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Math Help - fair die probability

  1. #1
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    fair die probability

    A fair die is rolled over and over until the sum of all the numbers obtained is greater than 4. What is the probability that this takes 1 roll or 2 rolls or 3 rolls or 4 rolls or 5 rolls, etc.?
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  2. #2
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    To complete task in one roll you have to get "5" or "6", so there are 2 possible outcomes that are correct, but there are total of 6 possible outcomes. So probability have to be some combination of these numbers, you can roll "5", so one number from six and also roll "6", so again one number from six. In total you can roll 2/6 numbers to be successful.

    In two rolls you can get 6*6 possible outcomes, how many times you can be succesful? just write it down and you will see the answer. Also for 3 and 4 rolls. Maybe start with 5 rolls and you will see it imediately.
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  3. #3
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    Quote Originally Posted by mhitch03 View Post
    A fair die is rolled over and over until the sum of all the numbers obtained is greater than 4. What is the probability that this takes 1 roll or 2 rolls or 3 rolls or 4 rolls or 5 rolls, etc.?
    For 1 roll you will need a 5 or 6, P_{1}({5}\cup{6})=2\times \frac{1}{6}=\frac{2}{6}

    For 2 rolls, your first roll must fail. To fail you must get either of these in the set { 1\cup2\cup3\cup4}, which gives P_{2,1}=\frac{4}{6}, given that it happened before proceeding to the subsequent roll. For the subsequent roll immediately to this, you will get the probability given that the first roll was X=1, 2, 3, or 4. This seems to be and open ended question for we do not know not what comes first; we only know the probability only one has already occured. The following is the list of all four possible first numbers denoted as X_{2,1}

    \begin{array}{ccc}X_{2,1}&S&p_{2,2}(X)\\<br />
1&{4,5,6}&{3/6}\\<br />
2&{3,4,5,6}&{4/6}\\<br />
3&{2,3,4,5,6}&{5/6}\\<br />
4&{1,2,3,4,5,6,}&1\end{array}

    Same for 3 rolls, 4 rolls, 5 rolls,..., n\in N, repeat the process.

    In your case, at worse, you will need 5 rolls, at which you get {1,1,1,1,1}

    Note: I have re-evaluated this post many times and concluded that it is labor intensive to answer this question. It appeared rather innocent when I first attempted, but as I comtemplated, I found it more of a busy work than brain work.
    Last edited by novice; October 23rd 2009 at 12:51 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    This is off a bit...........
    In your case, at worse, you will need 5 rolls, at which you get {1,1,1,1,1}

    In the case of five rolls you must have ...{1,1,1,1,anything}

    that would be (1/6)^4
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  5. #5
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    Quote Originally Posted by matheagle View Post
    This is off a bit...........
    In your case, at worse, you will need 5 rolls, at which you get {1,1,1,1,1}

    In the case of five rolls you must have ...{1,1,1,1,anything}

    that would be (1/6)^4
    Matheagle, may I strike a bargain with you? Here is my take:

    The reason for taking five rolls to succeed is that the first 4 events failed. Let q denotes the probability of failure, and p the probability of success.

    To fail the first, we need 1 or 2 or 3 or 4, from which q_1=\frac{4}{6}.

    To fail the second event, the probability will depend on what appeared on q_1. If indeed "1" appeared, 4 is the smallest number we must get subsequently. To fail, we need to get 1, 2, or 3, from which q_2=\frac{3}{6}.

    To fail the third event, if indeed "1" had just appear the second round, this time we need 1, or 2 to give us failure, from which q_3=\frac{2}{6}

    Segue, to fail the forth, q_4=\frac{1}{6}, and of course we will succeed on the fifth attemp, at which p_5=\frac{6}{6}.

    Consequently, were we to take five attempts to succeed, proveded that we have got a string of 1's in the last 4 attempts, p=q_1q_2q_3q_4p_5, where p=\left(\frac{4}{6}\right)\left(\frac{3}{6}\right)  \left(\frac{2}{6}\right)\left(\frac{1}{6}\right)\l  eft(\frac{6}{6}\right)=\frac{1}{324}
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