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Posers and Puzzles

Posers and Puzzles

  1. Standard member Nemesio
    Ursulakantor
    24 Jun '04 05:51 / 1 edit
    Dear Esteemed Community,
    I am a reasonably bright person (though not a particularly good chess player). I have fancied puzzles, trivia, and useless knowledge since as far back as I can remember. I did well on the math part of the SATs and the math and logic parts of the GREs. Though I work as a liturgical musician, I keep sharp with the very sorts of puzzles presented in this forum.

    AND I DON'T GET THE MONTY HALL PROBLEM. I've used the internet. I've worked it out. I've looked at it and stared and thought and contemplated. And it still seems that the odds are 1/2 once Monty shows the goat.

    Usually, when I don't get a puzzle (and I don't get a lot of the tough ones), if I look at the answer, I am able to work it out, learn a new way of thinking, and move on and apply it.

    Since I don't get the &^%$*^& Monty Hall problem, I certainly don't get this one with prisoners A, B and C. Can someone work this out for me?

    I've been here: http://mathworld.wolfram.com/MontyHallProblem.html

    and read (and reread, and rereread, etc. etc.) the explanation. And I get caught up here:

    But after Monty has eliminated one of the doors for you, you obviously do not improve your chances of winning to better than 1/3 by sticking with your original choice. If you now switch doors, however, there is a 2/3 chance you will win the car (counterintuitive though it seems).

    INDEED IT DOES SEEM COUNTERINTUITIVE. If he elimates the door, and asks if I want to switch, it is essentially the same as saying, "Ok, pick the door you REALLY want (the one you said first or the other one)." Obviously, two doors, one car, 1/2. Obviously???? Help!!

    My problems are analogous with the A, B, and C prisoner problem originally posed. Please help me. I am agonizing over this and I am sure it is easier than this web page is making it seem.

    I thank you all and am,

    Appreciatively yours,
    Nemesio
  2. 24 Jun '04 06:02
    Everyone I tried this on argues to their blue that the odds now become 1/2. But the way I got my head around this was to start with 1 million doors, my original choice is 1/1000000, once all other doors accept 1 are removed for my original chioce to be right it is still 1/1000000. So it would be madness to keep that door.......

    Is that helping?
  3. Standard member psychopath42
    Green Slime
    24 Jun '04 06:20
    Originally posted by nemesio

    But after Monty has eliminated one of the doors for you, you obviously do not improve your chances of winning to better than 1/3 by sticking with your original choice. If you now switch doors, however, there is a 2/3 chance you will win the car (counterintuitive though it seems).

    INDEED IT DOES SEEM COUNTERINTUITIVE. If he elimates the door, an ...[text shortened]... e you said first or the other one). Obviously, two doors, one car, 1/2. Obviously???? Help.
    what about:
    probability of A winning (originally) = 1/3
    probability of not A winning (originally) = 2/3

    Monty eliminates one of B or C as empty, and asks if you want to switch.
    does the probability of A winning become 1/2? It can't; it stays at 1/3.

    just because there are two doors left does not mean there is an equal probability of winning... due to conditional probability (the probabilities are conditioned on the fact that the contestent originally chose door A to begin with).
  4. Standard member Nemesio
    Ursulakantor
    24 Jun '04 06:27
    Originally posted by timebombted
    Everyone I tried this on argues to they're blue that the odds now become 1/2. But the way I got my head around this was to start with 1 million doors, my original choice is 1/1000000, once all other doors except 1 are removed for my original chioce to be right it is still 1/1000000. So it would be madness to keep that door.......

    Is that helping?
    No. It seems to reinforce my point, I think.

    If you have, let's say, 100 doors, your chance is 1% initially. Fair enough. But, if he keeps opening doors, showing goats, until there are two left (whether you switch throughout, occassionally, or never), one has a goat, one has a car. 1/2. I don't see how my chances remain 1% if I never switch and go to 99% if I do switch at the end.

    I guess I don't see the madness in keeping the door because the odds seem, to me, to change every time he gives new information. Basically he saying, "Pick a door, but now let me remove 98 of the 99 wrong answers, which door now?" Two doors, one prize, 1 in 2 shot? No?

    Ow ow ow ow ow.
  5. Standard member Nemesio
    Ursulakantor
    24 Jun '04 06:33 / 1 edit
    Originally posted by psychopath42
    Monty eliminates one of B or C as empty, and asks if you want to switch.
    does the probability of A winning become 1/2? It can't; it stays at 1/3.
    Does it become 1/2? Why not? I guess I don't see why the probablity of winning can't change whether or not I switch. I either have the right one or the wrong one; the other variable is taken out of the equation.

