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Geek Brain Teasers 80

Posted by michael
from the better-than-chocolate dept.
muce writes "A few days ago my cube mate entertained a lot of us engineers by presenting us with the famous Monty Hall problem. That problem sparked a day of strong debates, coding simulations, and ramped writing of equations on whiteboards. Since then we've been thirsting for more good geeky mathematical brain teasers to pass the time at work. Does anybody know of any good ones like the Monty Hall problem, or by chance is there a web page with a collection of them?"
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Geek Brain Teasers

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  • by Anonymous Coward
    Check out The Grey Labyrinth. Lot's of good puzzles. A top notch site.
  • your expenditure is $27, plus the collective $3 amongst you and your friends, so $27 + $3 = $30, not $27 + $2.

    Don't make the mistake of adding the bellhop's $2, because that's money that is already spent (included in your $27 cost). Actual Room cost($25) + bellhop's cut($2) = $27. So you can see why adding the $2 again is just plain stupid.

  • check out the puzzles faq (ftp from rtfm.mit.edu) lots of math puzzles, language puzzles, every sort of puzzle you could want.

    if you and two friends go to a hotel, the hotel manager charges you thirty dollars, you split it three ways, ten bucks each, right? then you go to your room, the manager realizes it was only 25, so he sends a bellhop with five dollars to your room. The bellhop steals two bucks, giving you each a dollar back, reducing your expenditure to nine dollars each, nine times three is twenty-seven, plus the two that the bellboy stole is twenty-nine dollars, where's the missing dollar?

  • http://techinterview.editthispage.com/

    This weblog is devoted solely to brain teasers (under the guise of tech interview questions) and started just a few weeks ago.
  • The first try is a 1/3 shot. The second is a 1/2 shot. So, the real question here is will the 1/2 chance match with your 1/3 chance.

    Let's add a point. You choose. He opens. You have a sudden bout of amnesia and forget which you chose. Being embarrased you say nothing.

    You choose from the remaining two. You are now correct fifty percent of the time.

    What are the chances that your second choice is the same as the first choice?



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  • This question was asked earlier in the comments. They beat me to it. :-)



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  • The second question was asked for an answer. It would have been asked even if none, or only one, had mud.



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  • 99 + 9/9

    Too easy, took about ten seconds.



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  • I wasn't arguing the logic. Rather the statement that the second question could not be asked. Every question can be asked, even if there is no answer. This one has an answer and is interesting to some, therefore it was asked. I just didn't think the second sentence of that poster was worded appropriately to convey the intended meaning.



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  • You'd think you had a 50/50 chance by knowing that you seem to have a choice between two doors. Because, should a second person come along not knowing of your choice, and choose between the two doors, his chances would have to be 50/50.

    Let's ask the question another way. Monty asks you to choose a door with no chance to switch. Your chances are 1/3. Even if he opens one of the wrong doors before he tells you if you won or lost, you cannot change, and therefore your chances are 1/3.

    Again, but this time Monty asks you to *remove* a door. In other words, you win if the door is *not* the one that you pick. Your chances are 2/3. Even if Monty opens up a losing door before he tells you the winner.

    Should you choose to stay or switch *before* you choose any doors, these two cases are stay and switch respectively, showing the odds are 1/3 for staying and 2/3 for switching.

    The thing that boggles the mind, is if someone else chose his chances are 50/50. Yes, this is true. But had he known the previous situation, his chances are the same as yours.

    It still seems not to make sense since how does opening the door affect your decision. So here is what helped me. When you make your first choice, you have a 1/3 chance of getting the car. So, in 1/3 of the cases, Monty can choose either of the other two doors to open, as they are both losers. In 2/3 of the cases (that is, when you choose a goat door) the door that monty open is chosen by your choice. That is, he can't choose your choice to open, nor can he open the one with the car. Thus, in 2/3 of the cases you *affected* the opening and cancellation of one of the doors. Thus bring it into the equation and allowing it to raise the cahnces of the other door.



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  • by Chacham (981) on Friday March 23, 2001 @06:30AM (#346108) Homepage Journal

    I posted this lat time Ask /. had a puzzle of boring Sunday a while back.

    Three smart kids are on a beach and all have mud on their foreheads. An old man comes over to them and asks for each of them to look at *both* their friends, and should one *or* both of their friends have mud on their foreheads, they should raise their hands. All three kids look at both of their friends, and seeing mud on both of their friends heads, they raise their hands.

