Puzzle

Discussion in 'Physics & Math' started by oxymoron, Jan 6, 2004.

  1. oxymoron Registered Senior Member

    Messages:
    454
    Quote: from ProCop
    What if you have two contestants: they stand each for his/her door. The quizman opens the 3rd door (empty). If the two remaining contestants switch their doors do they encrease their chances? If so people should switch their lotery tickets....

    I think you should take a course on Discrete Probability first. The probabilities in lottery is completely different to the probabilities in the game. The numbers that you choose DO NOT depend on the numbers chosen the week before nor do they depend on what someone else chose.

    Your choice in the game, however, DOES depend on the door that is opened. It is a conditional probability! (I'm not sure how many times I have to explain it?!)

    Go and read this site www.math.uah.edu/statold/games/games6.html

    If two contestants played then player A chooses door 1 and player B chooses door 2. If the car was behind neither door then the game ends! because the quiz man cannot open another door. In which case (which defies the rules of the game) you have no better chance of swapping doors.

    Say there was four doors and two contestants. They choose doors like above. The quizman can now open a door freely (door 4). Then for player A, if he swaps to either door 2 or door 3 (remember door 2 is held by player B which means player B would have to swap to door 3 if player A chose his door) he increases his chances of winning.

    Player A swaps from door 1 to door 2 - Player B MUST swap to door 3. Player B has a better chance of winning.
    Player A swaps from door 1 to door 3 - Player B can stay (same chance for him) or swap to door 1 (same chance again) - only player A has better chance.
    Player A stays (same chance) - Player B can swap to door 3 (better chance).
    Player A stays and Player B stays (both obviously have the same chance).

    If two people played this game then there is no scenario where both would have a better chance of winning. Apply this thought process to any number of doors/contestants. It is easy to visualise.
     
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  3. oxymoron Registered Senior Member

    Messages:
    454
    If there is someone who has a degree in probability/statistics could someone help. I am having a hard time explaining this properly. Thanks.

    Now here is a question from me.

    We all know that the escape velocity from the surface of the Earth is 11.2km/sec (I think it is 6.96ft/sec for the Imperial's here;-).

    The plan:
    We want to launch a small satellite into space (escape the gravitational pull so that it does not fall back). We have at our disposal a booster rocket which has enough fuel to climb to an altitude of 1000km. Hereafter the satellite package disconnects from the rocket and continues on its way. The catch is it is not very reliable at high speed so we cannot just launch it at full speed to guarantee an escape trajectory. Full speed is 12km/sec.

    So I ask you, what is the minimum initial velocity required for the booster rocket so that the satellite makes it safely into space?

    (Booster Rocket has constant acceleration).

    Please show some working.
    Have fun!
     
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  5. malkiri Registered Senior Member

    Messages:
    198
    I'll take a shot at it...we'll see if it comes out any clearer.
    Let's restate the problem slightly. Instead of Monty Hall opening a losing door and giving you the chance to switch, he instead opens no extra doors, but gives you the following choice: either open the door you originally selected, or open all of the other doors, and keep the prize if it's behind one of those doors. This is essentially the same thing...think about it for a moment if you don't buy it. Monty Hall always opens up all the other doors except for one. In the new problem, he's avoiding the work and instead making you open all those doors in addition to the one you switched to.

    If you agree that these problems are equivalent (I'm not sure I did a fantastic job of convincing you), then you can see that your chances are increased by switching. Do you stick with your original selection, which has a 1/3 chance of winning? Or choose to open all the other doors, which has a 2/3 chance of winning?
     
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  7. contrarian Registered Senior Member

    Messages:
    110
    I'll give this one more shot. Let's say you pick the first envelope. The odds that you have the 100 euros are 1/100, the odds that someone else has it is 99/100. I think we agree so far.

    Now when you throw out an empty envelope in the unpicked pile, do you change the probability that the original card you picked is the one with the euros? No, you don't, you are simply changing the probability that any given card that you did not pick is the one with the euros. (you may want to reread these sentences so you can see the difference between them. The probability that the one you chose is the winner will always be 1/100(unless you eliminate all the unpicked envelopes in which case the probability is 1). This may not seem logical but it is.

