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Math Help - A dice puzzle

  1. #1
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    A dice puzzle

    Hi,

    I'm not a university student (I was, but that was many years ago). Nonetheless, I'm trying to solve a probability puzzle, and this seems like the best place on the Internet to ask for help. The puzzle is as follows:

    --------------------------------

    N people roll two dice.

    After the first roll, half of the people are randomly eliminated, and the
    other half get to roll both dice a second time. This continues until all
    the people are eliminated. (Thus, N/2 people roll the dice exactly once;
    N/4 people roll them exactly twice, N/8 people roll them exactly three
    times, etc).

    When the process is complete, how many of the N people will have rolled
    double sixes at least twice? How many will have rolled double sixes at
    least three times?

    ---------------------------------

    Any insights you have would be greatly appreciated. I took probability courses years ago, but I've forgotten a lot of it, and I don't even think that problems like this are anything I ever learned to solve!

    Thanks very much.

    Abracadab

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  2. #2
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    Quote Originally Posted by abracadab View Post
    Hi,

    I'm not a university student (I was, but that was many years ago). Nonetheless, I'm trying to solve a probability puzzle, and this seems like the best place on the Internet to ask for help. The puzzle is as follows:

    --------------------------------

    N people roll two dice.

    After the first roll, half of the people are randomly eliminated, and the
    other half get to roll both dice a second time. This continues until all
    the people are eliminated. (Thus, N/2 people roll the dice exactly once;
    N/4 people roll them exactly twice, N/8 people roll them exactly three
    times, etc).

    When the process is complete, how many of the N people will have rolled
    double sixes at least twice? How many will have rolled double sixes at
    least three times?

    ---------------------------------

    Any insights you have would be greatly appreciated. I took probability courses years ago, but I've forgotten a lot of it, and I don't even think that problems like this are anything I ever learned to solve!

    Thanks very much.

    Abracadab
    A couple of suggestions:

    1. N needs to be a power of two for this to work: N = 2^m

    2. Start with a concrete value of N to get the feel of things. Perhaps N = 8 (or even N = 4).

    3. First consider at least once = 1 - Pr(no-one). Then at least twice = 1 - Pr(no-one or one).
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by abracadab View Post
    Hi,

    I'm not a university student (I was, but that was many years ago). Nonetheless, I'm trying to solve a probability puzzle, and this seems like the best place on the Internet to ask for help. The puzzle is as follows:

    --------------------------------

    N people roll two dice.

    After the first roll, half of the people are randomly eliminated, and the
    other half get to roll both dice a second time. This continues until all
    the people are eliminated. (Thus, N/2 people roll the dice exactly once;
    N/4 people roll them exactly twice, N/8 people roll them exactly three
    times, etc).

    When the process is complete, how many of the N people will have rolled
    double sixes at least twice? How many will have rolled double sixes at
    least three times?

    ---------------------------------

    Any insights you have would be greatly appreciated. I took probability courses years ago, but I've forgotten a lot of it, and I don't even think that problems like this are anything I ever learned to solve!

    Thanks very much.

    Abracadab
    As Mr Fantastic observes N must be a power of 2, then the number of times the die are rolled is:

    K=N+N/2+N/2^2+ ...+1=1+2+2^2+...+2^{\log_2(N)}

    The right most expression is a finite geometric series and so:

     <br />
K=\frac{1-2^{1+\log_2{N}}}{1-2}<br />

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    As Mr Fantastic observes N must be a power of 2, then the number of times the die are rolled is:

    K=N+N/2+N/2^2+ ...+1=1+2+2^2+...+2^{\log_2(N)}

    The right most expression is a finite geometric series and so:

     <br />
K=\frac{1-2^{1+\log_2{N}}}{1-2}<br />

    RonL
    I'm probably misunderstanding your reply CaptainB, but the question asks how many (expected number?) of the N people will have rolled double sixes at least twice etc., not how many (expected number?) of the total number of rolls are double sixes etc. (which is how I originally analysed it .....)
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