    Let's say I have a bag with three marbles in it, colored red, yellow and blue, and the goal is to get the red one. I reach in and clasp a marble and pull it out without looking. Now, someone reaches in the bag and pulls out a marble that isn't red from the bag. Given the choice, how is it better for me to switch. Either the one I have or the one I don't have is red. Seems like 1/2 chance.

    ow ow ow ow
  6. 24 Jun '04 06:39
    OK, to prove it to a friend I wrote a number between 1 and 10 on a piece of paper. She guessed a number..... I then eliminated 8 wrong numbers. As she believed its always 1/2 she stuck with her original choice. We ran this test 10 times and she failed to win the bunch of tulips once. If the odds were 1/2 you would expect 2-3 wins at least, and if you run this test enough you would expect closer to 50% win tulips 50% nothing right? But try this experiment and it will always be closer to 1/10 because thats the odds of the orginal chioce, the gameshow host eliminating 8 doors after your first choice has not effected the odds of your first chioce........

    Any better?
  7. Standard member psychopath42
    Green Slime
    24 Jun '04 06:45 / 3 edits
    how about a lottery example?

    say there are a million tickets, and only one winning ticket. (and say you cannot switch tickets)

    you take a ticket, and hold on to it. some authority slowly throws away all others but 1. then by your argument, your ticket has a 1/2 chance of winning.

    which means that half the time you should win the lottery! i like those odds!

    (there is a flaw with this reasoning, and it is due to conditional expectations... at the end of the above process, although there are two possible outcomes - either your ticket is a winner or the other ticket is a winner - the probability of occurance of these outcomes are *not equal*.)

    yes? no?
  8. Standard member Cribs
    Moderately Offensive
    24 Jun '04 07:22 / 8 edits
    Consider the game is played fifteen times. Suppose that your original guess
    is always A. Suppose that the prize is behind door A the first five times,
    behind door B the second five times, and behind door C the third five times.

    Now, suppose your strategy is to never switch. Then you will win the first five
    plays and lose the last ten plays. Your win ratio is 1/3.

    Alternatively, suppose your strategy is to always switch. You will lose the
    first five plays and win the last ten plays. Your win ratio is 2/3.

    Now, to explain why 1/2 seems like it should be correct solution, you have
    to consider a subtle variation on the problem. Suppose, Monty Hall does
    not know where the prize is. (This is as opposed to the traditional problem
    in which Monty has perfect information.) In this problem, if Monty opens a
    door with a goat, then you are correct: you have not gained any extra
    knowledge and your original guess will be right 1/2 the time.

    However, it is also possible in that formulation that when Monty opens the door,
    he will reveal the prize and the game is over. This is the key subtle difference: in the traditional problem, Monty is required to reveal a door without a prize. To put it another way, he is effectively sharing some of his complete information,
    and therefore the contestant is gaining information. This implies that the contestant's odds must increase by using this information - they do not remain at 50/50 as they would without extra information.

    So the key idea is that the contestant starts with zero information (i.e., all doors
    equally likely). Then Monty shares information with the contestant. If this were
    not the case (i.e., if Monty had no information to share) then 1/2 would be the
    correct solution.

    Finally, it is because of this subtle difference that the prisoner problem
    must be carefully specified with respect to the guard's information. In that
    thread people presented valid arguments for a solution of both 1/3 and 1/2,
    but they failed to recognize that both solutions are correct, depending on
    what the guard knows. If the guard has zero knowledge, then the prisoner's
    odds increase to 50/50 after hearing B's fate; if the guard has perfect
    knowldege, they remain at 1/3.

    The moral of the story is very important to grasp: probability is a measure
    of information; it is not a description of reality
    . In this case, the changing
    probabilities don't indicate that the prizes behind the doors jump around
    in different proportions once a door is opened; rather, the changing probabilites
    just reflect the contestant's increased information.

    Cribs
  9. Standard member Cribs
    Moderately Offensive
    24 Jun '04 08:12 / 2 edits
    Let's say I have a bag with three marbles in it, colored red, yellow and blue, and the goal is to get the red one. I reach in and clasp a marble and pull it out without looking. Now, someone reaches in the bag and pulls out a marble that isn't red from the bag. Given the c ne I have or the one I don't have is red. Seems like 1/2 chance.
    Consider this analysis of your example. You draw a ball and without looking at it, you know that 1/3 of the time it is red. That means that 2/3 of the time the red ball remains in the bag. That means that when your friend takes one away, a red
    marble remains in the bag 2/3 of the time. So putting all the information
    together, the red is in your hand 1/3 of the time and in the bag 2/3 of the time.
    Thus it is better to switch.