    The old man then offers a dollar to anyone who can answer his next question, and prove it. He asks if any of the kids know that they do, or do not, have mud on their foreheads, and if they can prove it. The kids look at each other and are bewildered for a bit. Suddenly, one of them screaches, "Oh!" and raises his hand. He then explains to the old man how he must have mud on his forehead and explains his reasoning. As his reasoning was excellent, the old man gives him the dollar.

    What was the boy's reasoning?



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  • Well, not quite, but tts an old problem, a stastical brain teaser, that (if you watch your James Burke) is partly responsible for the application of statistics to the French census durin the Napoleanic era.

    The original is "how many [black|red] cards are there in three face down cards; does the number seem to "change" when one of the cards is revealed?" The problem is from the 16th century (about the time of the application of color to what became the "standard" deck), and the solution came in the 18th century...with no computers involved. The solution uses basic probability mathematics that's still taught as-is in today's Prob&Stats classes for Math and CS Majors.

  • It depends. Is it one of the spidergoats [slashdot.org], that lactate a product that can be spun into silk?

    --
  • Three travelers stop at a hotel for the evening. The manager offers the travelers a single room at a cost of $30, to which the travelers agree, each paying the manager $10.

    Later that evening the manager realizes that he has over charged the travelers for their room: the price should have been $25 rather than $30. The manager calls the bellhop over, giving him $5 and instructing him to take the money up to three travelers as a refund. On his way up to the travelers' room the bellhop considers that there is no even way to divide $5 between three people, and decides, in the interest of civility, to keep $2 for himself as a tip, giving the travelers' only a $3 refund

    When the bellhop gets to the room, he gives the travelers their $3, which they divide equally amoung themselvs, $1 per person, and everyone is happy. Here, however, is the problem: Each traveler has now paid $9 ($10-$1=$9), meaning that they have paid a total of $27 (3*$9=$27) for the room. Add in the $2 that the bellhop kept as a tip and we have $29 ($27+$2=$29). But the three travelers originally paid $30! Where is the missing dollar?

  • Exactly true. When I first heard this puzzle I was told that the solution had to do with the order of mathematical operations and precedence, but I have also had the solution explained to me in terms of standard accounting practice. (something about not counting both debits and credits in the same group, but I'm no accountant so I don't remember the details) The real trick is in the telling. This didn't become clear to me, however, till I had to write the problem down, and I found that you had to carefully avoid any mention of the price after the refund (or the price claimed by the bellhop, to obscure his petty theft).

    The simplest way I have found to explain the error is to say that you should only count actual dollars that people in the problem are holding: $3 (in the travelers' hands) + $2 (in the bellhop's pocket) + $25 (in the manager's till) = $30 (originally paid for the room).

    Still, most people, when they first hear the puzzle, are quite flummoxed and have a very hard time explaining exactly what the error is.

    1. Pick a number between 1 and 100.
    2. Multiply by 4.
    3. Add 3.
    4. Square the result.
    5. Find the largest prime number less than the result and subtract it from the result.
    6. Spell out the result, count the number of letters, and add it to the result.
    7. If the result it greater than 23, subtract 23.
    8. If the result is a perfect square, add 4.
    9. Call the result n. Divide n by 7 and count the n'th digit to the right of the decimal point.
    The answer is 8
  • Try again. Use English.
    1. 1 or 2
    2. 4 or 8
    3. 7 or 11
    4. 49 or 121
    5. 2 or 8
    6. two has 3 letters + 2 = 5, or eight has 5 letters + 8 = 13
    7. 5 or 13
    8. 5 or 13
    9. 5/7 = 0.71428... or 13/7 = 1.8571428571428...
  • Try this variation and see if you change your mind:

    Monty shows you 1000 doors and asks you to pick the one you think hides a car. You pick door 94. Then Monty opens 998 doors. At each door, he checks the number on the door against a little scrap of paper he has. When Monty is done opening doors, he has revealed 998 goats, and left your selection, door 94, and one other door, door 672, closed.

    Monty reminds you that one of the two remaining closed doors hides the 999th goat, and the other hides a brand new car! You can keep your original choice, or change your choice to door 672.

    Should you switch?

    If yes, how is this different from the three door case?

  • Don't feed the trolls.
  • Do you ever try working to pass the time at work? ;-)

    -------------------------------------------
    I like nonsense, it wakes up the brain cells.
  • by Pseudonym (62607) on Saturday March 24, 2001 @06:43AM (#346120)

    Let's assume without loss of generality that you pick the first digit. That rules out cases 4 to 7. Knowing there's at least a 1 rules out case 0. The cases left are 001, 010 and 011. Therefore the odds that there are two 1's is 1/3.

