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Math Help - Probability of throwing two dice

  1. #1
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    Probability of throwing two dice

    • Consider throwing two dice


    What is the probability an 8 will be rolled before a 7 is rolled?
    What is the probability the player will either: roll a 7 or 11; or (if not) roll a 4, 5, 6, 8, 9 or 10 AND THEN repeat that same number before a 7 is rolled?
    Any help will be appreciated.
    Thank you,

    Consider
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  2. #2
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    hi August80,

    if you mean "what is the probability of rolling a combined 8 followed immediately by rolling a combined 7",

    then the probability of rolling an 8 is \frac{5}{36}

    while the probabilty of rolling a 7 is \frac{6}{36}

    as there are 5 ways to roll an 8 ....... 62 53 44 35 26
    and 6 ways to roll a 7 .................... 61 52 43 34 25 16

    multiply the probabities to find the probability of an 8 before a 7 (or a 7 before an 8)

    since the sequences are 6261 6252 6243 6234 6225 6216 and similarly for the other ways starting 5, 4, 3 and 2.

    That's 6(5)=30 ways out of 6^4

    which means the probability is \frac{30}{6^4}=\frac{5}{36}\ \frac{6}{36}=\frac{5}{6^3}
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  3. #3
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    Thank you very much Archie!

    Best,
    August80
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  4. #4
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    Quote Originally Posted by August80 View Post
    • Consider throwing two dice



    What is the probability the player will either: roll a 7 or 11; or (if not) roll a 4, 5, 6, 8, 9 or 10 AND THEN repeat that same number before a 7 is rolled?
    Any help will be appreciated.
    Thank you,

    Consider
    In rolling a 7 or 11, we sum their individual probabilities.

    7 = 16, 25, 34, 43, 52, 61
    11=65, 56

    So the probability of rolling a 7 or 11 is \frac{8}{36}=\frac{2}{9}


    For the last one, you are seeking the probability of the following sequences arising...

    4-4-7, 5-5-7, 6-6-7, 8-8-7, 9-9-7, 10-10-7

    For these the denominator will be 6^6

    As an example, 10 can arise in the following ways....
    64, 55, 46

    10-10 can arise in the following ways...
    6464, 6455, 6446,
    5564, 5555, 5546,
    4664, 4655, 4646

    Since 7 can arise in 6 ways, there are 6(3)(3)=54 ways to get 10-10-7

    so it's probabilty is \frac{54}{6^6} or \frac{9}{6^5}

    4 can occur as follows....
    13, 22, 31

    so you could try the rest as it's the same procedure and let us know how you get on if you like.
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  5. #5
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    Thanks Again Archie,

    I definetely will work on this and I'll let you know!
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  6. #6
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    Quote Originally Posted by August80 View Post
    • Consider throwing two dice



    What is the probability an 8 will be rolled before a 7 is rolled?
    What is the probability the player will either: roll a 7 or 11; or (if not) roll a 4, 5, 6, 8, 9 or 10 AND THEN repeat that same number before a 7 is rolled?
    Any help will be appreciated.
    Thank you,

    Consider
    The probability of rolling an 8 is 5/36; the probability of rolling a 7 is 6/36; and the probability of rolling something else (neither a 7 nor an 8) is 25/36.

    In order to roll an 8 before a 7, you must roll a sequence of zero or more "others" (neither 7s nor 8s), followed by an 8.

    The probability that the final 8 occurs on the first roll is 5/36.

    The probability that the final 8 occurs on the second roll is (25/36) (5/36).

    The probability that the final 8 occurs on the third roll is (25/36) (25/36) (5/36).

    And in general, the probability that the final 8 occurs on the nth roll is
    (25/36)^{n-1} \cdot (5/36).

    Summing over all the possibilities for the final 8 to occur (there are infinitely many of them), the total probability of rolling an 8 before a 7 is

    (5/36) + (25/36) (5/36) + (25/36)^2 (5/36) + (25/36)^3 (5/36) + \dots
    = (5/36) \; [1 + 25/36 + (25/36)^2 + (25/36)^3 + \dots]
    =\frac{5}{36}\cdot  \frac{1}{1 - 25/36}
    = \frac{5}{11}

    where we have used the formula for the sum of an infinite geometric series.
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  7. #7
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    Thank you very much awkward,
    I really appreciate your help.

    Best,
    August80
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  8. #8
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    Hello, August80!

    I totally agree with awkward.
    Here's my approach . . .


    Consider throwing two dice
    What is the probability an 8 will be rolled before a 7 is rolled?
    There are: . 6^2 = 36 possible outcomes with a pair of dice.

    There are 6 ways to roll a "7": . P(7) \,=\,\frac{6}{36}
    . . There are 5 ways to roll an "8": . P(8) \,=\,\frac{5}{36}
    . . There are 25 ways for any Other sum: . P(\text{Other}) \:=\:\frac{25}{36}


    He could get an "8" on his first roll.
    . . P(\text{8 on 1st roll}) \:=\:\frac{5}{36}

    He could get an "8" on his second roll.
    His first roll must be an Other.
    . . P(\text{8 on 2nd roll}) \:=\:\left(\frac{25}{36}\right)\cdot\frac{5}{36}

    He could get an "8" on his third roll.
    His first two rolls must be Others.
    . . P(\text{8 on 3rd roll}) \:=\:\left(\frac{25}{36}\right)^2\cdot\frac{5}{36}

    He could get an "8" on his fourth roll.
    His first three rolls must be Others.
    . . P(\text{8 on 4th roll}) \:=\:\left(\frac{25}{36}\right)^3\cdot\frac{5}{36}

    And so on . . . we see the pattern, right?


    Hence: . P(\text{8 before 7}) \;=\;\frac{5}{36} + \left(\frac{25}{36}\right) + \frac{5}{36}\left(\frac{25}{36}\right)^2 + \frac{5}{36}\left(\frac{25}{36}\right)^3 + \hdots

    . . . . . . . . . . . . . . . =\;\frac{5}{36}\underbrace{\bigg[1 + \frac{25}{36} + \left(\frac{25}{36}\right)^2 + \left(\frac{25}{36}\right)^3 + \hdots \bigg]}_{\text{geometric series}}

    The geometric series has the sum: . \frac{1}{1-\frac{25}{36}} \:=\:\frac{1}{\frac{11}{36}} \:=\:\frac{36}{11}


    Therefore: . P(\text{8 before 7}) \;=\;\frac{5}{36}\cdot\frac{36}{11} \;=\;\frac{5}{11}

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  9. #9
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    Thanks Soroban for your approach!

    I appreciate it!
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