What probability or weight should I assign to X so that it has a Y% chance of being one of the 6 items selected out of a population of 12? I have a population of 12 items, let's call them A through L. I'm going to be drawing 6 of those items out of the population without replacement. So you can't choose the same item twice.
What I know is that I want A to be selected as any one of the 6 items approximately 99% of the time, F to be selected as any one of the 6 items approximately 57% of the time, etc.. I'll post the full values below.
I'm using Python's NumPy library to sample without replacement, but I believe the methodology is similar to R. As each item is chosen from the population, its weight is distributed among the remaining items and then set to 0.
Kudos to @Glen_b for the example:

"If we're selecting 2 items from a pool of 3 items with first draw
  probabilities (0.1,0.4,0.5) and we select the third item first, the
  probabilities for the second item will be (0.2,0.8,0)"

What I'm trying to deduce is how to reverse engineer what weights would get sent to a sampling function to end up with the results below.
# Probabilities that each item is one of the 6 selected out of 12  

itemSelected = {
A = 0.99
B = 0.97
C = 0.95
D = 0.82
E = 0.68
F = 0.57
G = 0.48
H = 0.33
I = 0.11
J = 0.09
K = 0.01
L = 0.00
}

 A: At first, I thought a solution might require summing over all possible sequences of draws, with a resulting combinatorial explosion. But, thinking about the problem from a slightly different angle, there's actually an efficient way to approximate the solution. It only requires solving a simple system of linear equations.
Assuming the target probabilities given in the question and 6 draws without replacement, I calculated the item weights using the method described below. The weights for the first 11 items are:
$$w = [.2864, .2181, .1863, .1066, .0709, .0525, .0407, .0249, .0072, .0059, .0006]$$
The 12th item has zero weight and can be ignored, since it has zero probability of being drawn. Then, I simulated random draws using these weights, and calculated the empirical probabilities. They match the target probabilities fairly well:

Formulating the problem
Suppose there are $n$ unique items (which I'll refer to by number instead of letters). Let $w = [w_1, \dots, w_n]$ be weights for each item, which are assumed to be nonnegative and sum to one. Suppose we draw $k$ items without replacement. On the first draw, the probability of each item is given by its weight. On subsequent draws, the probability of each item is given by its weight divided by the summed weights of all remaining items (or zero if the the item has already been drawn). 
Let $X = [X_1, \dots, X_n]$ be a binary random vector indicating whether each item has been drawn after $k$ draws; $X_i = 1$ if item $i$ has been drawn, otherwise 0. Let $\mu = [\mu_1, \dots, \mu_n]$ denote the expected value of $X$. Then, $\mu_i$ can be interpreted as the probability that item $i$ is included in the final set of $k$ selected items. Therefore, the problem can be framed as follows: Find the weights $w$ that give rise to a specified set of values in $\mu$ (target probabilities).
The relationship between $w$ and $\mu$
The distribution of $X$ can be identified by drawing an analogy to a paricular ball-and-urn problem. Let each item correspond to a weighted, colored ball in an urn. Each ball is uniquely colored, so there are $n$ colors with 1 ball each. We draw $k$ balls from the urn without replacement, according to their weights $w$ (as in the data generating process described above). $X$ can be interpreted as the number of times we observe each color. Then, the distribution of $X$ is a special case of the multivariate version of Wallenius' noncentral hypergeometric distribution. This is a relatively obscure, but fascinating distribution; see the references below for details. The general form allows for multiple balls of each color, but things simplify slightly in our case, since balls are uniquely colored.
$\mu$ is the mean of this distribution. Unfortunately, it doesn't have a closed form expression. But, Manly (1974) proposed an approximation (which is also described clearly in Fog 2008). Under this approximation (in our simplified case), $\mu$ and $w$ are related as:
$$(1-\mu_1)^{\frac{1}{w_1}}
= (1-\mu_2)^{\frac{1}{w_2}}
= \cdots
= (1-\mu_n)^{\frac{1}{w_n}}$$
Solving for the weights
Take the reciprocal of the log of all sides:
$$\frac{w_1}{\log (1-\mu_1)}
= \frac{w_2}{\log (1-\mu_2)}
= \cdots
= \frac{w_n}{\log (1-\mu_n)}$$
This is a system of $n (n-1) / 2$ linear equations. That is, for each unique $(i,j)$ pair:
$$\frac{w_i}{\log (1-\mu_i)} - \frac{w_j}{\log (1-\mu_j)} = 0$$
We also know that $w$ must sum to one, which can be expressed as an additional linear equation:
$$\sum_i w_i = 1$$
The task is to solve these equations for $w$, given $\mu$. This can be done by writing the entire system in the form $A w = b$, where $A$ is a matrix (constructed using the given $\mu$) and $b$ is a vector. $A$ and $b$ can then be passed to a standard linear solver to obtain a numerical solution for $w$.
Note that elements of the given $\mu$ must sum to $k$ by the definition of the problem. Furthermore, they must be strictly positive; otherwise $\log(0)$ would appear in the above equations. This isn't a big deal, since $\mu_i=0$ would imply that item $i$ has zero probability of being drawn, and we might as well drop it from the problem.
References
Manly (1974). A Model for Certain Types of Selection Experiments.
Fog (2008). Calculation Methods for Wallenius' Noncentral Hypergeometric Distribution.
