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A probability question!

  1. Dec 4, 2008 #1
    A miner is trapped in mine containing 3 doors The first door will take him out of the tunnel in 2 hours the second door will take hin out of the tunnel in 5 hours and the third door will take him out of the tunnel in 7 hours. Assuming that the miner at all times is equally likely to choose any of the doors, what is the expected length of time until he reaches safety.

    Let X denote the time he reaches safety, Y denotes the initial choice

    E(X) = 1/3 [E(X|Y=1)+E(X|Y=2)+E(X|Y=3)]

    E(X|Y=1) = 2
    E(X|Y=2) = 5 + E(X) <- I DONNO WHY IT ADDS E(X)
    E(X|Y=3) = 7 + E(X) <- AGAIN...DONNO WHY...

    E(X) = 1/3[14+2E(X)]

    THUS, E(X) = 14.
  2. jcsd
  3. Dec 4, 2008 #2


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    Who is "IT" that "ADDS"? I don't know why either. As the problem is stated you shouldn't.
    The answer should be the weighted average of the times the weights being the probabilities which are all equal in this case giving a simple average of 14/3.
  4. Dec 4, 2008 #3
    oh im in trouble now...the answer is given by tutor ="=
    Last edited: Dec 4, 2008
  5. Dec 5, 2008 #4
    can anyone give me some advice?
  6. Dec 5, 2008 #5


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    The expected time is 14 hours? At worst it will take 7 hours! You are told that the times will be either 2, 5, or 7 hours! The "expected time" is, as jambaugh told you, an average of the possible times. An "average" is always between the largest and smallest values.

    The expected value is (1/3)(2)+ (1/3)(5)+ (1/3)(7). Since "the miner at all times is equally likely to choose any of the doors" this is just the average of the three times given.

    Your tutor may be under the impression that he must open three consecutive doors, adding the times for each, but that is obviously not true.
  7. Dec 5, 2008 #6
    no, the 5-hour and 7-hours door will lead him back to where he was
  8. Dec 5, 2008 #7


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    Ah, you didn't explain that before!

    So, if he picks the 2-hour door, he's out; so

    E(X|Y=1) = 2.

    But, if he picks the 5-hour door, he spends 5 hours walking, and then is right back where he started. At this point, he needs to pick another door, and since we are told that he is still equally likely to pick any door (even though he knows the door he picked last ime only led him back to where he was!), the expected additional amount of time he will spend walking is E(X). We must add to that the 5 hours he already spent. So,

    E(X|Y=2) = 5 + E(X).

    Same story for the 7-hour door, so

    E(X|Y=3) = 7 + E(X).
  9. Dec 5, 2008 #8


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    I suspected as much given the form of the solution. The best thing to do with one of these is to draw a decision tree.

    His first choice is one of the three, so begin with 3 forks one for each case. Two of those are dead-ends so he'll return and have to make one of 2 choices in each case. One of those leads him out the other leads him to another dead end but then he knows which way is out.

    Your tree should look like:

    Put probabilities of each choice at the legs of each fork.
    Multiply probabilities along the paths of the tree and keep track of the time needed for each path.
    Last edited: Dec 5, 2008
  10. Dec 5, 2008 #9


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    No, he's too stupid to mark the door he tried, and so he is (we are told) equally like to try each door every time. Thus your decision tree is infinite, and we'd have to sum infinite series to get the answer. (But this could be done.)
  11. Dec 5, 2008 #10


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    Uggg. OK but let's not be too insulting. Let's say he has no reliable way to mark the door in the dark . . . well no he could come up with something on the umpteenth try . . . the problem should be a rat in a maze... no they're smarter than that. They can at least remember. OK he's stupid.
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