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Math Help - Rolling two dice over and over

  1. #1
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    Rolling two dice over and over

    We have one 4 sided die and a 6 sided one. We are tossing them together and observing their sum.
    X: sum of the dice
    From a previous part
    P(4)= 3/24
    P(7)= 4/24

    Now we are throwing the dice over and over what is the probability that a sum of 4 appears before a sum of 7 ?
    How do I tackle this kind of a problem ?
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  2. #2
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    There are 3 outcomes at every trial, 4,7 and something else. I will denote "something else" by 0. It has a probability of 17/24.

    The only way that a 4 can appear before a 7 is if you roll it on the first go; or you roll 0 a few times and then get a 4. So you are interested in the followign events

    P(4)
    +P(0,4)
    +P(0,0,4)
    +P(0,0,0,4)
    .....



    which have probabilities
    P(4) = \frac{3}{24}

    P(0,4) = \frac{17}{24} \times \frac{3}{24}

    P(0,0,4) = \left(\frac{17}{24})^2 \times \frac{3}{24}

    P(0,0,0,4) = \left(\frac{17}{24})^3 \times \frac{3}{24}
    ...

    You can recognise the above as a geometric progression, with first term \frac{3}{24} ; and common ratio \frac{17}{24}. use the standard formula for the sum of an infinite geometric progression

    P(4~before~7) =\frac{\frac {3}{24}}{1-\frac{17}{24}} = \frac{3}{7}

    The fact that the answer is 3/7 may not be surprising. Can you see why?

    Spoiler:


    3/7 is the relative likelihood of 4 compared to 7.

    <br />
P(4|4~or~7) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{4}{24}} = \frac{3}{7}<br />


    so, at whatever point we stop rolling 0 we have a 3/7 chance of getting a 4.
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  3. #3
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    Hello, Hitman6267!

    Same solution as Springfan25 . . . slightly different wording.

    [After all this typing, I flatly refuse to delete this.]

    We have one 4-sided die and a 6-sided die.
    We toss them together and observe their sum.


    From a previous part

    . . \begin{array}{ccc}P(\text{sum of 4}) &=& \frac{3}{24} \\ \\[-3mm] P(\text{sum of 7}) &=& \frac{4}{24} \end{array}


    We throw the dice repeatedly.
    What is the probability that a sum of 4 appears before a sum of 7 ?

    We have these probabilities: . \begin{array}{ccc}P(\text{sum of 4}) &=& \frac{1}{8} \\ \\[-4mm]<br />
P(\text{sum of 7}) &=& \frac{1}{6} \\ \\[-4mm]<br />
P(\text{other}) &=& \frac{17}{24} \end{array}


    A sum of 4 could appear on the first roll:
    . . P(\text{4 on 1st roll}) \:=\:\frac{1}{8}

    A sum of 4 could appear on the second roll:
    . . P(\text{4 on 2nd roll}) \:=\:P(\text{other, 4}) \:=\:\left(\frac{17}{24}\right)\left(\frac{1}{8}\r  ight)

    A sum of 4 could appear on the third roll:
    . . {(\text{4 on 3rd roll}) \:=\:P(\text{other, other, 4}) \:=\:\left(\frac{17}{24}\right)^2\left(\frac{1}{8}  \right)

    A sum of 4 could appear on the fourth roll:
    . . P(\text{4 on 4th roll}) \:=\:P(\text{other, other, other, 4}) \:=\:\left(\frac{17}{24}\right)^3\left(\frac{1}{8}  \right)

    . . . and so on.


    Hence: . P(\text{4 appears before 7}) \;=\;\frac{1}{8} + \frac{1}{8}\left(\frac{17}{24}\right) + \frac{1}{8}\left(\frac{17}{24}\right)^2 + \frac{1}{8}\left(\frac{17}{24}\right)^3 + \hdots

    . . . . . . . . =\;\frac{1}{8}\underbrace{\bigg[1 + \tfrac{17}{24} + \left(\tfrac{17}{24}\right)^2 + \left(\tfrac{17}{24}\right)^3 + \hdots \bigg]}_{\text{geometric series}}

    The sum of the geometric series is: . \frac{1}{1-\frac{17}{24}} \:=\:\dfrac{1}{\frac{7}{24}} \:=\:\frac{24}{7}


    Therefore: . P(\text{4 before 7}) \:=\:\dfrac{1}{8}\cdot\dfrac{24}{7} \;=\;\dfrac{3}{7}

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  4. #4
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    Thank you
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