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Probability: chances of winning

  1. Sep 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider a game where a player rolls 2 dice. The player looses if the sum of the dice of the first roll is 2, 3 or 12, and wins if the sum of the dice is 7 or 11. If the outcome of the first roll is 4, 5, 6, 8, 9 or 10, the player continues to roll the dice until the initial outcome is rolled again (player wins) or the 7 appears (the player looses).
    Find the probability of winning the game.

    2. Relevant equations
    The probability of event E occurring on a sample space [tex]\Omega[/tex] is: P(E) = [tex]\frac{|E|}{|\Omega|}[/tex]

    3. The attempt at a solution
    The size of the sample space is |[tex]\Omega[/tex]| = [tex]6 * 6 = 36[/tex], since we have two six-sided dice, so a total of 36 possible outcomes on a given throw.

    I gather that if the player rolls a sum of 2, 3, or 12 on their first try, they automatically lose the game. Therefore, I let event [tex]E_{L}[/tex] denote the event that either of those sums is thrown. Therefore, [tex]E_{L}[/tex] is the set of all outcomes that result in one of those sums.

    [tex]E_{L}[/tex] = { (1,1), (1,1), (2,1), (1,2), (6,6), (6,6) }
    |[tex]E_{L}[/tex]| = 6

    Likewise, I let event [tex]E_{W}[/tex] denote that a player throws a 7 or 11 on their first try, meaning they automatically win.

    [tex]E_{W}[/tex] = { (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) }
    |[tex]E_{W}[/tex]| = 6

    I also worked out the size of events [tex]E_{4}[/tex], [tex]E_{5}[/tex], [tex]E_{6}[/tex], [tex]E_{7}[/tex], [tex]E_{8}[/tex], [tex]E_{9}[/tex], [tex]E_{10}[/tex], where each subscript represents the number of outcomes that could lead to that sum being thrown. I got:

    |[tex]E_{4}[/tex]| = 4
    |[tex]E_{5}[/tex]| = 4
    |[tex]E_{6}[/tex]| = 6
    |[tex]E_{7}[/tex]| = 6
    |[tex]E_{8}[/tex]| = 6
    |[tex]E_{9}[/tex]| = 4
    |[tex]E_{10}[/tex]| = 4

    Ok, having that done, the chances of automatically winning the game by throwing a 7 or 11 initially are:
    P([tex]E_{W}[/tex]) = [tex]\frac{6}{36}[/tex]

    This is where I am stuck, however. How do I handle the probabilities of winning by throwing a 4, 5, 6, 8, 9, 10 and then getting that same outcome again without getting a 7?
  2. jcsd
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