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Average number of attempts

  1. Aug 8, 2004 #1


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    Hi everyboby :smile:

    I need something very simple for a personnal project, but I'm not quite sure I got it right. Here it is:

    Suppose there are 5 boxes side-to-side. One of them contains an object (the probablity is equal for all boxes, 1-in-5). On average, how many attempts does it take to find the object?

    Here is what I figured:

    if the object is in box 1, you need 1 attempt;
    if the object is in box 2, you need 2 attemps;

    After finding the object, you 'randomize' the system and try to find the object again. After doing this 5 times, the object will have been in each of the boxes 1 time (let's say that the system is REALLY random, or that we did a great number of tests). We will then have made 1, 2, 3, 4 and 5 attempts (not necessarily in that order) out of 5 tests. So, on average, we have made
    \frac{{\left( {1 + 2 + 3 + 4 + 5} \right)}}{5} = 3

    Generalizing to N boxes, we have
    \left\langle n \right\rangle = \frac{1}{N}\sum\limits_{i = 1}^N i = \frac{{N + 1}}{2}

    So... is this right or am I wrong somewhere? It seems suspiciously simple. Anyway, thanks alot for your help :smile: (and sorry about my english...)
  2. jcsd
  3. Aug 8, 2004 #2


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    Looks okay to me. I may have reasoned slightly differently though :

    Assume you are going to pick in the order 1, 2, 3, ....
    The probablity that the object is in any particular box is 1/N.

    [tex] \left\langle n \right\rangle = \sum\limits_{i=1}^N{p_i~n_i} = \frac{1}{N} \frac{{N(N + 1)}}{2} [/tex]
    Last edited: Aug 8, 2004
  4. Aug 11, 2004 #3


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    Thanks for your quick reply :smile:

    I appreciate the insight you provided with your alternative approach. It seems so evident now...

    Thanks again
  5. Aug 18, 2004 #4


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    Well, it's quiet in here :zzz:

    Not that anybody cares, but I pursued my little quest and found an equation for a more general problem. I don't know why I'm posting this... just sharing, I guess.

    Suppose you have k objects in N boxes. On average, how many picks (n) will you need to find an object? Here's what I found:

    \left\langle n \right\rangle = \frac{{\sum\limits_{i = 1}^N {i \cdot \left( \begin{array}{l}
    N - i \\
    k - 1 \\
    \end{array} \right)} }}{{\left( \begin{array}{l}
    N \\
    k \\
    \end{array} \right)}} = \frac{{N + 1}}{{k + 1}}

    Thanks to my good friend Maple for the simplification of the messy factorials.
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