You don't specify an initial state. From your result for $A$, I infer that you're interested in the expected score in equilibrium (which is $\frac14$ per draw in $A$), not starting from a full deck (which would be slightly different).
In the general case, in a shuffled deck the $i+j$ non-reshuffle cards are equidistributed over the $k+1$ intervals formed by the $k$ reshuffle cards. Thus, on average we draw $\frac{i+j}{k+1}$ non-reshuffle cards before drawing a reshuffle card, and the expected score from these is $\frac i{k+1}$. Thus the expected score per draw is
$$
\frac{\frac i{k+1}}{\frac{i+j}{k+1}+1}=\frac i{i+j+k+1}
$$
(where $+1$ counts the reshuffle card). In your two specific cases, this yields
$$
\frac1{1+0+2+1}=\frac14
$$
and
$$
\frac1{1+10+2+1}=\frac1{14}\;,
$$
respectively.
Note that interestingly it doesn't matter how many of the non-scoring cards are zeros or reshuffles. As long as there's at least one reshuffle card, the expected score per draw is simply $\frac i{n+1}$, where $n$ is the total number of cards, compared to $\frac in$ if you loop through a deck without reshuffle cards.
Edit in response to a comment:
The ratio $\frac{\frac i{k+1}}{\frac{i+j}{k+1}+1}$ comes about as follows: Each time we draw from a full, shuffled deck up to and including a reshuffle card, the expected score is $\frac i{k+1}$ and the expected number of draws is $\frac{i+j}{k+1}+1$. For one such run, the expectation of the score per draw is hard to determine; it isn't simply the ratio of the two expectations. For instance, in your first example, we score $0/1$ with probability $\frac23$ and $1/2$ with probability $\frac13$, so the expectation of the score per draw is $\frac23\cdot0+\frac13\cdot\frac12=\frac16\ne\frac14$. However, if we perform a very large number $m$ of such runs, the number of draws will be close to $m$ times the expected number of draws per run, and the score will be close to $m$ times the expected score per run, so for all the runs together, the expectation of the score per draw is simply the ratio between the expected score per run and the expected number of draws per run.
There are $6$ single-Heart cards and $1$ double-Heart card, so to get at least $n$ successes in $x$ draws, we must either
- Draw at least $n$ single-Heart cards and no double-Heart card, or
- Draw the double-Heart card and get at least $n-2$ single-Heart cards in the other $x-1$ draws.
In the first case, if we are to draw $k$ single-Heart cards, there are $\binom 6k$ ways to choose them, and there are $\binom 9{x-k}$ ways to choose the remaining $x-k$ cards from among the $9$ non-Heart cards, so there are $$\sum_{k=n}^x{\binom 6k\binom 9{x-k}}$$ ways in all.
We can do the second case in much the same way. The double-Heart cards must be chosen, and then we have $\binom 6k\binom 9{x-1-k}$ ways to choose $k$ single-Hearts and $x-1-k$ non-Hearts, giving $$\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}$$
There are $\binom{16}{x}$ ways to draw $x$ cards, so the probability is
$$\frac1{\binom{16}{x}}\left(\sum_{k=n}^x{\binom 6k\binom 9{x-k}}+\sum_{k=n-2}^{x-1}{\binom 6k\binom 9{x-k-1}}\right)$$
Best Answer
Sure, inclusion-exclusion (as you've done) is one way to do it. It's nice to check by doing it another way when you can. In this case, consider the ways to choose $9$ cards without choosing all three of any one "rank". You must either have $9$ singletons, or $7$ singletons and a pair, or $5$ singletons and two pairs, or $3$ singletons and $3$ pairs, or one singleton and $4$ pairs. Each singleton or pair can be chosen in three ways from its "rank". So this amounts to $$ 3^9+\frac{9!}{7!}\cdot3^8+\frac{9!}{5!2!2!}\cdot3^7+\frac{9!}{3!3!3!}\cdot3^6+\frac{9!}{4!4!}\cdot3^5=3523257 $$ ways to not get three of any one "rank". Dividing this by the ${{27}\choose{9}}=4686825$ draws gives a $$ \frac{3523257}{4686825}\approx 75.174\% $$ chance of not getting three of a kind, which doesn't quite agree with your stated chance of getting one. You've done two small things wrong. First, when subtracting the double-counted cases (where you've drawn two sets of three and three leftovers), there are ${{9}\choose{2}}=36$ (not just $8$) ways to choose the two ranks. Second, when adding back in the double-counted cases from that set (where you've drawn three sets of three), there are ${{9}\choose{3}}=84$ possibilities (not just $7$). So you should get $$ \frac{9\cdot{{24}\choose{6}} - 36\cdot {{21}\choose{3}} + 84}{{27}\choose{9}}=\frac{1163568}{4686825}\approx 24.826\% ... $$ this is now in exact agreement with the other approach.