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Math Help - What did I do wrong?

  1. #1
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    Question What did I do wrong?

    A prize is randomly placed in one of 13 boxes, numbered from 1 to 13.You search for the prize by asking yes-no questions. Find the expected number of questions until you are sure about the location of the prize, under each of the following strategies.

    1. An enumeration strategy: you ask questions of the form 'is it in box k'.

    (1/13)*1+(12/13)*(1/12)*2+(12/13)*(11/12)*(1/11)*3+...+(12/13)*(11/12)*...*(1/2)*12=(1+2+...+12)/13=6 but for this answer, it says incorrect. what's wrong about my calculation?


    2. A bisection strategy: you eliminate as close to half of the remaining boxes as possible by asking questions of the form 'is it in a box numbered less than or equal to k?'.3*(3/13)+4*(10/13)=3.7692 which is correct.
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  2. #2
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    Quote Originally Posted by essedra View Post
    A prize is randomly placed in one of 13 boxes, numbered from 1 to 13.You search for the prize by asking yes-no questions. Find the expected number of questions until you are sure about the location of the prize, under each of the following strategies.

    1. An enumeration strategy: you ask questions of the form 'is it in box k'.

    (1/13)*1+(12/13)*(1/12)*2+(12/13)*(11/12)*(1/11)*3+...+(12/13)*(11/12)*...*(1/2)*12=(1+2+...+12)/13=6 but for this answer, it says incorrect. what's wrong about my calculation?
    Only your computation about the case when 12 questions are needed is wrong: this happens either when the prize is in the 12th box or in the 13th box, hence the probability of this case is \frac{2}{13}.

    (In this same way you could have said that k(<12) questions are needed iff the prize is in the k-th box, which happens with probability \frac{1}{13})
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  3. #3
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    Thank you for your kind response...
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