    BTW, did you try out the test I suggested? I think it would clear up this misconception fairly quickly.

    Cheers,

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  8. BigBlueHead Great Tealnoggin! Registered Senior Member

    Messages:
    1,996
    Oxy - the probability is set when the car and player are placed, not when the door is opened. One the car and player positions are chosen, the initial probabilities don't change, only the apparent ones.
     
  9. hockeywings Don't dance without music Registered Senior Member

    Messages:
    132
    I would agree with procop here.

    To put it simply if a bad door is ALWAYS opened, you should know from the start you have a 50/50 shot at it.

    NO matter which door you pick one of the remaining two will be able to and will be opened, leaving you with 2 possible doors. Only one of which is the car. 1 out of 2 is 50 percent.
     
  10. ProCop Valued Senior Member

    Messages:
    1,258
    Sorry contrarian, I do not have a pack of cards.

    Malkiri, read my roulete example on the previous page. If I had a chance to bet or on one number from 100 or 99 from 100 I would naturally choose the 99 option. But it is not the case here. It is being proposed that if I eliminate choices the remaining choises accummulate the chance capacity from the eliminated ones while my single choice (which I made at the first moment 1/100 remains that!!!) (while the other choices grow). The same with the Monty door. The door I choose inicially has chance 1/3 the remaining 2nd door took over the chance from the eliminated 3rd door. It has no logic in it because my roulete example or 100 envelope example show that my chance encreases to. It is obvious that something is wrong or in my 1/2 or in 2/3 reasoning. I wish Dinosaurus were around, he was very good at chance problems.
     
  11. malkiri Registered Senior Member

    Messages:
    198
    The difference between this problem and your roulette problem is that in the roulette problem, the croupier has as much knowledge as you about where the ball will land - none. In this problem, Monty knows where the prize is. You can leverage this knowledge.

    An analogous roulette variation would be if the croupier knew exactly where the ball would land...say, there's a magnet at that spot. He shuts off 98 extra holes, leaving the one you chose and another one. Knowing that he knows, you now decide to switch. It's not a 50/50 situation since it's not completely up to chance.

    Let me try another approach.
    I make my initial choice. The door I choose may be the winner (1/100), but it's probably a loser (99/100). Say it's the winner. Monty will open 98 losing doors and leave one extra losing door. Here, if I switch, I obviously miss out.
    Say instead I chose a loser. Monty opens 98 losing doors and leaves the winning door. Here, if I switch, I win.
    What would you choose to do if you didn't know whether your first choice was a winner or not?
     
  12. ProCop Valued Senior Member

    Messages:
    1,258
    I understand your argument. What I do not understand why 100 envelopes example obviously leads to different picture

    <i>, we stand in a room with 100 people. Everybody gets an envelope. One envelope contains 100 euro. The more people open their envelope and it's empty the more chance it is that we two have got it. Now only me and you have an not open envelope. You want to switch it (believing chance is 99/100 I have got it). </I>


    If monty (it doesn't matter if he knows - because I do not know) opens more and more empty doors/(enfvelopes) my doors chance encreases too (the last two doors or envelopes do not differ (it is 50:50). Would you try to swap your envelope with me to encrease your chance (haha if we swap we <b>both</b> would have encreased our chance 99/100 + 99/100 pure nonsence). I think the reason is that we believe the TV more then our own brains....
     
  13. hockeywings Don't dance without music Registered Senior Member

    Messages:
    132
    hmmmm i think i am seeing the other side, in the 100 choices example you start off with a 1/100 chance of winning and 99/100 chance of losing.

    Now if you take away 98 leavin you and a loser, then 99/100 you would be switching from a loser to a winner, aha, i see what you guys are saying now.

    Taken with the monte game you start with a 1/3 chance of winning and 2/3 chance of losing. one is taken away and you have 2/3 chance of switching from loser to winner and only a 1/3 chance of STAYING with a winner.

