Advanced Monty Hall Problem with N door and M cars

Discussion in 'Scientific Statistics Math' started by Jack, Feb 21, 2005.

  1. Jack

    Jack Guest

    Hello Everybody,
    I really need you help in solving an advanced Monty hall problem.
    Assuming that we have N door with M of these door have cars behind
    them and N-M of these door has $1 bill behind them.
    Assuming that the player made an initial selection (she picked a door
    D1), now.... Monty open a door that has a $1 bill behind .... now he
    asks her, if she want to stay with her initial choice or if she wants
    to switch?

    What is the probability of winning if she stay with her choice? Isn't
    it P(winning if she stay with her inital decision) = M/N?


    What is the probability of winning a car if she switched to different
    door?


    Thanks
    Jack
     
    Jack, Feb 21, 2005
    #1
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  2. Jack

    Anon. Guest

    If you understand the original Monty Hall problem, then you should be
    able to work this out. If necessary, just write down the different
    possibilities.

    I'm not giving you the answer - this might be homework.

    Bob

    --
    Bob O'Hara
    Department of Mathematics and Statistics
    P.O. Box 68 (Gustaf Hällströmin katu 2b)
    FIN-00014 University of Helsinki
    Finland

    Telephone: +358-9-191 51479
    Mobile: +358 50 599 0540
    Fax: +358-9-191 51400
    WWW: http://www.RNI.Helsinki.FI/~boh/
    Journal of Negative Results - EEB: www.jnr-eeb.org
     
    Anon., Feb 21, 2005
    #2
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  3. It is not enough to say that "Monty opens a losing door."
    Is that "losing" a description, or a requirement?

    The vital trick of the original Monty Hall problem was
    that Monty *always* opens the door that loses - when
    the other one is a winner.
    The P(winning) does not change if we don't
    know what motivates opening what door, but
    assume that there is SOMETHING.

    If it was totally random, then P(winning) for every door
    is adjusted by the "-1" in the computations.
     
    Richard Ulrich, Feb 21, 2005
    #3
  4. Jack

    Jack Guest

    I promise you that this is not a homework.
    And I actually calculated the probabilites ... but I am sure of them:

    P(staying)=M/N

    P(swtiching) = (M/N)*(N-1)/(N-2)

    Do you think that is true.


     
    Jack, Feb 21, 2005
    #4
  5. Jack

    Anon. Guest

    Yep, that's what I get too.

    Bob

    --
    Bob O'Hara
    Department of Mathematics and Statistics
    P.O. Box 68 (Gustaf Hällströmin katu 2b)
    FIN-00014 University of Helsinki
    Finland

    Telephone: +358-9-191 51479
    Mobile: +358 50 599 0540
    Fax: +358-9-191 51400
    WWW: http://www.RNI.Helsinki.FI/~boh/
    Journal of Negative Results - EEB: www.jnr-eeb.org
     
    Anon., Feb 22, 2005
    #5
  6. Richard Ulrich wrote in
    Are you really able to demonstrate that the motivation is important? My
    hunch is that you will not be able to show, either via simulation or
    logical proof, that the motivation of the door-opener is the important
    aspect of the problem.

    The crucial factor is that new information has been added to the
    situation after the initial selection by the contestant. **Why** that
    information was given should be immaterial.
     
    David Winsemius, Feb 22, 2005
    #6
  7. [snip, about 'revised' Monty Hall problem
    "Why" matters.

    Here is logical proof. Go back to the original problem,
    with 3 doors and one car, where Monty always shows you
    a losing door, before offering a switch. That rule matters.

    OPPOSITE RULE --
    If Monty ALWAYS open the door that HAS the car,
    if your choice did not, then "you lose" immediately
    when he reveals a car and kills the game, or "you lose"
    when you switch, because it will only happen when
    your first choice was correct.

    NO RULE --
    If Monty randomly picks between two doors, then NO new
    *relevant* information is added, and there is no gain
    in switching -- Monte Carlo simulation will illustrate that.
     
    Richard Ulrich, Feb 22, 2005
    #7
  8. Jack

    Reef Fish Guest

    No, it doesn't, GIVEN that the opened door has no car.

