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The Parcel Problem

  1. Nov 2, 2012 #1
    I've taken the day off to wait for a parcel at home and (because I'm a physics student and therefore have no ability to actually enjoy my day off) I got thinking about this as a problem.

    Suppose the delivery company is perfect, i.e. if they say they're going to deliver between 09:00 and 18:00 then they certainly will. If I need to go out to do something which takes 1 hour to do in the delivery time slot, when should I go?

    My first guess would be at 09:00. At 09:00 there is a 1/540 chance it will come during any given minute. If it hasn't come at 09:59 then it must come between 10:00 and 18:00 and so the probability increases to 1/480 per minute, until the last hour is 1/60 per minute, meaning that if it hasn't come by 16:59 I have to leave at 17:00 and will surely miss my parcel. (1/60 per minute * 60 minutes).

    But. I think there's a problem with that. Suppose we go back to the beginning. At 09:00 there is a 1/540 probability the parcel will come in any given minute. So at 4:59 there have been 480 minutes and there will be 60 minutes left. So the probability it will come before then is 8 times larger than the probability it will come in the last 60 minutes. Thinking this way would require me to plan to go out in the last hour.

    So my two answers are, the first hour or the last hour.

    Does anyone have any thoughts on this? Or would like to point out where my faulty logic is?

    Thank you.
     
  2. jcsd
  3. Nov 2, 2012 #2
    I would think that there is a 88% chance of them delivering during the time your not there will always be 88% if you pre-select the time however if you were to leave the minute you got the package you would have a much higher probability of leaving towards the end.
     
  4. Nov 2, 2012 #3
    You are confusing the decision made before 9:00 with the decision made some time after 9:00 given that it did not arrive yet. If you assume a uniform distribution then there is equal probability of it arriving during any two time intervals of the same length. If, however, you wait until 10:00 and it hasn't arrived then you are asking for the probability of the package arriving during a given interval given that it did not arrive in the interval between 9:00 and 10:00. So if you are deciding at 8:00 what time you should run your errand then there is a 1/9 probability that you miss your package. If you wait until 10:00 and it hasn't arrived then you have a new distribution which depends on the fact that it did not arrive yet. Now there is a 1/8 probability that you will miss it. However, the overall probability that you do not get your package is still 1/9.

    Let m=event that you miss package, 1=event that it comes in first hour

    P(m)=P(1)P(m/1)+P(not 1)P(m/not 1)

    =(1/9)(0)+(8/9)(1/8)

    =1/9
     
  5. Nov 2, 2012 #4

    Bacle2

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  6. Nov 2, 2012 #5
    Thanks Alan.

    I thought about the unexpected hanging but couldn't quite work out how it applied. Then I looked at Bayes' elegant formulae and got more confused.

    Anyway, my parcel arrived at 13:45.
     
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