Probability of picking from a sublattice

Question

Short version

The set of partitions of a four-element set forms a lattice. Suppose that I pick $n$ times from the set of tri- and bipartitions (i.e., the top element = quadripartition and the bottom element = monopartition are not possible choices). What is the probability that my choices form part of a sublattice corresponding to a tripartition?

Clarification

The way this problem arises in my work, it is possible to pick the same partition more than once (i.e., this isn't a lottery-style problem, where each successive choice drains the number of choices available). However, that isn't crucial: i.e., I'm happy with solutions that take each choice to be unique and irreplicable.

Detailed version

The set I am choosing from contains six tripartitions ($A, ..., F$), four bipartitions in which one partition element is a singleton ($a, b, c, d$), and three bipartitions in which the partition elements are of equal size ($x, y, z$). Each tripartition is the join of two “singleton” bipartitions and one “equal” bipartition:

{A, a, b, x}          {D, a, c, z}
{B, b, d, z}          {E, b, c, y}
{C, a, d, y}          {F, c, d, x}

Call these sets $\bar{A}, ..., \bar{F}$ respectively. What is the probability that, if I choose $n$ tri-/bipartitions, all are in one of $\bar{A}, ..., \bar{F}$?

Comment

For the version in which early choices restrict subsequent ones, is the answer just $$6\times \frac{{^4}\mathrm{C}_k}{{^{13}}\mathrm{C}_k} $$

for $k = 2, 3, 4$ (and $1$ if $k<2$ and $0$ if $k>4$)?

Answer 1

Without replacement: your answer is right. It's just dividing the number of ways to have $k$ choices all in one set divided by the number of ways to choose $k$ from $13$ objects.

With replacement: this can be calculated by similar reasoning.

In counting the numbers of choices it's easiest to assume that "order matters", so that, for example, we count choices, $b$ then $c$ separately to $c$ then $b$.

For the $k = 1$, the probability is obviously $1$. So we assume now $k \gt 1$.

\begin{eqnarray*} P(k \mbox{ choices in one set}) &=& \dfrac{\mbox{#Ways to choose $k$ from same set with repetition allowed}}{\mbox{#Ways to choose $k$ from 13 objects with repetition allowed}} \\ && \\ &=& 6 \times \dfrac{\mbox{#Ways to choose $k$ from a particular set with repetition allowed}}{\mbox{#Ways to choose $k$ from 13 objects with repetition allowed}} && \\ && \\ &=& 6 \times \dfrac{4^k}{13^k} \end{eqnarray*}