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Pair of dice probability

  1. Oct 20, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A and B alternate rolling a pair of dice, stopping either when A rolls the sum 9 or when B rolls the sum 6. Assuming that A rolls first, find the probability that the final roll is made by A.

    3. The attempt at a solution

    A rolls a sum 9 on each roll with prob 1/9
    B rolls a sum 6 on each roll with prob 5/36

    Given that A wins, he will win on an odd number of turns. (since A starts)

    Let E be the event that the game finishes on an odd number of turns
    Then P(E) = (1/9)(1-5/36)(1/9)(1-5/36)....

    Where do I go from here?
     
  2. jcsd
  3. Oct 20, 2012 #2

    tiny-tim

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    Hi CAF123! :smile:
    No, that's the probability that A wins on every throw, but generously pretends that he didn't, because he wants to let B win. o:)

    Try again. :smile:
     
  4. Oct 20, 2012 #3

    CAF123

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    Hi tinytim.
    Can you give me a hint to start?
     
  5. Oct 20, 2012 #4

    tiny-tim

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    Try ∑ P(Wn)

    where Wn is the event of A winning on his nth throw.
     
  6. Oct 20, 2012 #5

    CAF123

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    The event that A wins on his turn is just 1/9. What does this sum represent?
     
  7. Oct 20, 2012 #6

    tiny-tim

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    That's P(W1).

    What's P(W2) ?​
     
  8. Oct 20, 2012 #7

    CAF123

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    I thought that the probability of A winning on any of his turns is 1/9. Is this not correct?
    If not, why not? Surely whether A wins is dependent only on what he throws and not B's result.
    Or did I misunderstand something?
     
  9. Oct 20, 2012 #8

    tiny-tim

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    no, that's the probability of A winning on say his 10th turn, given that neither A nor B has already won

    1/9 = P(Wn | neither A nor B has already won before the nth turn)

    you want P(Wn)

    for example, P(W2) is 1/9 times the probability that neither A nor B won on their first throws …

    that's obviosuly less than 1/9 ! :wink:
     
  10. Oct 20, 2012 #9

    CAF123

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    Ok, I think I understand now.
    So P(W2) = (1/9)(1-1/9)(1-5/36)
    P(W3) = (1/9)(1-1/9)^2 (1-5/36)^2..

    Can I write this as [tex] \frac{1}{9} \sum_{i}^{∞} (\frac{8}{9})^i \sum_{i}^{∞} (\frac{31}{36})^i [/tex]
     
  11. Oct 20, 2012 #10

    HallsofIvy

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    In order that P win on the first turn, he must roll a 9. The probability of that is 1/9. In order that P win on the second turn, he must roll anything except a 9 on the first roll, B must roll anything but a 6, and P must roll a 9 on. The probability of that is (8/9)(31/36)(1/9)= (2/9)(31/9)(1/9)= 62/729.

    On any one turn, the probability that P does NOT roll a 9 and B does NOT roll a 6 is (8/9)(31/36)= 62/81. In order that P win on the nth turn both P and B must NOT roll the correct number the previous n- 1 turns and P must roll a 9 on the last turn- the probability of that is (62/81)n-1(1/9).
     
  12. Oct 20, 2012 #11

    tiny-tim

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    (type "\left(" and "\right)", and they come out the correct size :wink:)

    almost :smile:

    that's not the same as [itex] \frac{1}{9} \sum_{i}^{∞} \left(\frac{8}{9}\frac{31}{36}\right)^i [/itex] is it?

    (and starting at i = … ?)
     
  13. Oct 20, 2012 #12

    CAF123

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    i would start at 1 here. Many thanks again for your help. Can I ask, in general, what advice you would offer when tackling probability problems. It is by far the hardest course I am doing this semester and I feel I have difficulties starting the problems, what area of probability to apply etc.. Any advice would be appreciated - thanks.
     
  14. Oct 20, 2012 #13

    tiny-tim

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    (isn't it i = 0?)

    Probability problems are usually solved by rewriting the events in English first, so that you know clearly what the events are.

    Then you can start translating them into maths. :smile:
     
  15. Oct 20, 2012 #14

    Ray Vickson

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    Often it is best to forget formulas for a while and concentrate on understanding the nature of the "sample space" underlying the problem. In this case, it would help to write down the first few instances where A wins:
    Step 1: A wins---stop
    Step 1: A does not win; go to step 2
    Step 2: B does not win; go to step 3
    Step 3: A wins---stop
    Step 3: A does not win; go to step 4
    Step 4: B does not win; go to step 5
    Step 5: A wins--stop
    Step 5: A does not win; go to step 6
    etc., etc.

    For these first few steps it is easy enough to write out the probabilities associated with the outcomes "A wins", and you can use the revealed pattern to develop a formula for the entire game.

    RGV
     
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