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It's Math-Puzzle Time

2456

Posts

  • SuckafishSuckafish Registered User regular
    edited April 2008
    Absurdist wrote: »
    Nobody can solve my puzzlers. Do I win something?

    10, 0, 3, 20, 16, 51, 32, 67, 74, ?
    Spoiler:

    Absurdist wrote: »

    1, 4, 7, 12, 15, 18, 21, 24, 27, ?
    Spoiler:

  • SuckafishSuckafish Registered User regular
    edited April 2008
    Here is a fairly easy one.

    If my three were a four, and my one where a three, what I am would be nine less than half what I'd be.
    I am a three digit, whole number. What am I?

  • SavantSavant Registered User regular
    edited April 2008
    Suckafish wrote: »
    Here is a fairly easy one.

    If my three were a four, and my one where a three, what I am would be nine less than half what I'd be.
    I am a three digit, whole number. What am I?
    Spoiler:

  • ScooterScooter Registered User regular
    edited April 2008
    It has multiple answers, doesn't it?

  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    Suckafish wrote: »
    Absurdist wrote: »
    Nobody can solve my puzzlers. Do I win something?

    10, 0, 3, 20, 16, 51, 32, 67, 74, ?
    Spoiler:

    Absurdist wrote: »

    1, 4, 7, 12, 15, 18, 21, 24, 27, ?
    Spoiler:
    And either way, it's not a math puzzle.

    It's an easy game to hate
  • FunkyWaltDoggFunkyWaltDogg Registered User regular
    edited April 2008
    Absurdist wrote: »
    Apothe0sis wrote: »
    Absurdist wrote: »
    Apothe0sis wrote: »
    This is most assuredly not my series, and math geeks will probably see why this is funny.

    2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

    Usually when they jump right from small numbers into enormous numbers it means that the series is a list of numbers with an unusual property.
    Spoiler:
    You probably have to recognise the series. It's concievably solvable, but you'd have to have an Erdos number of 0 or something to do so.

    REAL SPOILER
    Spoiler:

    I'm just happy that I know what an Erdos number is. :)

    Speaking of, wouldn't anyone with an Erdos number of 0 be dead, and thus unable to solve the problem?

    Burnage wrote:
    FWD is very good at this game.
  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    Scooter wrote: »
    It has multiple answers, doesn't it?
    Not for 3 digits, as far as I can tell.

    More complex explanation:
    Spoiler:


    Also, Paul Erdös is the only person with an Erdös number of 0, and he is dead. There are a lot of people still alive with an Erdös number of 1, however. And Natalie Portman has an Erdös number of 5.

    It's an easy game to hate
  • ScooterScooter Registered User regular
    edited April 2008
    Spoiler:

  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    Scooter wrote: »
    Spoiler:
    I don't think that the third digit is allowed to change. That's the limitation that keeps it down to six possible additions.

    It's an easy game to hate
  • ElJeffeElJeffe Super Moderator, Moderator, ClubPA mod
    edited April 2008
    Scooter wrote: »
    It has multiple answers, doesn't it?
    Not for 3 digits, as far as I can tell.

    More complex explanation:
    Spoiler:


    Also, Paul Erdös is the only person with an Erdös number of 0, and he is dead. There are a lot of people still alive with an Erdös number of 1, however. And Natalie Portman has an Erdös number of 5.
    Spoiler:

    I always like math problems that favor reason over brute force.

    [While watching popcorn in the microwave]
    Maddie: "Look Riley, the bag's as big as your head now!"
    Riley: "Hahaha, yeah!"
    Maddie: "Look, now it's as big as your butt!"
    Riley: "Omigosh, it looks just like my butt!"
  • MoridinMoridin Registered User regular
    edited April 2008
    A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

    On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

    The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

    "I can see someone who has blue eyes."

    Who leaves the island, and on what night?


    There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

    And lastly, the answer is not "no one leaves."