    The contestent has chosen a door which MAY or MAY NOT be a winner.
    If Monty ALWAYS shows a losing door, before offering a switch,
    then whether the contestent switch or not, the probability of winning
    a car is 1/2, because the only relevant information is that one of
    the three doors has been ruled out to be the winner, and there are
    two doors left: one has the car, and the other doesn't.
    Irrelevant. The fact that it didn't show a car ruled out this rule.
    Randomly open another door. Since that door has no car, the
    probability
    of winning by switching (or not switching) is again 1/2, as in the
    first
    case.

    -- Bob.
     
    Reef Fish, Feb 22, 2005
    #8
  9. Jack

    Bruce Weaver Guest


    In a simulation, if the contestant switches on half of the trials, and
    stays on the other half, p(win) is indeed 0.5. This strategy is
    tantamount to randomly selecting one of the two doors that remain after
    one has been eliminated. But it is *not* the same as consistently
    staying, or consistently switching. If the contestant always stays with
    the original choice, p(win) = 1/3 in the long run; and if the contestant
    always switches, p(win) = 2/3. (It's no coincidence that the average of
    1/3 and 2/3 is 1/2.)

    Here's another way to think about it. Suppose that before opening any
    doors, Monty offers to let you exchange your original choice for BOTH of
    the other doors (and you get the prize if it is behind either of them).
    Would you opt for two doors, or stick with your original choice? How
    does this differ from the normal procedure? How does the normal
    procedure change your decision?
     
    Bruce Weaver, Feb 22, 2005
    #9
  10. I'll try again.
    Even though Bruce seems to say it fine,
    I thought my first attempt was fine, too.

    On 22 Feb 2005 08:23:33 -0800, "Reef Fish"

    [ ... ]
    Three doors.

    If the Guesser is right, which was 1/3, then
    Monty will show us one of the losers.

    If the Guesser was wrong, which was 2/3, then
    Monty will show us the loser of the two.

    Here is an identical problem, except for semantics:
    Monty lets you choose two doors. (What is your chance
    that one is a winner?)
    THEN he opens one of your doors which is a loser
    (picking randomly if they are both losers).

    What is the chance, now, that one of your two doors
    was a winner? How is any information added to
    that original "2/3"? "Should you change?"
     
    Richard Ulrich, Feb 22, 2005
    #10
  11. Jack

    Henry Guest

    I thought both were fine, except you asked a question rather than
    answering it.
    In the random Monty door case (as opposed to the "Classic" losing
    Monty door case), you do have more information (you know what is
    behind that door, prize or not) but it doesn't help you decide whether
    to change.
     
    Henry, Feb 23, 2005
    #11
  12. Jack

    Reef Fish Guest

    Actually what Bruce said is not exactly right either. No simulation
    is required to understand that the conditional probability of a
    change or no-change to win the car will BOTH be 1/2, on the given
    trial.
    Three doors, but one that Monty opens is ALWAYS a losing door, no
    matter
    whether the Guesser has chosen the right or the wrong one. Therefore,
    if the Guesser randomly switches or stays, his probability of winning
    the car is 1/2. But the random switching is an unnecessary red
    herring.
    The original chance of winning was "1/3". But rare events DO happen.
    So, the Guesser could have chosen the door with the car, even though
    the chance was only 1/3. Monty Hall merely changed the CONDITIONAL
    probability of winning to 1/2. On a single trial basis, it matters
    not whether he changes or not. The conditional probability of either
    door being the winner will be 1/2.

    -- Bob.
     
    Reef Fish, Feb 23, 2005
    #12
  13. Jack

    Duncan Smith Guest

    Try plugging the values into Bayes Rule. Given the rules of the game, the
    marginal probability of Monty opening a door without a car; and the
    probability of Monty opening a door without a car GIVEN that there is a car
    behind the chosen door, are both 1. So the posterior probability that the
    chosen door hides a car is equal to the prior probability, 1/3. The
    probability that the opened door hides the car is 0, which leaves 2/3 for
    the remaining door.

    In the random case the marginal probability of Monty opening a door without
    a car is 2/3, hence the different result. Monty's motivation is crucial.

