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Thread: Probability involving money

  1. #1
    Eso is offline
    Oct 2009

    Probability involving money

    Suppose you have $3600 to spend on the following items:
    A  $2200
    B  $1270
    C  $980
    D  $980
    E  $980
    F  $2120
    G  $2120
    H  $2000
    I  $2000
    I am trying to find the overall probability of purchasing any particular item. The order it was purchased doesn't matter. Items can not be purchased twice, they are purchased in random order, and items are purchased until no more items can be afforded.

    For example, suppose item F is randomly chosen. $3600 - 2120 = $1480, therefore items B, C, D, and E can still be purchased. Once either of these items are purchased, no more items may be purchased.

    Alternatively, suppose item C is randomly chosen. $3600 - 980 = $2620, therefore all items except for C can still be purchased. If B is randomly chosen next, $2620 - $1270 = $1350, therefore items D and E can still be purchased.

    I have created a computer program that determines every possible combination of items that may be purchased through an iterative process. The intent is that it will also be able to determine the probability of each item being picked at least once as it does so. I have attached the output of the program in excel format (item A is highlighted showing each combination it appears in).
    Attached Files Attached Files
    Last edited by Eso; Oct 13th 2009 at 04:43 PM.
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  2. #2
    May 2009

    try this

    It seems like you have done all the work.. At each stage there is a probability associated with each purchase. Since there are no conditions at the first stage, the probability of picking a particular item is 1/9. Then the next stage is determined by the number of remaining items (depending on how much you spent on the first round). So if F is chosen first, there are 4 things left (B,C,D, and E) so your second round has a 1/4 chance...
    So you've picked one of these four and now you can't continue. .. The probability of getting to where you are now is simply (1/9)x(1/4)=0.027 . . . so there is a 2.7% chance that the items you picked happened in that order.

    Your next example = (1/9)x(1/8)x(1/2) once EITHER D OR E has been chosen =0.0069 = 0.69% chance of those three purchases in that order.

    Hope this helps
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