Subset Probability to Element Probability Is there any way to match (or map) from Subset Propabilities to Element Probabilities?
Suppose that John may select x-sized subsets from a population of N items. 
In every subset he has exactly x items. 
Let subset A has for example {item1, item3, item7} and subset B has {item2, item3, item9}.
How can I compute the probability of selecting merely item1, item2, etc???
(From the above example, we can see that item3 is in both subsets, so the Pr[item3]=1, correct?)
Thanks for your help
 A: For $i=1,\dots,N$ define $\mathcal{S}_{i}=\left\{ A\mid A\text{ contains item }i\right\} $.
Then $Pr\left[\text{item }i\right]=\sum_{A\in\mathcal{S}_{i}}Pr\left[A\right]$.
This if John elects one subset. Here $Pr[A]$  stands for the probability that John elects subset $A$.
A: This is reminiscent of Rota's version of Möbius inversion formula... Here, one is given the probability $P(I)$ of every subset $I$ of size $n$ of some set of size $N$, with $N\geqslant1$ and $0\leqslant n\leqslant N$, and one looks for the probability $p_i$ of each individual element $i$ of the set. 
If $n=0$ or $n=N$, we are given nothing hence the probabilities $p_i$  are unknown, except if $N=1$. 
In the non degenerate case $1\leqslant n\leqslant N-1$, fix some $i$ in the set and consider the sum $$s(i)=\sum_IP(I)\,[|I|=n]\,[i\in I].$$ Then, $$s(i)=\sum_I\sum_jp_j\,[j\in I]\,[|I|=n]\,[i\in I]=\alpha p_i+\beta\sum_{j\ne i}p_j,$$ with $$\alpha=\sum_I[|I|=n]\,[i\in I],\qquad\beta=\sum_I[j\in I]\,[|I|=n]\,[i\in I].$$ One sees that $$\alpha={N-1\choose n-1},\qquad\beta={N-2\choose n-2},\qquad\sum_{j\ne i}p_j=1-p_i,$$ hence a simple way to recover $p_i$ from the collection $\{P(I)\mid |I|=n\}$ is to consider $$p_i=\frac{s(i)-\beta}{\alpha-\beta}.$$ Note that $\alpha$ and $\beta$ depend on $(N,n)$ only, and that, if $n=1$, then $(\alpha,\beta)=(1,0)$, which can help as a confirmation... (do you see why?) Another example/confirmation: if $N=4$, $n=2$, and the set is $\{1,2,3,4\}$, then $(\alpha,\beta)=(3,1)$ hence $$p_1=\frac{P(\{1,2\})+P(\{1,3\})+P(\{1,4\})-1}2.$$