    Shamelessly stolen from http://www.xkcd.com/blue_eyes.html

    No one ever believes the reasoning behind the answer to this at first. It's like the whole "which door do you look behind" game-show puzzle in terms of the answer being counterintuitive. Speaking of...I should look that one up, too.

    sig10008eq.png
  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    Sounds like you still used some form of brute force, Jeffe - testing the 9, then the 8. I only tested a few more, for thoroughness. :D

    It's an easy game to hate
  • ElJeffeElJeffe Super Moderator, Moderator, ClubPA mod
    edited April 2008
    Sounds like you still used some form of brute force, Jeffe - testing the 9, then the 8. I only tested a few more, for thoroughness. :D

    Well, I used a little brute force. But only after I'd whittled down the pool of options to a few possibilities. The alternative was figuring out how to set up a system of equations and solving for X, Y and Z. My way took about 3 minutes.

    [While watching popcorn in the microwave]
    Maddie: "Look Riley, the bag's as big as your head now!"
    Riley: "Hahaha, yeah!"
    Maddie: "Look, now it's as big as your butt!"
    Riley: "Omigosh, it looks just like my butt!"
  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    I love math!

    It's an easy game to hate
  • SmasherSmasher Starting to get dizzy Registered User regular
    edited April 2008
    Moridin wrote: »
    Shamelessly stolen from http://www.xkcd.com/blue_eyes.html

    No one ever believes the reasoning behind the answer to this at first. It's like the whole "which door do you look behind" game-show puzzle in terms of the answer being counterintuitive. Speaking of...I should look that one up, too.

    That one's called the Monty Hall problem. I'll save you the trouble and post it:

    You're asked to choose among three closed doors. Behind one door is a car, and behind the other two are goats. After you make your pick (though the door you picked is not opened, so you don't know what's behind it), the host opens one of the two doors you didn't pick. The host knows which door the car is behind, and he will always open a door with a goat behind it. He then gives you the option of switching your pick to the other unopened door. Assuming you want the car rather than a goat, should you switch, and why or why not?

  • enlightenedbumenlightenedbum Ann Arbor, MichiganRegistered User regular
    edited April 2008
    Smasher wrote: »
    Moridin wrote: »
    Shamelessly stolen from http://www.xkcd.com/blue_eyes.html

    No one ever believes the reasoning behind the answer to this at first. It's like the whole "which door do you look behind" game-show puzzle in terms of the answer being counterintuitive. Speaking of...I should look that one up, too.

    That one's called the Monty Hall problem. I'll save you the trouble and post it:

    You're asked to choose among three closed doors. Behind one door is a car, and behind the other two are goats. After you make your pick (though the door you picked is not opened, so you don't know what's behind it), the host opens one of the two doors you didn't pick. The host knows which door the car is behind, and he will always open a door with a goat behind it. He then gives you the option of switching your pick to the other unopened door. Assuming you want the car rather than a goat, should you switch, and why or why not?

    Argument start! Though simulations prove the following is correct:
    Spoiler:

    Interestingly to me at least, Deal or No Deal is built on a similar principle, but allows you to open the biggest money case. I've never decided on an optimum strategy for that show, though it might be an interesting problem in itself.

  • MoridinMoridin Registered User regular
    edited April 2008

    Interestingly to me at least, Deal or No Deal is built on a similar principle, but allows you to open the biggest money case. I've never decided on an optimum strategy for that show, though it might be an interesting problem in itself.

    Eh, deal or no deal is an exercise in expected value. And I'm pretty sure what the banker offers as a deal is always a bit less than the expected value, too.

    sig10008eq.png
  • enlightenedbumenlightenedbum Ann Arbor, MichiganRegistered User regular
    edited April 2008
    Moridin wrote: »

    Interestingly to me at least, Deal or No Deal is built on a similar principle, but allows you to open the biggest money case. I've never decided on an optimum strategy for that show, though it might be an interesting problem in itself.

    Eh, deal or no deal is an exercise in expected value. And I'm pretty sure what the banker offers as a deal is always a bit less than the expected value, too.

    It's actually not, I did do that math at one point trying to figure out what they were doing to calculate it, figuring as you did. It's weighted by how many cases are left. More cases = over expected value and as the number decreases it goes below.

