would have to artificially bloat up my working by 200-300% just to meet someone else's standards; and I still don't get arbitrary rules nearly two decades later.)

    When I read what you wrote, it has the same whiff - the smell of incredulous mediocrity; the inability to properly recognise a level of ability you can't approach. I'm sure you'll get bright people who can follow arbitrary rules terribly well and are desperate to prove how good they are - but I'm equally sure you won't get anyone who is that good; even if they did apply, I'm fairly confident you'd turn them down.

  • (cs) in reply to chrismcb
    chrismcb:
    Anonymous:
    I interviewed for MS once and was asked a few of these questions. I probably failed miserably, but not because I am incapable of answering the questions. Answering these is just not in my frame of reference when I am interviewing. I prepare carefully for the subject of the job, and I guess I'm blindsided when concentrating on having a good interview and then suddenly thrown a question out of context. I think this would be the equivalent of walking over to a software engineer's cube and asking them the appropriate temperature at which meat can be assumed to be fully cooked. Why the hell would most engineers know that?

    I did, however, ask why the person was asking me the question. The response was, "I want to see how you think."

    Well....I have since come up with the exact answer I will give the next time (if ever...I doubt I'd ever apply at MS again) I am asked this question:

    If you are asking this question to see how I think, then there can be no right or wrong answer. Unless you have specific criteria in which you will give me some sort of "grade" by which to compare me to other candidates that are asked the same question, what you are really asking for is my opinion. In my opinion, a question that involves solving a puzzle is hardly relevant to providing a software solution to a non-arbitrary, real world business problem. But since you asked, here is what I think:

    1. Did the people that architected, approved the release of, and implemented that disaster called ActiveX prove that they were exceptional people by answering this type of question? Would you hire them now if you could see the future? If I let you watch me write an ActiveX control right now that pwnz0r3d your box, would you hire me?

    2. How about the people that came up with COM/COM+ as the future of cross platform development? I guess that little performance problem called marshaling that is done WITH EVERY SINGLE METHOD CALL sort of killed it, eh? Not to mention that it never actually went mainstream with those bogus cross platform problems.

    3. Microsoft Bob. Enough said.

    4. If this interview method lets you only choose brilliant people, why is it that so many security holes are found in your software that involve simple buffer overruns?

    5. Did the developers and QA people responsible for letting BizTalk 2004 out the door when it would crash when receiving a file larger than 2GB (or was it 3GB?) prove they were k-k00l, aw3s0m3, and l33t people by answering these questions?

    I could seriously find a lot more points I think. If your interviewer, imho, can't figure out if you're skilled from just asking you questions about relevant past experience or even questions relevant to the job position, they are the wrong interviewer.

    No Hire. Candidated can't tackle ambiquious problem, instead points out other peoples failure.

    But seriously, the company is hiring you to solve problems. Not on how well you can lie on a resume. What? You didn't lie on the resume, how do we know without asking you questions?

    On the contrary, he perfectly illustrated what's wrong with the riddle method. You're right that the riddles would be a good way of determining whether a candidate is suited to solving problems, but software engineering is not the same as shuffling foxes or coins or hats or condoms. The capacity to solve those riddles has little or no intersection with the capacity to design and write effective, usable, stable software for end users.

    Imagine how much Microsoft would benefit if, instead of asking riddles, they showed the candidate a piece of code with a buffer overflow vulnerability in it, and asked the candidate what was wrong with it.

  • (cs)
    the guy who builds a water-displacement scale/barge, taxis a 747 to the docks, and then weights the jumbo jet using that, instead of simply calling Boeing in the first place

    I've seen some code written by that guy!

  • Ed (unregistered) in reply to anon

    Yes, but those aren't the only possiblities for Alice. You are correct in saying that, should Alice see two Red hats or two Blue hats, she'll have a 50/50 chance of getting the right answer.

    HOWEVER.

    The case that she sees one Blue AND one Red hat is just as likly to happen, and in that case, the winning percentage of the group, as a whole, is a gurrenteed 100%. If she sees one Blue and one Red, and she's wearing Red, the Blue person will answer correctly. If she sees one of each and she's wearing Blue, the Red person will answer correctly.

  • KM (unregistered) in reply to anon
    anon:
    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.
    You missed the part where all three hats are assigned by independent events. The probability that Alice got a red hat is unrelated to whether or not Bob/Carol got a red hat. P(A=red, B=red, C=red) is 1/8, but P(A=red, B=red, C=red | B=red, C=red) is 1/2.

    Rephrased... Initial possibilities: {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}. All are equally likely. Let's take just the situations where Alice sees two red hats in front of her. The two ways this can happen are BRR and RRR, which (as stated before) are equally likely.

