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Thread: Request about a problem with a die

  1. #1
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    Request about a problem with a die

    Dear members. I recently started a statistics course, but I'm having a hard time with some problems like the one I'm requesting help from you. Suppose a fair die with faces 1 to 6 is rolled 20 times. Now we define the random variable: the maximum value observed from the 20 rolls. What is the probability mass function and cumulated probability distribution?

    My first attempt was to define the range of this variable: {1, 2, 3, 4, 5, 6} but I'm not sure if the probabilities to be assigned to each value are all equal to 1/6.

    Any help with explanation about how to construct these distributions will be appreciated.

    Regards.
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  2. #2
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    Re: Request about a problem with a die

    Let's calculate the probability of each maximum roll:

    For a maximum of 1, it means that all 20 rolls show a 1.
    $$\left(\dfrac{1}{6}\right)^{20}$$

    For a maximum of 2, it means that all 20 rolls show a maximum of 2, and at least one of the rolls was a 2 (or not every roll was a 1):
    $$\left(\dfrac{2}{6}\right)^{20}-\left(\dfrac{1}{6}\right)^{20}$$

    For a maximum of 3, it means that all 20 rolls show a maximum of 2, and at least one of the rolls was a 3 (or not every roll was a 1 or 2):
    $$\left(\dfrac{3}{6}\right)^{20}-\left(\dfrac{2}{6}\right)^{20}$$

    Then for 4, 5, and 6, we have:

    $$4: \left(\dfrac{4}{6}\right)^{20} - \left(\dfrac{3}{6}\right)^{20}$$

    $$5: \left(\dfrac{5}{6}\right)^{20} - \left(\dfrac{4}{6}\right)^{20}$$

    $$6: \left(\dfrac{6}{6}\right)^{20} - \left(\dfrac{5}{6}\right)^{20}$$

    Can you finish from there?
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  3. #3
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    Re: Request about a problem with a die

    So if I understood correctly SlipEternal, in the third line: for a maximum of 3 it means that all 20 rolls show a maximum of 3, but because it is possible to get a 1 or a 2 also, you compute that probability with 3/6 raised to 20, right? Of course to compute the requested probability we need o substract the cases where the die face is 1 or 2, so that's the reason of substracting 2/6 raised to 20. I computed all the probabilities and I got the sum of 1, so everything seems perfect!. Thanks for your help!
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  4. #4
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    Re: Request about a problem with a die

    Quote Originally Posted by rbriceno View Post
    So if I understood correctly SlipEternal, in the third line: for a maximum of 3 it means that all 20 rolls show a maximum of 3, but because it is possible to get a 1 or a 2 also, you compute that probability with 3/6 raised to 20, right? Of course to compute the requested probability we need o substract the cases where the die face is 1 or 2, so that's the reason of substracting 2/6 raised to 20. I computed all the probabilities and I got the sum of 1, so everything seems perfect!. Thanks for your help!
    Not quite. Each individual roll can be a 1, 2, or 3. So, out of 6 faces of the die, we can have any of 3 faces show up. So, each die roll has a 3/6 chance of rolling so that the maximum is no greater than 3. There are 20 rolls, so we multiply the probability of a single roll 20 times. But, if we only force the dice to roll no higher than a 3, it is possible we can get 20 consecutive rolls where the dice roll only 1's and 2's (that is included in that probability of $\left(\dfrac{3}{6}\right)^{20}$. So, we subtract off the probability where every die roll is a 1 or a 2.
    Thanks from rbriceno
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