  • ElJeffeElJeffe Super Moderator, Moderator, ClubPA mod
    edited April 2008
    I started a thread on Deal Or No Deal shortly after it debuted to discuss optimum strategies. It didn't get much interest.

    [While watching popcorn in the microwave]
    Maddie: "Look Riley, the bag's as big as your head now!"
    Riley: "Hahaha, yeah!"
    Maddie: "Look, now it's as big as your butt!"
    Riley: "Omigosh, it looks just like my butt!"
  • musanmanmusanman Registered User regular
    edited April 2008
    I taught the monty hall problem today in Algebra.

    In other news, I need to know why that eye problem is tricky...

    can't he see 100 blue eye'd people? All the time? That doesn't help them know their own eye color. I don't think I understand the problem

    sic2sig.jpg
  • LaudaniLaudani Registered User
    edited April 2008
    Absurdist wrote: »
    Here is another tricky one!

    What's the next number in this sequence?

    10, 0, 3, 20, 16, 51, 32, 67, 74, ?


    Wow thats a killer, it must be like "C(x) = C(x-1)+C(x-2)" or some weird sequence.

    Good Puzzle!

  • SuckafishSuckafish Registered User regular
    edited April 2008
    Moridin wrote: »
    ...
    "I can see someone who has blue eyes."

    Who leaves the island, and on what night?
    Spoiler:

  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    I really enjoyed the blue-eyes problem - if you want to follow other peoples' train of thought on it, I'd check this thread out first.

    It's an easy game to hate
  • SavantSavant Registered User regular
    edited April 2008
    I think I just figured out the blue eyes one. It's pretty weird and unintuitive.

    Edit: Yep, I was right. You just have to approach it the right way and it falls out.

  • GooeyGooey Registered User regular
    edited April 2008
    ElJeffe wrote: »
    I started a thread on Deal Or No Deal shortly after it debuted to discuss optimum strategies. It didn't get much interest.

    I find it amazing that people will get an offer of 250,000 or whatever and turn it down because they are convinced they have the 1,000,000 case. If I were on a show like that I would go long enough to get rid of my debt, if not take a huge chunk out of it, and then quit.

    919UOwT.png
  • ElJeffeElJeffe Super Moderator, Moderator, ClubPA mod
    edited April 2008
    Gooey wrote: »
    I find it amazing that people will get an offer of 250,000 or whatever and turn it down because they are convinced they have the 1,000,000 case. If I were on a show like that I would go long enough to get rid of my debt, if not take a huge chunk out of it, and then quit.

    Most people probably think the same thing, before being wooed by HUGE MONIES (that they probably won't get, but whatever, details).

    [While watching popcorn in the microwave]
    Maddie: "Look Riley, the bag's as big as your head now!"
    Riley: "Hahaha, yeah!"
    Maddie: "Look, now it's as big as your butt!"
    Riley: "Omigosh, it looks just like my butt!"
  • AroducAroduc regular
    edited April 2008
    One little counterintuitive riddle that I always find interesting to see how people react is this:

    You're offered two briefcases. One contains twice as much money as the other. You pick briefcase A, and it's shown to have $X in it. You're then offered the chance to switch to briefcase B. Should you?

  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    Logically, you should. You stand to lose 50% of your winnings with an equivalent chance to increase them by 100%.

    It's an easy game to hate
  • SavantSavant Registered User regular
    edited April 2008
    Aroduc wrote: »
    One little counterintuitive riddle that I always find interesting to see how people react is this:

    You're offered two briefcases. One contains twice as much money as the other. You pick briefcase A, and it's shown to have $X in it. You're then offered the chance to switch to briefcase B. Should you?
    Spoiler:

  • ElJeffeElJeffe Super Moderator, Moderator, ClubPA mod
    edited April 2008
    Logically, you should. You stand to lose 50% of your winnings with an equivalent chance to increase them by 100%.

    This is true whether you open the case or not. You could just say, "Given this unopened case, I stand to either gain X or lose X/2 by switching."

