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Math Help - Probability question with a twist

  1. #1
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    Probability question with a twist

    This question starts off seemingly simple but theres a twist at the end which I can't get my head around, it was in my statistics end of year paper at university.

    In the game of "craps", a player rolls two dice independently. The score of each die is 1, 2, 3, 4, 5, 6, and each score can be assumed to have the same probability. The outcome of each roll of the two dice is the sum of their scores.

    Now the question asked was:

    If the outcome of the first roll is 8, the player continues by rolling the two dice again and again until the outcome is either 7 or 8. He then wins if this outcome is 8, and loses if it is 7. What is the probability that the player rolls 8 in the first roll, AND goes on to win the game?

    The last part got me because I cant find a way to work out how many time the dice will be rolled, but I'm probably going down the wrong alley here... help would be greatly appreciated
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  2. #2
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    Quote Originally Posted by holaboo View Post
    the question asked was:
    If the outcome of the first roll is 8, the player continues by rolling the two dice again and again until the outcome is either 7 or 8. He then wins if this outcome is 8, and loses if it is 7. What is the probability that the player rolls 8 in the first roll, AND goes on to win the game?
    Is it clear to you that it takes at least two turns to win?

    The probability that a player does win in exactly two turns is \left( {\frac{5}{{36}}} \right)^2 . Do you see why?

    The probability that a player does win in exactly k turns is \left( {\frac{5}{{36}}} \right)^2\left( {\frac{25}{{36}}} \right)^{k-2 }. Do you see why?

    If so, you have to add all that up.
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  3. #3
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    The dice can be rolled many, many times. So? Go forward with that.

    Roll 1 - 8
    Roll 2 - it's an 8 or a 7 or anything else. Anything but 7 or 8 coninues.

    What's p(neither 7 nor 8)? Then look up the Negative Binomial Distribution
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    Quote Originally Posted by Plato View Post
    Is it clear to you that it takes at least two turns to win?

    The probability that a player does win in exactly two turns is \left( {\frac{5}{{36}}} \right)^2 . Do you see why?

    The probability that a player does win in exactly k turns is \left( {\frac{5}{{36}}} \right)^2\left( {\frac{25}{{36}}} \right)^{k-2 }. Do you see why?

    If so, you have to add all that up.
    Aha I see, I actually got that far in the exam, but couldnt get the negative binomial distribution part.

    Thanks guys!
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    Quote Originally Posted by holaboo View Post
    Aha I see, I actually got that far in the exam, but couldnt get the negative binomial distribution part.
    Can you find \sum\limits_{k = 2}^\infty  {\left( {\frac{5}{{36}}} \right)^2 \left( {\frac{{25}}{{36}}} \right)^{k - 2} }~?
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  6. #6
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    Quote Originally Posted by Plato View Post
    Can you find \sum\limits_{k = 2}^\infty  {\left( {\frac{5}{{36}}} \right)^2 \left( {\frac{{25}}{{36}}} \right)^{k - 2} }~?
    Actually no, I got up to the step with (25/36)^k-2 but didnt know where to go from there.

    I get confused whenever I see the infinity sign.... especially when evaluating integrals...

    How would you evaluate this sum?
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    Quote Originally Posted by holaboo View Post
    Actually no, I got up to the step with (25/36)^k-2 but didnt know where to go from there.
    I get confused whenever I see the infinity sign.... especially when evaluating integrals...
    How would you evaluate this sum?
    This has absolutely nothing to do integrals.
    It is all about geometric series.
    If |r|<1 then \sum\limits_{k = J}^\infty  {Ar^k }  = \frac{{Ar^J }}{{1 - r}} where J\in \mathbb{Z}~\&~A\in \mathbb{R}.
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  8. #8
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    I think there is a slick way of avoiding a geometric series, the following just being a heuristic argument. First, condition on rolling an 8. We will roll until we hit a 7 or 8. So, conditional on rolling the 8 on the first roll, the probability of winning is the probability that we roll an 8 before we roll a 7; intuitively, this should be equal to P(Rolling an 8 on the second roll) / P(Rolling a 7 or 8 on the second roll) because the only thing that matters is the first roll where we roll either a 7 or 8; on this roll, the probability of it being an 8 should be the probability of rolling an 8 given that we roll either a 7 or 8.

    Thus, the probability of winning conditional on rolling an 8 on the first roll would be \frac{5 / 36}{5/36 + 6/36} = 5/11. Then multiply by the probability of rolling an 8 on the first roll and get \frac 5 {11} * \frac{5}{36} = \frac{25}{396}. This agrees with using the geometric series.
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