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Interpret random choice of coin as equal prior probabilities, and the tosses as independent conditional on the chosen coin. Let $H_2$ denote two heads. The conditional probabilities are $P(H_2\mid A)=1/16$ and $P(H_2\mid B)=9/16$. For posterior coin selection from repeated tosses, Bayes theorem givesThe tosses need not be unconditionally independent, because the same hidden coin is used twice. If the coin-choice prior were instead $P(B)=\pi$, the answer would be $9\pi /(1+8\pi )$; equal choice gives the stated $9/10$.

For nondecreasing uniform samples in the lottery, all $n^r$ ordered samples have equal probability by the discrete uniform distribution and independence. A nondecreasing sample is uniquely specified by its multiplicities $k_1,\dots ,k_n\ge 0$, with $k_1+\cdots +k_n=r$. The stars and bars argument places $n-1$ separators among $r$ identical markers, giving $\left(\genfrac{}{}{0ex}{}{r+n-1}{n-1}\right)$ multiplicity vectors. There is exactly one nondecreasing ordered sample for each such vector, so the probability isThis counts samples with ties correctly; multiplying a single strictly increasing count by $r!$ would not handle the repeated values.