    Get it now?

  • A couple of years ago my daughter brought this one home from junior high (none of the junior high kids worked it out in the limited time), and I eventually turned it into a little lecture on organizing search techniques for the lunch-time talk series here at work.

    You stop at the local Seven-Eleven and pick out four items. The sum of the four prices is $7.11. The product of the four prices is also 7.11. What are the prices of the four items? Is the solution unique? Are there other sum/product values with solutions? For those who insist on making things more complicated than necessary: there are no taxes, the prices are all integral in cents, and the product is exactly 7.11, not just close and rounded off.

  • There were some pretty heated discussions here a while ago, until we wrote down all the possible outcomes. It was a relief to find a mathematician agreed:

    http://math.ucsd.edu/~crypto/Monty/montybg.html [ucsd.edu]

  • Arguments and code simulations? Why? Are your co-workers mentally weak? This is an open and shut case of logical analysis.

    At first glance, there are three doors, two goats, and one car. However, this is misleading.

    In any run of the problem, one of the unchosen doors is shown to be a goat. You cannot choose that door, and both it and its associated goat never enter into the statistical problem.

    The actual problem has only two doors, one goat, one car. Plain old 50/50 chance. There is no gain to be had by changing doors.

    Well, unless the door is hung a bit high and you can see goat legs under the door you have chosen.
  • Heh.. I was expecting a standard magicians lead routine. Hadn't seen that as a 'online' version. How many people would remember more than one card, so what danger is there in not changing them all? :)

    I'd seen the real world equivalent. Took me one try to beat it, and only because I cheated. (Cards you have just laid out in front of you in intentionally spilled beer on do NOT suddenly become dry and fresh. You can always smell the beer)

    Take the five changed cards, and invert them on the bottom of the deck. If you know how to tab them into a cut, I'd shuffle a time or two and make them arrive at the bottom. Pick through the deck for them, in front of the patsy. Have the pick made, then tell them to shuffle the six and place them back on top of the deck, removing any chance the order will ring a bell with the patsy. Tell them to think about it, while you grab the deck and deal the changed cards to them after a quick inversion.

    Oh, and three socks.
  • [sigh] I see how there were arguments now.. :)

    If I think about it that way, it's still a 1/2 problem since the second goat is a known quantity.

    Think about it like this. You are presented with three doors, one open displaying a goat, *before* you make your choice. (When the door is opened is meaningless, as it always displays a goat) You cannot choose the open door, obviously, so there are only 2 to choose from.

  • Heh.. Why worry about it, when the answer can easily be gained through social engineering at no cost?

    'Man, three kids.. How do you handle the sibling rivalry?' should produce at least one childs name and the gender of the other two. If it doesn't, eg, you get a result like 'Oh, it is/is not terribly bad.' you have to go fishing again. If you had siblings, or have at two children of the same sex, try the empathy ploy. Detail how when you/your children had to share a room, the infighting was worse, and ask if keeping them seperated could be the reason/solution.

    Leading, generic talk can get you anywhere.
  • Goddamnit.

    You're right, I'm wrong.

    [/me bashes his head against the table.]

    I hearby apologize to everyone I flamed. I was wrong. For lack of a better programming language at my disposal, here's an Office macro that proves not only am I an idiot, I'd only get the car half as much as Rev Snow.

    Sub goat_problem()
    Dim chosen As Integer
    Dim you As Integer
    Dim looper As Integer
    Dim swi As Integer
    Dim norm As Integer
    For looper = 1 To 20000 Step 1
    chosen = Int((Rnd() * 3) + 1)
    you = Int((Rnd() * 3) + 1)
    If (chosen = you) Then norm = norm + 1
    If (Not (chosen = you)) Then swi = swi + 1
    Next looper
    looper = MsgBox("goat " & swi & " " & "notgoat " & norm & " " & swi / norm, vbOKOnly, "goat?")
    End Sub
  • Nope. A decreased chance. Seeing a girl's bedroom is as good as seeing the girl. There are a myriad of excuses to have the hockey schedule on the fridge, from 'My wife picks up the neighbor's kid when she picks the girls from ballet' to 'My twelve year old, Daphne, was so good they moved her into boys AAA Pee-Wee this year.'.
  • Because the odds of each case occouring are not evenly distributed. Your logic would hold if the story was changed so that you sumble into the room of a child of one sex, and find a schedule for a child of the other. But it doesn't. The room was a girl's, so it is more likely that there are two girls. (by the authors logic)

    -Peter


    "There is no number '1.'"
  • Couldn't the kid just scrape some mud off his head and show it to the guy?
  • One of my favorites is, how many people would have to be in a room before the odds of at least two of them having the same birthday would be greater than 50-50? The answer (if I'm doing my math right here) is only 23 - smaller than you'd think. Not a hard problem, but one that catches a lot of people.
  • Umm...

