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    This can be calculated exactly

    For example:

    [0] = 1 (always starts here)

    [1] = P(1) (probability of rolling a 1) = 1/6

    [2] = P(2) + P(1)*P(1) = 1/6 + 1/36 (you can get here by rolling a 2 or rolling 2x1s)

    ...

    It's a combinatorial problem and should have a solution as a function of n