    Duncan

    [snip]
     
    Duncan Smith, Feb 23, 2005
    #13
  14. Richard Ulrich wrote in
    Which was not what the OP posited. He specified a losing door being
    opened.
    Not true. After the opening we now know what is behind one of the doors.
    Following the conditional exclusion of all those events where a random
    opening disclosed a car (as specified by the OP), the shift in probabilities would be the same as
    if the opener had knowledge and intent. In your simulation, removing the
    branches with a winning door opening should leave you with the same
    choice set as you would have with "intentional" opening of a losing
    door. What matters is the change in knowledge after the opening, not the
    intent that went into the opening.
     
    David Winsemius, Feb 23, 2005
    #14
  15. Jack

    Duncan Smith Guest

    probabilities would be the same as
    A difference being that, in the non-random case, the excluded events (where
    Monty's opening reveals a car) have zero probability. In the random case,
    one of the excluded events has non-zero probability. i.e. The probability
    that the chosen door does not hide the car and Monty reveals the car is 1/3.

    Duncan
     
    Duncan Smith, Feb 23, 2005
    #15
  16. Jack

    Reef Fish Guest

    I have shown why in the post you snipped completely except for the
    preceding paragraph.
    Don't be such a pedant. You're just repeating the same fallacious
    argument given by Ulrich. The problem is so simple that you're
    stumbling all over some Bayes Rule you heard about once upon the
    time. The problem is nothing but a simple case of a CONDITIONAL
    probability GIVEN the fact that Monty always opens a door without
    a car.
    -- Bob.
     
    Reef Fish, Feb 23, 2005
    #16
  17. Duncan Smith wrote in
    But the premise of the OP's question was that a car was NOT revealed.
     
    David Winsemius, Feb 23, 2005
    #17
  18. Jack

    Duncan Smith Guest

    No you haven't.
    Yes it is simple, (1/1) * (1/3) = 1/3.

    Duncan
     
    Duncan Smith, Feb 23, 2005
    #18
  19. Jack

    Duncan Smith Guest

    In the non-random case the contestant knows a priori that a car will not be
    revealed, so there are only two possible events,

    contestant chooses door hiding car, Monty reveals no prize, p = 1/3
    contestant does not choose door hiding car, Monty reveals no prize, p = 2/3

    Subsequent conditioning doesn't exclude either of these events (and the
    posterior probability of having chosen the correct door is equal to the
    prior, 1/3).

    In the random case the contestant does *not* know a priori that a car will
    not be revealed, and there are three possible events,

    contestant chooses door hiding car, Monty reveals no prize, p = 1/3
    contestant does not choose door hiding car, Monty reveals a prize, p = 1/3
    contestant does not choose door hiding car, Monty reveals no prize, p = 1/3

    Subsequent conditioning excludes the second event and results in
    probabilities of 1/2 for the remaining events.

    The prior knowledge of the contestant is different in each case, so despite
    the evidence being the same, the posteriors turn out to be different.

    Duncan
     
    Duncan Smith, Feb 23, 2005
    #19
  20. snipped prior text but will insert here the OP's positing of the problem:
    Duncan Smith wrote in
    The problem posed by the OP stated that a non-car door was opened after
    the contestant choice had been made. That means your second event in your
    random case description was excluded by the manner in which the problem
    was posed. The prior probability that the contestant chose the car was
    still 1/3, so the probability that the car is behind the other door
    becomes 2/3 after non-car-door-opening by whatever mechanism, and the
    logical choice remains to switch.

    My sole point is that the mechanism by which the non-car-door-opening
    occurs is immaterial. If the problem is restricted to analyzing only
    those situations where a non-prize door is opened, the probabilities and
    knowledge of the contestant at the two distinct states of information
    remain equivalent. The analysis focus should be on the state of knowledge
    of the contestant and her logical choices given the evolution of his
    information. The contestant does not need to know that Monty always opens
    a losing door, only that on this occasion that a losing door was
    disclosed AFTER the first decision (which set the priors).
     
    David Winsemius, Feb 23, 2005
    #20
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