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24 February, 09:57

In a certain board game, a player rolls two fair six-sided dice until the player rolls doubles (where the value on each die is the same). The probability of rolling doubles with one roll of two fair six-sided dice is 1/6. What is the probability that it takes three rolls until the player rolls doubles?

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  1. 24 February, 10:44
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    (1/6) (5:6) ^2
  2. 24 February, 12:33
    0
    The probability is 25/216. Approximately 0.116

    Step-by-step explanation:

    You need

    - The first throw to fail (probability 1-1/6 = 5/6)

    - The second throw to fail (probability 5/5)

    - The third throw to be a success (probability 1/6)

    Since each throw is independent of the others, we have to multiply all probabilities to obtain the total probability of the event. Thus, the probability of requiring 3 rolls until getting doubles is

    5/6 * 5/6 * 1/6 = 25/216 = 0.115741

    This problem can also be solved with sophisticated theory;

    the random variable which counts the number of tries until the first success is a geometric distribution. The only parameter of the distribution is the probability of success p. If X is geometric with parameter p, then the probability of X being equal to k (in other words, requiring k tries for a success) is

    P (X = k) = (1-p) ^ (k-1) * p

    If p = 1/6, then

    P (X = 2) = (1-1/6) ^2 * (1/6) = (5/6) ²*1/6 = 25/216
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