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We have two dice. One is fair, but the other is skewed with the probability of getting a 6 being \(\frac{1}{2}.\)

We roll the dice for a total ...

\[ \displaystyle \sum_{n=1}^\infty \frac { (H_n )^2}{2^n} = \frac {\pi ^a}{b} + \log ^2 (c) \]

Let \(H_n\) denote the \(n^{\text{th}} \) harmonic number such that the above ...

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