The general case of $M$ candies drawn in equiprobable groups of $K$ is solved at Expected number of times a set of 10 integers (selected from 1-100) is selected before all 100 are seen. The expected number of required draws is
$$
\sum_{j=1}^M(-1)^{j-1}\binom Mj\frac1{1-\frac{\binom{M-j}K}{\binom MK}}\;.
$$
For $M=5$, $K=2$, this is
\begin{eqnarray}
&&\sum_{j=1}^5(-1)^{j-1}\binom 5j\frac1{1-\frac{\binom{5-j}2}{\binom 52}}
\\&=&
5\cdot\frac1{1-\frac{\binom42}{\binom52}}-10\cdot\frac1{1-\frac{\binom32}{\binom52}}+10\cdot\frac1{1-\frac{\binom22}{\binom52}}-5\cdot\frac1{1-\frac{\binom12}{\binom52}}+1\cdot\frac1{1-\frac{\binom02}{\binom52}}
\\
&=&
10\left(\frac5{10-6}-\frac{10}{10-3}+\frac{10}{10-1}-\frac5{10-0}+\frac1{10-0}\right)
\\
&=&
\frac{671}{126}\;,
\end{eqnarray}
in agreement with the existing two answers.
As far as I can tell, the case of unequal probabilities isn’t well-defined in your question. You write that “you are given the probability of selecting any given candy”, but that doesn’t define a unique distribution over the tuples of candies being selected: There are $\binom MK$ different probabilities $p_t$ for the tuples $t$ and only $M$ probabilities for selecting any given candy, and I don’t see a canonical resolution of this ambiguity that you might have intended to imply.
If you’re instead given the probabilities $p_t$ of selecting the tuples, then by inclusion–exclusion the expected number of draws required is
$$
\sum_{\emptyset\ne S\subset C}(-1)^{|S|-1}\frac1{\sum_{t\cap S\ne\emptyset}p_t}\;,
$$
where the outer sum runs over all subsets $S$ of the set $C$ of candies and the sum in the denominator runs over all tuples $t$ that intersect $S$. In the equiprobable case, $p_t=\binom MK^{-1}$, there are $\binom Mj$ subsets of size $j$, and for each subset of size $j$ there are $\binom MK-\binom{M-j}K$ tuples that intersect it, so we recover the result above.
As regards your additional “food for thought” question, I take it that you mean the following scenario: There is an unknown number $M$ of candies in a bag. You make $N$ draws of $K$ candies each, where $N$ is predetermined and doesn’t depend on the results of the draws. I assume that you can distinguish the candies, so “the data you collect” is the identity of the candies drawn in each draw.
In order to estimate $M$ from these results, you need to assume a prior for $M$. If you don’t know anything about $M$, a reasonable choice might be an improper uniform prior over $\mathbb N$ (improper because it can’t be normalized, so it’s not a probability distribution).
Another point to note is that given the number $D$ of different candies observed, the details of which of these candies you observed in which draw contain no additional information about $M$; that is, $D$ is a sufficient statistic for $M$.
The case $K=1$ of individual draws is treated in the Magic 8 Ball Problem. The results are quite interesting. The improper uniform prior leads to a normalizable posterior probability distribution for $M$ only if $D\le N-2$, that is, if at least two draws give you a candy that you’d seen before; and the expected value of $M$ under this posterior distribution is finite only if $D\le N-3$. To generalize this to $K\gt1$, let’s look at what changes in that analysis. (I’m referring to the above post but using your variables to keep the notation in this post consistent.) For $K=1$, the probability that $N$ draws yield $D$ distinct candies was
$$
\frac{D!}{M^N}\binom MD\left\{N\atop D\right\}\;,
$$
where $\left\{N\atop D\right\}$ is a Stirling number of the second kind that counts the partitions of a set of $N$ elements into $D$ non-empty subsets, so $D!\left\{N\atop D\right\}$ counts the ways to distribute $N$ elements into $D$ distinguishable non-empty bins. For $K\gt1$, the corresponding probability contains a corresponding count of the ways to distribute $K$-tuples of balls into distinguishable non-empty bins (with each $K$-tuple going into $K$ different bins). But this count cancelled out in the analysis because it didn’t depend on $M$, and the same happens for $K\gt1$, so we don’t have to worry about actually counting these configurations; all we need is the $M$ dependence of the probability. For that, we still get the factor $\binom MD$ for the number of ways to choose the $D$ distinct candies out of the $M$ in the bag, and instead of the probability $M^{-N}$ of drawing a particular sequence of candies, we now get the probability $\binom MK^{-N}$ of drawing a particular sequence of $K$-tuples of candies. Thus, the posterior distribution of $M$ given $D$ (assuming the uniform prior) is
$$
P(M=m\mid D=d)=\frac{\binom mK^{-N}\binom md}{\sum_{m=d}^\infty\binom mK^{-N}\binom md}\;.
$$
We get an analogous phenomenon as for $K=1$: the series for the normalization constant in the denominator only converges if $D\le NK-2$ (that is, if we see at least two fewer candies than we maximally could have in $N$ $K$-tuples), and the expected value
$$
E(M\mid D=d)=\frac{\sum_{m=d}^\infty m\binom mK^{-N}\binom md}{\sum_{m=d}^\infty\binom mK^{-N}\binom md}
$$
is only finite if $D\le NK-3$. As in the $K=1$ case, you might want to use the mode rather than the expected value to estimate $M$.
Note that (as one might have expected) for small $K$ the result isn’t all that different than if you’d drawn $NK$ candies one by one. For instance, in your example $K=2$, we’ve merely replaced $m^2$ by $m(m-1)$. (The factor $\frac12$ cancels.) Still, for small numbers of candies, it makes a slight difference. For example, if you draw $6$ candies individually and only see $2$ different ones, the expected value of $M$ is
\begin{eqnarray}
E(M\mid D=2)
&=&
\frac{\sum_{m=2}^\infty m\cdot m^{-6}\binom m2}{\sum_{m=2}^\infty m^{-6}\binom m2}
\\
&=&
\frac{90\zeta(3)-\pi^4}{\pi^4-90\zeta(5)}
\\
&\approx&
2.64\;,
\end{eqnarray}
with a probability of
\begin{eqnarray}
P(M=2\mid D=2)
&=&
\frac{2^{-6}}{\sum_{m=2}^\infty m^{-6}\binom m2}
\\
&=&
\frac{45}{16\left(\pi^4-90\zeta(5)\right)}
\\
&\approx&
69\%
\end{eqnarray}
concentrated at $M=2$, whereas if you only see $2$ different candies in drawing $3$ pairs, it’s
\begin{eqnarray}
E(M\mid D=2)
&=&
\frac{\sum_{m=2}^\infty m\binom m2^{-3}\binom m2}{\sum_{m=2}^\infty\binom m2^{-3}\binom m2}
\\
&=&
\frac{\sum_{m=2}^\infty m\binom m2^{-2}}{\sum_{m=2}^\infty\binom m2^{-2}}
\\
&=&
\frac12\cdot\frac{\pi^2-6}{\pi^2-9}
\\
&\approx&
2.22\;,
\end{eqnarray}
with a probability of
\begin{eqnarray}
P(M=2\mid D=2)
&=&
\frac1{\sum_{m=2}^\infty\binom m2^{-3}\binom m2}
\\
&=&
\frac3{4\left(\pi^2-9\right)}
\\
&\approx&
86\%
\end{eqnarray}
concentrated at $M=2$, so you’re considerably more confident that there are actually just those two candies in the bag.
