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Thread: Probability question

  1. #1
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    Probability question

    Dear all, I have a probability question:
    Tim throws two identical fair dice simultaneously. Each die has six faces: two faces numbered 1, two faces numbered 2, and two faces numbered 3. Tim plays a game with his friend Bill, who also has two dice numbered in the same way.
    Let X denote the largest number shown on the four dice.
    Find the probability that a) X=2 and b) X=3
    Thank you!
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  2. #2
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    Re: Probability question

    Quote Originally Posted by bbear123 View Post
    Tim throws two identical fair dice simultaneously. Each die has six faces: two faces numbered 1, two faces numbered 2, and two faces numbered 3. Tim plays a game with his friend Bill, who also has two dice numbered in the same way. Let X denote the largest number shown on the four dice. Find the probability that a) X=2 and b) X=3
    This is not an easy counting question.
    Lets work on $X=3$. That can happen by outcomes: $\{(3,3,3,3);~(3,3,3,a);~(3,3,a,a);~(3,a,a,a)\}$ where $a$ is either $1\text{ or }2$.

    Now the outcome $(3,3,a,a)$ can happen in six difference ways each of those having probability ${\left( {\dfrac{1}{3}} \right)^2}{\left( {\dfrac{2}{3}} \right)^2}$

    Doing that for each of those four outcome types and adding we get $\mathscr{P}\left( {X = 3} \right) = \dfrac{{65}}{{81}}$
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  3. #3
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    Re: Probability question

    The case of $X=2$ is similar to the previous post.

    The possible rolls are $(2,1,1,1),~(2,2,1,1),~(2,2,2,1),~(2,2,2,2)$

    There are 15 possible rolls, you can verify this, and thus

    $P[X=2]=\dfrac{4}{15}$

    I can't say I agree with Plato's solution for $X=3$

    Of the 15 possible rolls I get that $P[X=1]=\dfrac {1}{15},~P[X=2]=\dfrac{4}{15},~P[X=3]=\dfrac{2}{3}$

    $\left(
    \begin{array}{cccc}
    1 & 1 & 1 & 1 \\
    1 & 1 & 1 & 2 \\
    1 & 1 & 1 & 3 \\
    1 & 1 & 2 & 2 \\
    1 & 1 & 2 & 3 \\
    1 & 1 & 3 & 3 \\
    1 & 2 & 2 & 2 \\
    1 & 2 & 2 & 3 \\
    1 & 2 & 3 & 3 \\
    1 & 3 & 3 & 3 \\
    2 & 2 & 2 & 2 \\
    2 & 2 & 2 & 3 \\
    2 & 2 & 3 & 3 \\
    2 & 3 & 3 & 3 \\
    3 & 3 & 3 & 3 \\
    \end{array}
    \right),\left(
    \begin{array}{cccc}
    1 & 1 & 1 & 1 \\
    \end{array}
    \right),\left(
    \begin{array}{cccc}
    1 & 1 & 1 & 2 \\
    1 & 1 & 2 & 2 \\
    1 & 2 & 2 & 2 \\
    2 & 2 & 2 & 2 \\
    \end{array}
    \right),\left(
    \begin{array}{cccc}
    1 & 1 & 1 & 3 \\
    1 & 1 & 2 & 3 \\
    1 & 1 & 3 & 3 \\
    1 & 2 & 2 & 3 \\
    1 & 2 & 3 & 3 \\
    1 & 3 & 3 & 3 \\
    2 & 2 & 2 & 3 \\
    2 & 2 & 3 & 3 \\
    2 & 3 & 3 & 3 \\
    3 & 3 & 3 & 3 \\
    \end{array}
    \right)$
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  4. #4
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    Re: Probability question

    Quote Originally Posted by romsek View Post
    The case of $X=2$ is similar to the previous post.
    I can't say I agree with Plato's solution for $X=3$
    You did not take in account the possible permutations of the dice.
    For example: $X=2$ the case $(1,1,1,2)$ can occur in four ways.
    All of $(1,1,1,2),(1,1,2,1),(1,2,1,1),(2,1,1,1)$ must be counted for $X=2$