    That doesn't matter. You're fixing one person as the one who sees two identical hats. You're restricting the sample space incorrectly.

    Take all the possibilities and look at whether the strategy leads to a win or loss in each case. How many lead to a win? How many lead to a loss?

  • (cs) in reply to akatherder
    akatherder:
    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.

    Yes, but in the UK we have people who steal road signs and traffic cones... or in one instance I was aware of, a set of portable traffic lights. It's a trophy, not an acquisition.

    It's also why roads next to universities should never be worked on during term time...

  • Mikey Benny (unregistered) in reply to akatherder

    Correct, you have a 75% chance of winning with that strategy.

    For all of you "50%" freaks out there: notice that there are six correct answers (the middle six each have one correct answer) and six incorrect answers The first and last answers have three incorrect answers apiece. So, half the guesses are right, and half the guesses are wrong. The strategy just distributes those right and wrong guesses optimally so that you either completely fail, or do the minimum to win.

    Brilliant if you ask me.

  • tastey (unregistered) in reply to M.G.
    M.G.:
    It reminds me of the people who can't accept that .99999999... (repeats forever) is exactly equal to 1.
    Right, just like 1 over infinity equal nullity! And for the record .999999... (repeating ad infinitum) does not equal 1, but is close enough for all practical (and most impractical) purposes.
  • kungfu (unregistered) in reply to chrismcb
    chrismcb:
    Let me mention some of my stories. I used to interview college students on campus. One question I asked was "How many blades of grass are on campus."
    What would you have said to the OCD-autistic-savant (whom you did not know to be OCD-autistic-savant) who simply immediately answered your question with a specific number (and upon further queries simply stated he had counted them earlier that day).
  • dreadlocks (unregistered) in reply to chrismcb
    chrismcb:
    I start my interviews off by asking someone to tell me about a favorite project they worked on. This guy told me about an XML parser he just wrote. Then I asked him to write code for strcmp. Not only could he not write strcmp, he didn't understand the concept of COMPARING TWO STRINGS! How do you write an XML parser, if you can't understand how to compare strings?
    In some languages, string comparisons are handled for you .. perhaps the XML parser was written in such a language and the individual never had to learn about string comparisons.
  • (cs) in reply to tastey
    tastey:
    Right, just like 1 over infinity equal nullity! And for the record .999999... (repeating ad infinitum) does not equal 1, but is close enough for all practical (and most impractical) purposes.
    You may want to confront that with a typical "Introduction to algebra" or similar book at your local bookshop (yes - that one with ~1000 pages, taking as much space as 3 other books). It disagrees with you. Not all numbers have unique representation in decimal system.
  • Ed (unregistered)

    Incidently, the hats question is very similar to the Monty Hall problem.

    You're on "Let's Make a Deal" (the TV game show). The host, Monty Hall, shows you 3 doors. Behind one of the doors is a brand new car. Behind the other 2 are goats. You choose a door, and you win whatever's behind the door you choose.

    Now, obviously you want the car... unless you really have a thing for goats. I won't go there though.

    Anyway, Monty Hall obviously knows which door the car is behind before the game starts. After you choose a door. Monty will open one of the doors you didn't choose, revealing a goat. (if you picked the door with the car, he chooses which door to reveal at random. If you chose a door with a goat, he'll show the other door with a goat behind it)

    After he reveals an incorrect door, he asks you if you want to switch to the remaining unrevealed door, or stay with your origional choice.

    Do you switch?

    ANSWER: Yes. If you switch doors at that point, you have a 2/3 chance of winning the car. Discuss amounst yourselves why this is so.

  • (cs) in reply to anon
    anon:
    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.
    You missed the part where all three hats are assigned by independent events. The probability that Alice got a red hat is unrelated to whether or not Bob/Carol got a red hat. P(A=red, B=red, C=red) is 1/8, but P(A=red, B=red, C=red | B=red, C=red) is 1/2.

    Rephrased... Initial possibilities: {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}. All are equally likely. Let's take just the situations where Alice sees two red hats in front of her. The two ways this can happen are BRR and RRR, which (as stated before) are equally likely.

    I like Tom Woolf's answer. It's (captcha) yummy.

    Actually... by reading the comments here, I think I have discovered the true value of brain teasers for these interviews: So many people don't read the ****ing requirements before proposing a solution.

    Ok, I know someone made this point earlier, but let me try to underscore this again. Remember, no one guesses their color unless the other two hats are the same color. Then, they pick the opposite color. So...