    The question is: what were the odds of seeing the hockey card? If it was a certainty, then if every boy plays hockey, we're done: there can only be one boy (i.e. it's one boy, 2 girls a 100%)

    But if the odds of seeing the card are not 1, then surely it's more likely that a card is observed if there's 2 boys rather than 1?

    I guess I'm trying to make the same argument for an increased likelihood of there being two boys as you make for an increased likelihood of there being two girls.

  • No, there has to a stipulation as to the minimum number of muddied foreheads. Otherwise there's information to be had from seeing the other kids, and so no way to reason it out.

    If at least one kid is known to have mud, then if a kid sees exactly one muddied forehead then he knows his is also muddy, because if it weren't then the muddied kid he sees would see two kids without mud and know that he is the one. Since he's not saying anthing, he doesn't see two clean foreheads.

  • When attempting to explain this problem I've had the most success with:

    If you trade, the only time you don't get the car is when you picked it in the first place, which was obviously a 1/3 chance. So trading gets the car 2/3 of the time.

  • The website for the Car Talk [cars.com] NPR radio program has all of their past puzzlers [cars.com], some of which are car-related but most are not. They range from the completely obvious to the completely impossible. The Monty Hall puzzler is among them. They're available either as RealAudio or as html (transcripts of the radio segments). Definitely go check them out.

  • a few years back in a CS class I was assigned the sadistic task "write a program in C++ whose output is its own source code"

    I never figured it out but apparently it is possible.

  • This [xs4all.nl] page has a good selection of puzzles.
  • If boy 1 saw that boy 2 was clean and boy 3 was muddy, and boy 3 had his hand up, that means that one of himself or boy 2 has mud. Since boy 2 was clean (he could see that) he must be muddy.
    ----
  • The way I like to think of it is that hte switch is like asking: "Do you want that door, or do you want EITHER OF THE OTHER TWO DOORS" - which means the 2/3 seem much more correct.
    ----
  • For looper = 1 To 20000 Step 1

    Step 1

    gosh you like typing don't you
    .oO0Oo.
  • handA = (mudB or mudC);
    handB = (mudA or mudC);

    Since mudA and mudB are both true, the value of mudC is indeterminate. An individual, therefore, would not know the value of mudC (his own forehead muddiness), solely from handA, handB, mudA, and mudB. However, when the kid saw that at least one of the others was sufferring from the indeterminacy of his own forehead muddiness, the only reason is that that kid is also seeing the same problem of two hands and two muddy foreheads. Seeing this, the kid pronounced that his forehead must certainly also be muddy. QED.
  • >The subtlety in the statistical argument derives from the fact that the host is giving you more information half way through the game

    Yeah. Actually, the question, as usually posed, DOES result in a 50/50 choice. Ie, if someone says to you:

    `ok, there are 3 doors, 1 is open, and has a goat behind it, the other 2 are closed, and one of those 2 has a prize behind it, which do you chose`

    the answer is it doesnt matter - its 50/50.
    You have to include the initial choice and subsequence opening of the doors for it to be counter-intuitive. I think its THAT which confuses people.
    It did me, anyway!
  • If the anouncer open the door with the goat , why not just take the goat (who needs a car when you have a goat).
    Just asking.

    --
  • You're saying that if I show you the first as heads, then the second as tails that the odds of the third flip being heads is *higher* than it being tails? I don't buy it.

    No. That's different than the scenario I described. Flip three coins. You're told that there is at least one head, and you're shown one coin at random, and it is a tail. Now, what is the probability that there are two tails?

    This is very different than being told "coin 1 is a head, and coin 2 is a tail." Do you see why?
  • No, try again. You have a binary number of 3 digits. You are have two pieces of information:
    1. You choose one digit randomly, and are told it is a 0.
    2. You are told that there is at least one 1.
    Notice that the two pieces of information are not of the same "quality". Now, what are the odds that there are two 0's? What are the odds that there are two 1's?
  • Let's look at a simplified version of the possibilities. We'll only consider the case where the prize is behind door #1.