    Am i correct with what you guys were saying or am i just confusing myself?
     
  14. contrarian Registered Senior Member

    Messages:
    110
    This is the flaw in your reasoning. Your chance to win does not increase when you remove an empty envelope.

    Cheers,

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    P.S. you've got it, hockeywings

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  15. hockeywings Don't dance without music Registered Senior Member

    Messages:
    132
    so it basically rests with your beginning chance of losing haha, that is kinda odd
     
  16. malkiri Registered Senior Member

    Messages:
    198
    The 100 envelopes example as you originally posed it is not the same because it had no Monty Hall - it was just a bunch of people with no knowledge about the state of all the envelopes. Say those 98 other people all opened their envelopes. There's a good chance the winner will be in that group. In the Monty Hall problem, the winning door will never be in the group that's opened. In the envelope example, I agree that there's no point in swapping, since there's no reason for you to assume any other envelope is better than yours.

    Thus it does matter that Monty knows, even if you don't know. He will never open a winning door. You will have selected a losing or winning door to begin with (99/100 and 1/100, respectively), and he will leave you a winning or losing door to switch to, on purpose, depending on which you originally chose.
     
  17. ProCop Valued Senior Member

    Messages:
    1,258
    I think we do not know what Monty realy wants. He may wont you to loose to save the car for the next show.

    I have two groups of envelopes. G1 1 envelope G2 99 envelopes.
    Somewhere in these 100 envelopes is the 100 euros. I do not know where. At first moment I would choose G2 - obvious.

    But if 98 envelopes are removed one by one opened and empty at random by monty then the chance is 50:50 (for my original envelope and the last one of G2).

    If Monty knows which envelope holds the 100 and leaves it as last one (from G2) then the switch is obvious (again). But is such a form of game a chance game?

    But after thinking it through I think it is best strategy/chance to switch the door. Thanks all for their effort.
     
    Last edited: Jan 27, 2004
  18. malkiri Registered Senior Member

    Messages:
    198
    The circumstances in the original game show were indeed more complex than this problem. This problem is a simplified and well defined version, in which we can assume that Monty will always behave as described. If he starts doing other things, we're in deeper trouble than we already are.
    The point about knowledge is that Monty holds otherwise hidden knowledge about the problem, not we know what Monty will do (despite the fact that in this problem we do).

    Again, I don't have a problem with the original version of this. It's simply not the same as the Monty Hall problem. Adding in Monty, his knowlegde, and our conscious decision to leverage it makes it the same, so if Monty opens up 98 incorrect envelopes, knowing where the winner lies, the remaining chance is not 50/50.

    Is it pure chance? No. Are elements of it left to chance? Yes. Again, this is my point...it's no longer pure chance because we can leverage Monty's knowledge to obtain a favorable result.

    Glad we were able to convince you

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  19. oxymoron Registered Senior Member

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    454
    By the way, I have yet to come up with an answer for the magic square problem. I think I am going to have to give up or at least concede a hint.
     
  20. contrarian Registered Senior Member

    Messages:
    110
    Hi oxymoron

    Given the following magic square

    a b c
    d e f
    g h i

    Such that
    abc=def=ghi=adg=beh=cfi=ceg=aei

    IOW the products of the rows, columns and diagonals are the same and all the numbers are different, we can say that all the products =e^3

    IOW,
    abc=def=ghi=adg=beh=cfi=ceg=aei=e^3

    I can show you the derivation if you like, hopefully this should get you a little closer.

    Cheers,

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  21. hockeywings Don't dance without music Registered Senior Member

    Messages:
    132
    hmmm me thinks me has it

    2 36 3

    9 6 4

    12 1 36

    It didnt have to have equal rows and stuff did it?
     
  22. hockeywings Don't dance without music Registered Senior Member

    Messages:
    132
    PS very good hint
     
  23. hockeywings Don't dance without music Registered Senior Member

    Messages:
    132
    whoops typed it out wrong


    2 36 3

    9 6 4

    12 1 18
     

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