    It's true that Alice's chance of picking the correct color is 50/50. It's true that Bob's chance of picking the correct color is 50/50. It's true that Carol's chance of picking the correct color is 50/50. However, the strategy is such that the condition under which any one of them guesses the wrong color is the same as the condition under which the other two will guess the wrong color (specifically, BBB or RRR). In any of the other 6 cases, only one person will see the same color for the other two hats and will guess correctly (while the other two will see opposite colors and pass). In other words, while any individual still has a 50/50 chance of guessing correctly, all the incorrect guesses occur in the same 2 scenarios (BBB or RRR).

    I hope this post has at least a 50% chance of clearing things up...

  • tastey (unregistered) in reply to viraptor
    viraptor:
    tastey:
    Right, just like 1 over infinity equal nullity! And for the record .999999... (repeating ad infinitum) does not equal 1, but is close enough for all practical (and most impractical) purposes.
    You may want to confront that with a typical "Introduction to algebra" or similar book at your local bookshop (yes - that one with ~1000 pages, taking as much space as 3 other books). It disagrees with you. Not all numbers have unique representation in decimal system.
    .9999 (repeated) does not equal 1 for the exact same reason 1/infinity does not equal zero
  • Konstantin Surkov (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%

    Amazingly, this is correct. Even more amazingly, the probability to win the prize is 75% nevertheless.

  • (cs) in reply to UncleMidriff
    UncleMidriff:
    Undefined Reference:
    Nope, Min 2, Max 3.

    {...snip completely documented method...}

    I get Min=2, Max=4, Avg=3.25 {...snip a big pile of waffle...} If I am wrong, please call me on it.
    That might have been a reasonable thing to say if you were pointing out an error in the post you were replying to. But you didn't do that. You contradicted him for no good reason, waffled a lot about situations without ever actually enumerating the possibilities, completely failed to show (or even attempt to show) any error in the parent's methodology, and presented your suboptimal, inadequately documented, possibly incorrect solution as somehow superseding the complete, fully-documented, optimal solution you totally ignored.

    It's this mindset which will ensure that creationism ends up being taught in science classes, despite not being science, because "it's a valid alternative". It's anti-intellectualism coated with a veneer of faux humility and a disingenuous claim to egalitarianism. And it really shouldn't be allowed to show its face in public.

    So read and understand - when your solution is less optimal than solution X, you have shown that you are aware of X, and you have not demonstrated a flaw in X, you should be learning from X, not trying to present your solution as somehow "just as good really", or "worth airing anyway", and ESPECIALLY not a successor to X! Share your thoughts, sure - but quoting someone else's solution before presenting your own implies a criticism of that solution, so you'd damn well better actually make such a criticism.

  • (cs) in reply to tastey
    tastey:
    viraptor:
    tastey:
    Right, just like 1 over infinity equal nullity! And for the record .999999... (repeating ad infinitum) does not equal 1, but is close enough for all practical (and most impractical) purposes.
    You may want to confront that with a typical "Introduction to algebra" or similar book at your local bookshop (yes - that one with ~1000 pages, taking as much space as 3 other books). It disagrees with you. Not all numbers have unique representation in decimal system.
    .9999 (repeated) does not equal 1 for the exact same reason 1/infinity does not equal zero
    I didn't discover it - if you think that's wrong, tell it to all the mathematicians writing those books. But I think, you won't convince them by just disagreeing. 2 alg. books I own say the same - one is from '70s and one from 2k3 - each having one chapter for decimal and other systems - so I'm pretty sure I will believe them and not you for now. Go ahead - publish your research about unique representation in decimal - maybe someone will be interested ;)
  • Alex (unregistered) in reply to tastey
    tastey:
    viraptor:
    tastey:
    Right, just like 1 over infinity equal nullity! And for the record .999999... (repeating ad infinitum) does not equal 1, but is close enough for all practical (and most impractical) purposes.
    You may want to confront that with a typical "Introduction to algebra" or similar book at your local bookshop (yes - that one with ~1000 pages, taking as much space as 3 other books). It disagrees with you. Not all numbers have unique representation in decimal system.
    .9999 (repeated) does not equal 1 for the exact same reason 1/infinity does not equal zero

    All right. Consider a = 0.999999 (repeated) for ease of notation. In the case it's not equal to 1 (and I suppose you believe it's smaller) then 1 - a is a strictly positive number.

    Problem: Prove that b > 0 (you can't).

    Also, see http://en.wikipedia.org/wiki/0.999

  • Mikey Benny (unregistered) in reply to tastey
    tastey:
    M.G.:
    It reminds me of the people who can't accept that .99999999... (repeats forever) is exactly equal to 1.
    Right, just like 1 over infinity equal nullity! And for the record .999999... (repeating ad infinitum) does not equal 1, but is close enough for all practical (and most impractical) purposes.

