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Math Help - Die experiment

  1. #1
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    Die experiment

    Can someone please help me out for this question.

    Consider an experiment where we roll a single die and observe the # of dots. Thus,m our sample space is S = 1,2,3,4,5,6 . Suppose that the die is not fair and that P (i) = k/i for i = 1,2,3,4,5,6. What is the value of k?

    I don't understand what it means by the die not being fair. What would be the steps to solve such a problem? Would it be something along hte lines of:

    P (i) = k/6

    since it says we observe the # of dots do we just add the dots together?

    P (i) = 21/6?


    I'm just making a guess here but I'm not sure how to even start with this.

    Thanks.
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  2. #2
    Grand Panjandrum
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    Re: Die experiment

    Quote Originally Posted by agent2421 View Post
    Can someone please help me out for this question.

    Consider an experiment where we roll a single die and observe the # of dots. Thus,m our sample space is S = 1,2,3,4,5,6 . Suppose that the die is not fair and that P (i) = k/i for i = 1,2,3,4,5,6. What is the value of k?

    I don't understand what it means by the die not being fair. What would be the steps to solve such a problem? Would it be something along hte lines of:

    P (i) = k/6

    since it says we observe the # of dots do we just add the dots together?

    P (i) = 21/6?


    I'm just making a guess here but I'm not sure how to even start with this.

    Thanks.
    For a fair die the probability of the outcomes are equaly likely. A die is not fair if the probability of each outcome is not the same. Since one of 1, .., 6 must be the result of rolling the die for a fair die p(i)=1/6, i=1..6.

    This die has P(1)=k/1, P(2)=k/2, p(3)=k/3), p(4)=k/4, p(5)=k/5, p(6)=k/6

    but when you roll the die the result must be one 1, 2, 3, 4, 5, 6 so:

    p(1)+p(2)+..+p(6)=1

    and you may use this to determine what value k must have.

    CB
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  3. #3
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    Re: Die experiment

    I'm not sure if this is correct or not but I'm assuming you mean for a fair die each outcome will have a 1/6 probability. Since this is not fair it will be distributed through all the numbers meaning:

    P (1)= k/1 + P (2) = k/2 + P (3) = k/3 + P(4) = k/4 + P(5) = k/5 + P (6) = k/6

    Therefore

    P (1) = 1/1 + P (2) = 2/2 + P (3) = 3/3 + P (4) = 4/4 + P (5) = 5/5 + P (6) = 6/6

    would that be right or no? The problem is that k is different in each one so I think i'm still misunderstanding what you are trying to say.


    Thanks.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Die experiment

    You've been told that P(i) = k/i. Therefore:

    P(1) = k
    P(2) = k/2
    P(3) = k/3
    P(4) = k/4
    P(5) = k/5
    P(6) = k/6

    Since these 6 possibilities are the only ones that exist, you know that
    P(1) + P(2)+ P(3) + P(4) + P(5) + P(6) = 1.

    Now you can solve for k.
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  5. #5
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    Re: Die experiment

    Actually I didn't read that it all has to equal to 1. So just by plugging in numbers i came to the conclusion that k has to be somewhere around 0.4.

    So i did:

    0.4/1 + 0.4/2 + 0.4/3 + 0.4/4 + 0.4/5 + 0.4/6 = 0.98 so it's fairly close to 1... Is there an actual way of doing the math though to make it 1? 0.4082 reaches a bit over 1 but is there a way or formula to get a concrete answer... I'm pretty sure this is on the right track but once again it's just by plugging in numbers to try and make it equal to 1.
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  6. #6
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    Re: Die experiment

    Quote Originally Posted by agent2421 View Post
    Actually I didn't read that it all has to equal to 1. So just by plugging in numbers i came to the conclusion that k has to be somewhere around 0.4.
    Really \sum\limits_{n = 1}^6 {\frac{1}{n}}  = \frac{{49}}{{20}} thus k=\frac{20}{49}.
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  7. #7
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    Re: Die experiment

    I'm sorry but where do you get the 49 from? I feel really stupid for not understanding this right away lol.
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  8. #8
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    Re: Die experiment

    Quote Originally Posted by agent2421 View Post
    I'm sorry but where do you get the 49 from? I feel really stupid for not understanding this right away lol.
    Can you add
    \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}  {4}+\frac{1}{5}+\frac{1}{6}\right)~?
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