    Now, consider a variation on your problem in which your friend does not
    look and may or may not have taken the red one out. In this case, the red
    is in your hand 1/3 of the time, in the bag 1/3 of the time, and in your friend's
    hand 1/3 of the time. Thus your odds are 50/50 between keeping your original
    choice and swapping with the one in the bag.

    This reinforces the idea of my last post. Specifying the helper's knowledge
    is crucial. Once again we see that both 1/3 and 1/2 are valid solutions,
    depending on what the helper knows.

    Probability is nothing more than a measure of information.

    Cribs
  10. Standard member Nemesio
    Ursulakantor
    24 Jun '04 08:14
    I am beginning to see the Monty Problem; I am going to work my brain on it. But I am still struggling then with the prisoner one.

    So, in the prisoner problem, no complete information is given by the guard? That is, when the guard answers "All right, B will be executed tomorrow," he has not given any "complete information;" of course either B or C is going to be executed tomorrow. It isn't relevant to A which other prisoner is getting executed?

    Ok, the king puts one of the following slips of paper in a box, shows it to the guard, and seals it until the execution. The slips either have AB, AC, or BC on it.

    If the guard says "the slip had a B on it," he could be refering to slip AB or BC.

    B's chance of survival is 0%, period.
    There are two guys left. Why isn't A's chance 1/2?

    If the guard says "the slip had a C on it," he could be refering to slip AC or BC.

    C's chance is 0%, but A's is not 1/2?

    I, of course, see my own paradox; we already knew that the guard was going to say either B or C and we know that A's chance wasn't 1/2 to begin with. So where am I going wrong?

    Thanks to all who have been trying to help me.
  11. Standard member Nemesio
    Ursulakantor
    24 Jun '04 08:18
    I am of course assuming that the guard knows what is on the slip and that he tells the truth about it.
  12. Standard member Cribs
    Moderately Offensive
    24 Jun '04 08:19
    Originally posted by nemesio
    I am beginning to see the Monty Problem; I am going to work my brain on it. But I am still struggling then with the prisoner one.

    So, in the prisoner problem, no complete information is given by the guard? That is, when the guard answers "All right, B will be executed tomorrow," he has not given any "complete information;" of course either B or C is ...[text shortened]... 1/2 to begin with. So where am I going wrong?

    Thanks to all who have been trying to help me.
    You are not wrong. This analysis is perfect and yields the perfect solution of 1/2.
    But you are analysing the version of the problem in which the guard has zero information. That is, he had to draw the paper.

    A different analysis is required if the guard already knows who will be executed.
    In that case, the analysis is identical to Monty Hall.

    Cribs
  13. Standard member Cribs
    Moderately Offensive
    24 Jun '04 08:24
    Let me elaborate on what I mean by zero information and complete information.

    It is best illustrated by example. Consider the traditional Monty Hall problem
    in which Monty knows what is behind each door. What if Monty tries to guess
    where the prize is? His chances of getting it right are 100%. That is complete
    information. The contestant has zero information, which means that all doors
    are equally likely, for a success rate of 33%.

    Partial information would yield a success rate somewhere between 33% and 100%. The process of reavealing one bad door, gives the contestant
    sufficient parital information to improve his success rate to 66%. Better than
    where he started (zero knowledge) but not quite as good as Monty (complete
    knowledge).

    Cribs
  14. Standard member Nemesio
    Ursulakantor
    24 Jun '04 08:26
    Originally posted by Cribs
    You are not wrong. This analysis is perfect and yields the perfect solution of 1/2.
    But you are analysing the version of the problem in which the guard has zero information. That is, he had to draw the paper.

    A different analysis is required if the guard already knows who will be executed.
    In that case, the analysis is identical to Monty Hall.

    Cribs
    I'm sorry to waste your time. Could you please, as simply as you did with the Monty Hall problem, explain what you mean by zero and complete information; that is, why one analysis applies to one situation and a different one to another? I think this is where the (substantial) gap in my understanding/knowledge must lie.

    I thank you all again for all the help.

    Nemesio
  15. Standard member Cribs
    Moderately Offensive
    24 Jun '04 08:27 / 1 edit
    Originally posted by nemesio
    I am of course assuming that the guard knows what is on the slip and that he tells the truth about it.
    This is not what I mean by complete knowledge. Complete knowledge
    would mean that he wouldn't have to draw the slip in the first place.
    In the version you analyze, it is still a zero knowledge problem, and
    the correct solution is 1/2.

    Cribs