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Probability - 4 envelopes contain 4 diff amounts. open one by one

  1. Sep 6, 2008 #1
    1. The problem statement, all variables and given/known data
    4 envelopes contain 4 different amounts of money. you are allowed to open them one by one, each time deciding whether to keep the amount or discard it and open another envelope. once an amount is discarded, you are not allowed to go back and get it later. compute the probability that you get the largest amount under the following strategy:

    you open the first envelope, note that it contains the amount x, discard it and take the next amount which is larger than x (if no such amount shows up, you must take the last envelope)

    2. Relevant equations

    3. The attempt at a solution
    yeah... i'm completely lost on this problem. the only progress i've made is that the total number of outcomes must be 24, since it's a permutation problem without replacement so it should be (4)3 = 4 x 3 x 2 = 24

    i'm not sure how you would find the number of favorable outcomes. any hints?

    the book says the answer should be 11/24
  2. jcsd
  3. Sep 6, 2008 #2


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    I'm assuming 'favorable outcome' means you get the envelope with the largest amount. One way is to just write down all 24 arrangements of envelopes and then try the strategy. You can take some shortcuts, like if the largest amount is in the first envelope you always lose, if it's in the second you always win. Think about what happens if the largest amount is in the third and fourth envelopes. Hmm. I seem to keep getting 10/24.
  4. Sep 6, 2008 #3


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    I get 11/24 when I do the problem. Also, I find it easier to break the problem down into the following 4 cases. 1) Largest value in the first envelope (always lose); 2) second largest value in the first envelope (always win), 3) thrid largest value in the first envelope, 4) smallest value in the first envelope.
  5. Sep 6, 2008 #4

    D H

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    b0it0i - Follow Dick's logic, but not his accounting. The book is right.
  6. Sep 6, 2008 #5


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    Ok, so now I'm the dumb one. I didn't mean to say I thought 10/24 was right, I was just confused why I wasn't seeing it. Hence the "Hmm". Now I do. I was missing the 3124 case. Thanks for straightening me out.
    Last edited: Sep 6, 2008
  7. Sep 7, 2008 #6
    i see, my professor also gave the hint to list all possible outcomes, which i forgot to include.

    so i did so, and denoted each outcome as such 1234 (1 = 1st envelope, etc.)

    if 1 is the largest amount, you can take off all the combinations with 1 in the front, since you discard the first envelope (that's 6 gone)

    but you can save the combinations with 1 in the second position, because since 1 is the largest amount, it will always be greater than whatever was in the first envelope, (so that's 6 favorable: 2134, 2143, 3124, 3142, 4123, 4132)

    for the sake of argument, if 1 was the largest amount, 2 was the second largest, 3 was the third, and 4 was the fourth largest,

    you can exclude combinations: 3214, since you would take envelope 2, which is not the largest amount = fail
    3421, same reasoning. and so on

    the only combinations, left, that provide a favorable outcome is
    2314, 2431, 2341, 2413, 3412
    so basically you are looking for combinations where the 2nd and 3rd positions (that show up before the 1) are less than the 1st position

    and that's why there's 11 favorable outcomes

    thanks for the tip
    Last edited: Sep 7, 2008
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