    I understand the "logic", but I would like to see a simulation run to empirically demonstrate this, because it's nor merely counterintuitive, it's nonsensical.

    edit: I guess the problem is that if it's true that you're better off swapping to case B after picking case A, it's also true that you're better off swapping to case A after picking case B. Thus, you're better off ending up with case B and also better off ending up with case A which are mutually exclusive statements.

    [While watching popcorn in the microwave]
    Maddie: "Look Riley, the bag's as big as your head now!"
    Riley: "Hahaha, yeah!"
    Maddie: "Look, now it's as big as your butt!"
    Riley: "Omigosh, it looks just like my butt!"
  • SavantSavant Registered User regular
    edited April 2008
    ElJeffe wrote: »
    Logically, you should. You stand to lose 50% of your winnings with an equivalent chance to increase them by 100%.

    This is true whether you open the case or not. You could just say, "Given this unopened case, I stand to either gain X or lose X/2 by switching."

    I understand the "logic", but I would like to see a simulation run to empirically demonstrate this, because it's nor merely counterintuitive, it's nonsensical.

    edit: I guess the problem is that if it's true that you're better off swapping to case B after picking case A, it's also true that you're better off swapping to case A after picking case B. Thus, you're better off ending up with case B and also better off ending up with case A which are mutually exclusive statements.

    Actually, now that I think about it I was wrong.
    Spoiler:

  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    ElJeffe wrote: »
    edit: I guess the problem is that if it's true that you're better off swapping to case B after picking case A, it's also true that you're better off swapping to case A after picking case B. Thus, you're better off ending up with case B and also better off ending up with case A which are mutually exclusive statements.
    This seems to be a problem with wording. You are not truly "better off" switching if you already have the high case, but with the limited information available you have a better risk-benefit ratio by switching, even if ultimately it is not a good choice.

    This actually seems similar to the blue-eyes problem in one respect: you begin with a catalyst action already fulfilled. In the blue-eyes problem, the guru has declared the existence of blue eyes, beginning the problem. In this one, you've already picked one case, giving you a starting position and dilemma - stay or switch. If you hadn't chosen anything yet, then the two cases are functionally identical and your probability is 1/2 either way.


    Savant, you're saying that an equivalent value Y is at risk in either direction. The problem is that it depends on whether or not your case is the more valuable one. If X is the high case, X = 2Y and you're risking Y for a perceived chance at 4Y. If X is the low case, X = Y and you're risking Y/2 for a perceived chance at 2Y. But no matter what, you're risking X/2 for 2X, and how Y and X are related is ultimately irrelevant. You've simply changed the variable's name.

    It's an easy game to hate
  • SmasherSmasher Starting to get dizzy Registered User regular
    edited April 2008
    It seems counterintuitive because there's two separate situations you could be in. Let's say the briefcase has $100 in it. In one situation the other briefcase has $200, while in the other situation it has $50. Since the two situations are presumably equally likely, your expected value is higher for switching than it is for staying.

    Let's contrast this problem with a similar one that has a different answer. Suppose you're told the briefcases have Y and 2Y dollars in them. In this case you'll either gain or lose Y dollars by switching, and so your expected value for switching is 0 (ie, it doesn't matter if you do or not). This holds whether you know the value of Y or not.

    The difference between the two scenarios is that in the latter one you know the value of both briefcases, while in the former you don't. This frequently causes confusion when people treat the first one as if it were the second one, which it is not.

    BTW, don't look at ratios, they'll mess you up. All that matters is the absolute gain or loss you get from switching.

  • AroducAroduc regular
    edited April 2008
    Hee hee... I told you people would give interesting answers. Some completely wrong ones in there, but it makes the brain do strange things.

  • SavantSavant Registered User regular
    edited April 2008
    Smasher wrote: »
    It seems counterintuitive because there's two separate situations you could be in. Let's say the briefcase has $100 in it. In one situation the other briefcase has $200, while in the other situation it has $50. Since the two situations are presumably equally likely, your expected value is higher for switching than it is for staying.