    No, (.999...) = 1 (exactly 1)

    You can convert a repeating decimal to a fraction by taking the repeating portion of a decimal as the numerator and [10^(numdigits)-1] as the denominator:

    .333... = 333/999 = 1/3 .444... = 444/999 = 4/9 .555... = 555/999 = 5/9 .666... = 666/999 = 2/3 .777... = 777/999 = 7/9 .888... = 888/999 = 8/9 .999... = 999/999 = 1

  • Obi Wan (unregistered)

    I finally understand why Microsoft products work the way they do! Their software is written by people who are specialists at providing overly complex solutions to nonexistent problems! Elementary!

    Case in point: ribbons in Office 2007. Menus removed. Why? See above.

  • (cs) in reply to Jen Larkin
    Jen Larkin:
    I know someone who encountered this who simply got up and walked out, then when the company called back he told them that they clearly were not interested in hiring someone competent.

    I'm not sure if such companies are truly not interested in hiring someone competent, or if they simply have no idea what competence is or how to assess it.

    But either way, it tells you something useful about the quality of your potential colleagues. ;)

  • (cs) in reply to Mikey Benny
    Mikey Benny:
    tastey:
    M.G.:
    It reminds me of the people who can't accept that .99999999... (repeats forever) is exactly equal to 1.
    Right, just like 1 over infinity equal nullity! And for the record .999999... (repeating ad infinitum) does not equal 1, but is close enough for all practical (and most impractical) purposes.

    No, (.999...) = 1 (exactly 1)

    You can convert a repeating decimal to a fraction by taking the repeating portion of a decimal as the numerator and [10^(numdigits)-1] as the denominator:

    .333... = 333/999 = 1/3 .444... = 444/999 = 4/9 .555... = 555/999 = 5/9 .666... = 666/999 = 2/3 .777... = 777/999 = 7/9 .888... = 888/999 = 8/9 .999... = 999/999 = 1

    Oh boy... this is a fun day.

    so...

    let x = .999... 10x = 9.999... 9x = 9 x = 1

    Or, for you more abstract thinkers, define x(n) as 0.999...9 (n 9's) consider lim((x(n)) as n -> infinity...

    Ain't infinity fun?

  • Shinobu (unregistered)

    This has been noted before, but you have convinced me too that asking riddles on job interviews is actually a good idea.

  • (cs) in reply to Eggtastic
    Eggtastic:
    If someone knows the answer, they can just make a show of working it out according to any one of the detailed scripts available in every single bookstore... if someone just blurts out the answer, you know your test failed, but only an idiot would do that.

    Or someone who values honesty and integrity above keeping up appearances and stroking egos.

    ...oh, I see where you're going with that one.

  • (cs) in reply to unknown
    unknown:
    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

    I've always wondered what riding in the short bus feels like. How do you people like it? Do they have little stewardesses to wipe up your drool?

  • (cs) in reply to YourMoFoFriend
    YourMoFoFriend:
    But did the "walk outie" show his competence or his "jerkness"?
    Or his sense of self-preservation?
    I've worked in enough small startups to know that one WILL encounter an unexpected problem that has nothing to do with ones particular job description, and that one WILL have to solve it.
    Yeah, that can be lots of fun. Having to do it for some arsehole sat smugly in front of you who couldn't distinguish a decent programmer from a pint of milk is less fun, and any startup who uses such a person to assess new hires will be a shutdown pretty soon.
    Do you want to hire someone that did walk out on an interview?
    In a fucking heartbeat. Would you want to hire the person who wished they had?
  • Theo (unregistered) in reply to gwenhwyfaer
    bunch of users:
    Hat problem...
    Easy. The answer is red.
  • (cs) in reply to Anonymous Coward
    Anonymous Coward:
    Anything that ... makes interviews with soulless corporations a little more fun is a good thing in my book.

    Oh, that'll be from the same corporate definition of "fun" that gives us Hawaiian-shirt Fridays and "I Broke the Build" hats, I take it?

  • Shim (unregistered) in reply to Grandpa
    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.

    No communication of any sort is allowed, except for an initial strategy session before the game begins. Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass. The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly.

    What strategy would you use?

    There are five pages of comments, which I'm not going to read.

    But my strategy would be to have Phil look at his reflection in Bob's glasses and call out the color of his own hat.

  • (cs) in reply to Grandpa
    Grandpa:
    ...No communication of any sort is allowed, except for an initial strategy session before the game begins...

    What strategy would you use?

    Mount a mirror in the room during the initial strategy session. It's not prohibited, and looking at one's own reflection cannot possibly be construed as communication.

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