    You pick door 1, Monty opens door 2, you stay, you win.
    You pick door 1, Monty opens door 2, you switch to door 3, you lose.
    You pick door 1, Monty opens door 3, you stay, you win.
    You pick door 1, Monty opens door 3, you switch to door 2, you lose.
    You pick door 2, Monty opens door 3, you stay, you lose.
    You pick door 2, Monty opens door 3, you switch to door 1, you win.
    You pick door 3, Monty opens door 2, you stay, you lose.
    You pick door 3, Monty opens door 2, you switch to door 1, you win.

    (Note that there are 4 additional cases where Monty would open door 1, but he can't since the prize is there. I suspect that this is where some people make their mistake.)

    Now, out of those 8 cases, you win 4 times or 50%. In the cases you stay with your original door, you win 2 out of 4 (50%). If you always switch, you win 2 out of 4 (50%).

    If you expanded this to all 3 prize locations, you would get 12 out of 24, 6 out 12, and 6 out of 12, or 50% chance no matter what your strategy was.

    I think most people make the mistake of not fully considering the case where you pick the prize door in the first place. There are two possible branches from there since Monty can open either incorrect door. If you don't take that into account, then you end up thinking that you only win 1 of 3 by staying but win 2 of 3 by switching.

    Like we said, the first door choice really doesn't count for anything. The only choice that matters is the final door you chose. Since you're picking between 2 doors and 1 has the prize, that leads to a simple 50% probability.

  • After sleeping on this, I changed my mind. Count me in the "it's better to switch" camp instead of the "50-50" camp.

    In the 8 cases I listed, you can drop the first two. When you select the correct door at first, it doesn't matter which door Monty opens. He has two choices that are equivalent from your point of view (between two doors, each of which doesn't have a prize behind it). His choice doesn't impact the probability of you finding the prize.

  • by jon_adair (142541) on Thursday March 22, 2001 @09:23PM (#346148) Homepage

    As a side note, at one time I heard an explanation of her world record IQ that makes me discount her as the world record holder.

    Supposedly IQ scores depend on the sample size taking the test. If a relatively small number of people take the test, even scoring a perfect score can only rate an IQ a few points above the average.

    So the story I heard is that the test she set the record on was one of the most widely given standardized IQ tests. I believe she is solidly in the "baby boomer" set and I think this was a school test given years and years ago.

    Since no test since then has been given to more people, even a perfect score on every IQ test since then wouldn't result in breaking her record.

    Of course, that's just the information I heard. It could be wrong, but it does sound plausable.

  • Now, out of those 8 cases, you win 4 times or 50%. In the cases you stay with your original door, you win 2 out of 4 (50%). If you always switch, you win 2 out of 4 (50%).

    That's right - by cases. You forget, however, that those cases do not have the same probability. There are four cases in which you picked door 1 initially, but only two for each of the other doors. This means that you already found the right door in half the cases; unless you know from the beginning that door 1 is the right one, your guess can't be that good.

    Therefore, if you want to stay with your cases approach, you have to introduce duplicates for all door 2 and door 3 cases, resulting in six cases with switching; in four cases, you win. Regarding the six cases without switching, you are successful in only two of them.

    Now we again end up thinking that you only win 1 of 3 by staying but win 2 of 3 by switching, don't we? :-)

  • I am obfuscating the answer so as to let you think about it first: floor(log(7.39))/floor(10^(0.779))

    Assuming you mean ln and not log, your answer is wrong. :) Draw up a table.

  • The actual problem has only two doors, one goat, one car. Plain old 50/50 chance. There is no gain to be had by changing doors.

    Bzzt.

    Think of it this way: the host is saying, in effect, you can keep your original door, or you can have both of the other doors, one of which (the goat) I will show you.

    So, assuming the host *always* opens a door, it's better to switch.

    For your next exercise: what if Monty opens a door only half the time? :)

  • Other than that I stand by the posted solution.

    I still think you are wrong. :)

    The possibilities for the three kids:
    1. BBB
    2. BBG
    3. BGB
    4. BGG
    5. GBB
    6. GBG
    7. GGB
    8. GGG

    We know there is at least one boy and one girl, so we throw out cases 1 and 7. In 3 of the remaining 6 cases, there are 2 boys, so the probability is 1/2.

    Even if we say that the BBG combination is the same as GBB (etc), then we are left with BBB, BGG, GBB, GGG, and the answer is the same.