    Consider rolling just two dice in the usual game where we add both together.
    $\begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{( 1,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2 ,4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3, 4)}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4 )}&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4) }&{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)} &{(6,5)}&{(6,6)}\end{array}$
    To get the sum $7$ we count the pairs $(1,6),(1,6),(2,5),(5,2),(3,4),(4,3)$ to get $\mathscr{P}(S=7)=\dfrac{6}{36}$.
    Last edited by Plato; Feb 26th 2018 at 06:04 AM.
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  5. #5
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    Re: Probability question

    I would do complements. We want at least one die to show a 3. So, first calculate the none of the dice show a 3:

    $\left( \dfrac{2}{3} \right)^4$

    Take 1 minus that, and that is the probability that at least one die shows a 3 (and the maximum number shown is a 3):

    $P[X=3] = 1 - \left( \dfrac{2}{3} \right)^4 = \dfrac{65}{81}$ as Plato expected.

    Next, the probability that X=1 is given by:

    $P[X=1] = \left( \dfrac{1}{3} \right)^4 = \dfrac{1}{81}$.

    This leaves the probability that X=2 as the probability that $X \neq 1$ and $X \neq 3$:

    $P[X=2] = 1 - \dfrac{1}{81} - \dfrac{65}{81} = \dfrac{15}{81} = \dfrac{5}{27}$

    All of this agrees with what Plato is saying. romsek, you are looking only at outcomes. If each outcome were equally likely, then you are correct with your assessment. But, as Plato eluded to, the outcomes are not equally likely.
    Last edited by SlipEternal; Feb 26th 2018 at 06:47 AM.
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  6. #6
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    Re: Probability question

    Quote Originally Posted by Plato View Post
    You did not take in account the possible permutations of the dice.
    For example: $X=2$ the case $(1,1,1,2)$ can occur in four ways.
    All of $(1,1,1,2),(1,1,2,1),(1,2,1,1),(2,1,1,1)$ must be counted for $X=2$

    Consider rolling just two dice in the usual game where we add both together.
    $\begin{array}{*{20}{c}}{(1,1)}&{(1,2)}&{(1,3)}&{( 1,4)}&{(1,5)}&{(1,6)}\\{(2,1)}&{(2,2)}&{(2,3)}&{(2 ,4)}&{(2,5)}&{(2,6)}\\{(3,1)}&{(3,2)}&{(3,3)}&{(3, 4)}&{(3,5)}&{(3,6)}\\{(4,1)}&{(4,2)}&{(4,3)}&{(4,4 )}&{(4,5)}&{(4,6)}\\{(5,1)}&{(5,2)}&{(5,3)}&{(5,4) }&{(5,5)}&{(5,6)}\\{(6,1)}&{(6,2)}&{(6,3)}&{(6,4)} &{(6,5)}&{(6,6)}\end{array}$
    To get the sum $7$ we count the pairs $(1,6),(1,6),(2,5),(5,2),(3,4),(4,3)$ to get $\mathscr{P}(S=7)=\dfrac{6}{36}$.
    since when do we care about order with dice rolls?
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  7. #7
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    Re: Probability question

    Quote Originally Posted by romsek View Post
    since when do we care about order with dice rolls?
    We don't care about order with dice rolls. But, there are two distinct dice being rolled. We only care about outcomes. So, in Plato's example, he is looking at the sum of the dice. We don't care the order the dice were rolled. But, the different outcomes have different probabilities that can be found by considering the order of the dice. So, here are the outcomes we care about (we don't mention any order of the dice, as their order was removed by calculating the probabilities of each outcome):

    Code:
    Outcome   Probability
       2         1/36
       3         2/36
       4         3/36
       5         4/36
       6         5/36
       7         6/36
       8         5/36
       9         4/36
       10        3/36
       11        2/36
       12        1/36
    Next, let's consider probabilities for each of the outcomes you found:

    Code:
     Outcome   Probability
    (1,1,1,1)     1/81
    (1,1,1,2)     4/81
    (1,1,1,3)     4/81
    (1,1,2,2)     6/81
    (1,1,2,3)     12/81
    (1,1,3,3)     6/81
    (1,2,2,2)     4/81
    (1,2,2,3)     12/81
    (1,2,3,3)     12/81
    (1,3,3,3)     4/81
    (2,2,2,2)     1/81
    (2,2,2,3)     4/81
    (2,2,3,3)     6/81
    (2,3,3,3)     4/81
    (3,3,3,3)     1/81
    Different outcomes have different probabilities of occurring.
    Last edited by SlipEternal; Feb 26th 2018 at 09:03 AM.
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