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Because $ D $ and $ H $ are independent:
P(H = D) = \sum_{k=1}^{6} P(H = k) \cdot P(D = k) = \sum_{k=1}^{6} \binom{6}{k} \left(\frac{1}{2}\right)^6 \cdot \frac{1}{6}.
Factor out constants:

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