    What's wrong with this logic? I really want to know. :)

  • the bellhop's two dollars doesn't count as money that the travellers spent. So the equation isn't (9*3)+2 != (10*3). The money the bellhop keeps is money that the hotel thought was refunded to the travellers but was, in effect, stolen. So the equation is... 9*3 = (10*3) - (5-2) which balances. You can consider the bellhop part of the hotel, not part of the travellers group, then the apparent paradox evaporates. It's the misleading wording of the puzzle that leads you to believe that the bellhop's take should be considered part of the travellers contribution.
  • ok, so if we write out all the possible outcomes, we are left with 8 cases. Then you say that there is at least one heads, so we can cross out the TTT scenerio. This leaves us with two cases where there are two tails, and three cases with two heads. So, we pick a coin at random (7/21 being tails, 14/21 heads).. and it's a tail..
    So, although that was unlikely, that leaves us with 6 possible scenerios. Which is right back where we started.

    I think you've got me nicely confused here.

    Besides, how is seeing a girl's room any different from seeing a boy's schedule? Assuming the question is fair, that leaves us with 1 boy FOR SURE, and 1 girl for SURE.

    I guess i'm just confused.
  • The URL cited in the article has a JavaScript-driven simulation so you can see what results you get yourself. I just made an on-line version that lets you see total results from everybody who's played. Each page of my simulation has a link to let you see the source code. See my Monty Hall Puzzle [webappcabaret.com].
  • I believe you are correct that when trying to match a curve/distribution, statisticians will be more comfortable to assign a person a high IQ if they have a large sample of people.

    Where you are wrong is in simply discounting the achievement and mental discipline needed, since if there was a large sample of persons the test was calibrated on, this means the girl scored above all these sampled.
    But see it another way: If the girl had been in the sample used to calibrate the test before, we all would be considered to be less intelligent :-)

    Still remember, the IQ is just a number which is assigned by a mathematical, but not necessarily scientific model. For example, isn't the IQ distribution supposed to be a gauss curve ? If so, it should to be symmetric about the mean.
    That it is not suggests that the tests have been refined in a way that makes it easier to detect smart persons(IQ 100-300), while neglecting dumb persons(IQ 0-100). (ObPC: Excuses if you scored less than 100, you might still be disciplined enough or a mad enough scientist to be cool ).

  • Origonaly the hotel was paid $30
    Minus the 5 dollar refund the hotel has $25
    You paid have paid $27, $2 to the bellboy, and $25 to the hotel.

    $25 + $2 = $27 / 3 = $9 each.

    There is no missing dollar, as you can see.

  • Yeah, you're right. I saw this one in a Scientific American a few years ago, but I'm sure it's older then that.

  • 3 x 10 = 30 - paid
    30 - 5 = 25 - actually paid after refund of 5
    25 + 3 = 28 - what they paid plus refund
    28 + 2 = 30 - what was paid out plus the bellboys take
    each person didn't really pay 9 dollars, rather 28/3 or 9.333...

    I'm afraid that you're terribly mistaken because the pockets of the guests are missing 33 cents apiece if this solution is correct. You have not accounted for the missing dollar.

    Try this:

    3x10 paid out
    5 removed = 25 paid out
    25/3 = 8.33 paid each.
    8.33 + 1 returned = 9.33 accounted for each. The 66 cents missing each is the 2$ that the bellboy has.

    The guests didn't pay 9.33 each, they paid 8.33 each and got $1 each back. Those 33 "missing" cents apiece are in the hotel's cash register.

    Cyclopatra

    "We can't all, and some of us don't." -- Eeyore

  • My friend pointed out that while in the kitchen, you have a greater chance of seeing a boy's hockey schedule in the exact same sense that you stumble upon a girl's room. Therefor, 1/2...

    You can't control random environments, but you can easily make some vast oversimplifications of them with meaningless results ;)
  • Let's take another approach to this problem. And in this case we take 10 doors.

    When asked for my choice I say "I pick door number 2" out of the 10 doors.
    I have 10% chance of making the good choice.
    On the second step of the game I'm asked if I still choose door 2 or if I take the other one (the remainder of the 9 others). In other word I'm asked if I want or if I don't want door 2.

    Now, what if the games allows me to point directly the unique door I don't want to open?
    I say "I don't choose door 2".
    This will mean I'll choose a group of 9 doors which has a 90% chance of containing the prize. Removing 8 empty doors from this set won't change (since there is at least 8 empty doors) my 90% of making the good choice at the beginning.