    Let's contrast this problem with a similar one that has a different answer. Suppose you're told the briefcases have Y and 2Y dollars in them. In this case you'll either gain or lose Y dollars by switching, and so your expected value for switching is 0 (ie, it doesn't matter if you do or not). This holds whether you know the value of Y or not.

    The difference between the two scenarios is that in the latter one you know the value of both briefcases, while in the former you don't. This frequently causes confusion when people treat the first one as if it were the second one, which it is not.

    BTW, don't look at ratios, they'll mess you up. All that matters is the absolute gain or loss you get from switching.

    Yeah, this problem is the second case. There are two briefcases at the start, and one has twice as much money as the other. So they start at Y and 2Y, but you just don't know which is which or what Y is. When you open one you know that X=Y or X=2Y, but that has no effect on how much you stand to gain or lose by switching between them. You just know you will gain or lose X/2 or X by switching, but you don't have an equal chance of losing or gaining those two values. It is set that it is one or the other, you just don't know which.

  • ElJeffeElJeffe Super Moderator, Moderator, ClubPA mod
    edited April 2008
    I'm going to go all thought-experiment on this one. Pretend you write a computer simulation to model the act of picking a case, then deciding whether or not to switch. In the first case, we write it to not show the computer person how much is in the case. Since the argument is that switching is statistically better, we'll run two simulations - one in which the computer always switches, and one in which the computer never switches.

    Functionally, it'll work as follows:

    Case I:
    - Computer chooses A or B.
    - Computer asked if he wants to switch.
    - Computer keeps initial choice.

    Case II:
    - Computer chooses A or B.
    - Computer asked if he wants to switch.
    - Computer changes.

    I think it's pretty clear here that picking A then switching to B is indistinguishable from picking B at the outset, and picking B then switching to A is indistinguishable from picking A at the outset - there's no reason why it wouldn't be. If we ran this simulation, we would find no statistical discrepancy in how much money each computer person expected to get.

    Okay, now let's recreate the situation in the puzzle. We add a step in there whereby we "show" the computer the contents of his initial choice. It's easy to see that adding such a step would have no effect on the internal logic of the first simulation - if nothing functionally changes in the simulation, then the simulation would have to yield identical results. ie, there would be no statistical change between keeping the initial choice and switching.

    [While watching popcorn in the microwave]
    Maddie: "Look Riley, the bag's as big as your head now!"
    Riley: "Hahaha, yeah!"
    Maddie: "Look, now it's as big as your butt!"
    Riley: "Omigosh, it looks just like my butt!"
  • SithDrummerSithDrummer Registered User regular
    edited April 2008
    @ Savant: That only works if you maintain that X is a variable, but once you've opened the case, it's not.

    @ ElJeffe: Your thought experiment begins with the choice of A or B, which has already been made in the original question. Choosing to switch or stay when given a tangible piece of the Y/2Y is the dilemma.

    It's an easy game to hate
  • SavantSavant Registered User regular
    edited April 2008
    t Savant: That only works if you maintain that X is a variable, but once you've opened the case, it's not.

    Opening the case has no effect on how much money is in them, or what the probability of choosing the bigger case is. It just changes how much you know about the possible amounts of money in the cases.

    If you started out with a briefcase with X dollars, and then were offered to switch with equal chances of doubling or halving your money, then you would always want to switch. But you don't have equal chances of doubling or halving in this problem. It is set with probability 1 what you will gain or lose by switching.

  • Smug DucklingSmug Duckling Registered User regular
    edited April 2008
    Moridin wrote: »
    Shamelessly stolen from http://www.xkcd.com/blue_eyes.html

    Didn't read through yet, so I don't know if someone's done this already:
    Spoiler:

    smugduckling,pc,days.png
  • Marty81Marty81 Registered User regular
    edited April 2008
    Apothe0sis wrote: »
    2, 5, 877, 27644437, 35742549198872617291353508656626642567, 359334085968622831041960188598043661065388726959079837 ... ?

    :^:
    Spoiler:

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