    The second step of eliminating empty doors from the remaining 9 is a disguise. The real question is "do you choose this group of 1 door, or this group of 9 doors".
    Presented this way the answer is obvious. ;)

  • Your example holds, if you assume you pick correctly on the first guess 50% of the time, which is a flawed assumption (unless the game show is rigged, of course)

    Try looking at it this way: Monty knows which door the prize is behind. If your first pick is a door with no prize behind it, he can choose to open either of the other doors. If your first pick isn't the door with the prize behind it, he has to open the other door with no prize -- he doesn't have a choice. At the point where you pick a door (say door 1) and Monty is about to open a door there are three possibilities

    Possibility 1: Prize is behind door 1: 50% chance Monty will open either

    Possibility 2: Prize is behind door 2: 100% chance Monty will open door door 3. 0% chance Monty will open door 2

    Possibility 3: Prize is behind door 3: 0% chance Monty will open door 3. 100% chance Monty will open door 2.

    Each possibility has a 33% chance of happening. With Possibility 1, if you stay, you win and if you switch, you lose. With Possibility 2, if you stay, you lose and if you switch, you win. With Possibility 3, if you stay, you lose and if you switch, you win. Therefore the probability of switching and winning is 2/3.
  • This makes no sense, please post the bayesian rule you're using.

    The way I see it, in a collection of 3 binary numbers these are all the possibilities.

    case 0: 000
    case 1: 001
    case 2: 010
    case 3: 011
    case 4: 100
    case 5: 101
    case 6: 110
    case 7: 111

    Case 0 is thrown out because "You are told that there is at least one 1."

    Case 7 is removed because "You choose one digit randomly, and are told it is a 0."

    We can divide the remaining cases into two sets:
    Set A={001, 010, 100}
    Set B={110, 101, 011}

    The probability of there being two 0's = 1/2.
    The probability of there being two 1's = 1/2.

    Please explain your reasoning.
  • Could you deobfuscate the answer? Using log(7.39) = 2.0001 and log(6.011) = 1.793 I am stuck getting two, which is obviously not the right answer...
  • 3 x 10 = 30 - paid 30 - 5 = 25 - actually paid after refund of 5 25 + 3 = 28 - what they paid plus refund 28 + 2 = 30 - what was paid out plus the bellboys take each person didn't really pay 9 dollars, rather 28/3 or 9.333...
  • it didn't go anywhere, it's still there...

    another way to look at it is:

    hotel/motel got 30
    hotel/motel gave back 5
    bellboy got 2
    customers got 3

    hotel/motel has 25
    customers have 3
    bellboy has 2

    25 + 3 + 2 = 30

    either way you look at it, there's no money missing.
  • same numbers, just a different way of looking at them...and yes, I'll stick to both solutions...
  • make four 9's (9 9 9 9) equal 100 -- you can do whatever you like to the four 9's, add, multiply, put them in any order you want, but you must use only the four 9's
  • This is an open and shut case of logical analysis.

    Yes, everyone thinks that, and comes to the wrong conclusion,as you did. There's a discussion of the answer here [editthispage.com] - if you switch your choice your chances of winning are 2/3, if you stick, 1/3. The subtlety in the statistical argument derives from the fact that the host is giving you more information half way through the game, which the simplistic statistical interpretation you give doesn't take into account. But if you're still not convinced, the best way to provide evidence is to run a Monte Carlo simulation of the game, or to calculate every possible combination - which is what guy's co-workers were doing.

  • You have a non-zero binary number of 3 digits. You read in from left to right. The first number you encounter is a 1. The second number is a 0. What are the odds that the value of this binary number having the value 5?

    Your rule does not hold up in truly random environments. It's attraction is in semi-random environments like process lists and splay trees where it is likely that a certain process will be hit more often than another. Transmeta would know all about this. :-)

    Dancin Santa
  • Your example is bogus because though the likelihood of hitting the number with the greatest representation is greater than hitting one of lesser representation, it is equally likely for a 3 digit binary number to be 100 or 101 (assuming that you encountered the first two digits in this order).

    Flip a coin 3 times and record the results.

    You're saying that if I show you the first as heads, then the second as tails that the odds of the third flip being heads is *higher* than it being tails? I don't buy it.

    The odds of any child being male is 50%. Just because you encountered a female room first means very little when you consider that a house is not randomly laid out. "Take the first right" when "Take the first left" is correct could accidentally lead a guest to the girl's room.

    Let's say that you had a 10 digit binary number. If the first number you reveal is a 0 and the second is a 1, are you saying that the third number that you reveal is more likely to be a 0?

    Dancin Santa
  • There's a little duplication in the results, but this one is pretty quick (for some values of quick).

    • #!perl -w

    • for my $w (1..712) {
      • for my $x ($w..(712 - $w)) {

      • for my $y ($x..(712-($w + $x))) {
        my $z = 711 - ($w + $x + $y);
        printf ("%.2f :%.2f :%.2f :%.2f \n", $w/100,$x/100,$y/100,$z/100) if ($w*$x*$y*$z == 711000000);
      }
    }
    }

    Dancin Santa
    Of course the lameness filter thinks that I have too many lame characters. Maybe if I nip here and tuck there?
  • Yes, it's called a 'quine'. Check out Google [google.com].

    Dancin Santa
  • ... is 1/3, not the obvious 1/2.

    The key is that it is more likely that you stumble into a girl's room if there are two girls and one boy. This is easy to show using Bayes' rule.

    Because the hockey schedule is in a common location, you get no information from it other than there is at least one boy.

    I remember there was spirited debate in class over this one ...

  • I re-ran the numbers, and I would like to make the clarification: Every boy plays hockey with probability 1. (I'm Canadian after all.) Other than that I stand by the posted solution. The details are left as an exercise for the reader. :-)

  • Here's a variant on the Monty Hall problem that I got from a computer science prof, which illustrates arguments regarding Bayesian inference.

    Your friend invites you to his house. You know a priori that he has three children, but you have no information on their genders.

    While looking for the bathroom, you accidentally stumble into one of the children's rooms, and find that it is obviously a girl's room.

    While in the kitchen, you see a schedule for a boys' hockey team.

    So, you know that there is at least one boy and one girl. The question is: what is the probability that there are two boys and one girl?

    I am obfuscating the answer so as to let you think about it first: floor(log(7.39))/floor(10^(0.779))

  • I believe that, due to Monty's knowledge of which door conceals the car, your best bet is to switch doors. An explanation (a lengthy, not entirely coherent explanation) is below.

    My understanding of this problem is as follows: Monty chooses one door which to open. The other two are similar in that they are not chosen by him, but they are dissimilar in why they are not chosen by him. The unchosen door which you first select is unchosen simply because it is the door which you selected (and Monty cannot reveal to you whether your first selection was correct or there is no question of whether to switch doors). The remaining unchosen door was not chosen because the chosen door had a goat behind it. To put it another way, by not choosing this door, Monty wasn't forced to show you the car (although, it is possible that he wouldn't have been forced to show you the car, even if he'd not chosen the chosen door instead of this one).

    So Monty not choosing the door you initially selected gave you no new information about what's behind it (as he would have done it regardless of the location of the car). However, Monty not choosing the other door does give you new information: namely, that not choosing it didn't result in you seeing the car through the chosen door. (If one of the two doors which you didn't initially select is randomly eliminated -- equivalent to your random selection of a door when the game began --, it cannot, likewise, be said that Monty not choosing the door which you selected will result in Monty not revealing the location of the car. That is, if you initially select a goat door, and the other goat door is randomly eliminated, Monty not choosing your door does result in Monty revealing the location of the car.)

    Mathematically, your first selection is from three doors, one of which contains a car -- hence, there is a one-in-three chance that you'll select the door behind which is the car. Monty's choice of which door to open is between the remaining two. When making this decision, he may be faced with three car-goat arrangements (for simplicity, I'm just going to call the remaining doors "left" and "right," since there must be one of each): car left, goat right; goat left, car right; or goat left, goat right. Each of these is equally probable. In either of the first two, Monty must select the one door behind which is the goat, which means that the door not chosen must be the car; this is the case in two thirds of the possible arrangements. In the remaining case, Monty can select either goat door and the remaining door is also a goat. So in two of the three arrangements (or two thirds of them) the door not chosen based on Monty's knowledge of the location of the car actually conceals the car. The other one third of the time, Monty's knowledge is not needed and your first selection is, in fact, the correct one.

    2/3 + 1/3 = 1, so the car is still behind one and only one of the doors. But it is twice as likely to be behind the door which you didn't initially select. Monty tells you that by always selecting a non-car door to open. (If Monty selected a door at random, one third of the time he would select the car door and you'd have no chance of winning; two thirds of the time he'd select a goat door and then both doors would possess fifty-fifty odds. [0 + 1/2 + 1/2] / 3 = 1/3, so each door would have a one-in-three chance before he opened one of them and a one-in-two chance after he opened a non-goat door. That is the situation which you described.)

    The whole idea here is that you have more information about the door which you didn't initially select.

The first version always gets thrown away.

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