Probability of spherical particles touching one another in a cylindrical column I was wondering if you could help me out with a little problem, as my research is getting rather time restricted at the moment, and due to my limited mathematical background, any help would be greatly appreciated, even if it was just to point me in the right direction.
I have an experiment that involves diluting a solid (S1) in another solid (S2), where particles of S1 cannot be touching one another, in a cylindrical column. It is assumed that the particles of S1 and S2 are spherical in shape and all of the same diameter. The cylindrical column in which they will be packed has a diameter of 12.58 mm, and a length/height of 9.86 mm. The particles, let's say, are all 7.0 um in diameter.
I believe that I have calculated a way to dilute the solid: 24% of S1 with 76% of S2. I will then mix these two using a reciprocal shaker before packing inside the column. However...
... How can I calculate the probability that a particle of S1 will be touching another S1 particle in the column? It would be nice to say that the if I put the S1 and S2 particles in my column as above, there is a ?% chance that an S1 particle will be touching another S1 particle (hopefully, this number will be low).
I hope I have made my question clear. If not, I am happy to expand further, and any help would be very much appreciated (even if it's just pointing me to a theorem that I should consider, etc.).
Thank you very much for your time :)
 A: OK, I think I've come up with a useful probability model that is not too complicated.
Objective
Estimate the fraction of $S_1$ particles that are touching other $S_1$ particles. Assuming some mixture of two particle types $S_1,S_2$.

Modelling Assumptions


*

*Packed media is homogenous and infinite in extent. Requires that the number of particles that touch the edge of the cylinder is much less than the number that are on the interior of the packed material. 

*S1 and S2 particles are spheres with identical radii

*There are no attractions between particles

Model Formulation
Definitions


*

*$V$ is an arbitrary volume of packed media and $N_1,N_2$ be the number of $S_1,S_2$ particles in that volume, respectively. Let $N=N_1+N_2$ and $f=\frac{N_1}{N}$.

*$\tau_{N}$ is the fraction of $S_1$ particles in $V$ that are touching another $S_1$ particle

*$P_N$ is the number of touching pairs of particles in the packed media

*$X(\omega_i)$ is a random variable that indicates the particle type for the $i^{th}$ particle ($\omega_i$) in $V$. Specifically, $X(\omega_i)=1$ if $\omega_i=S_1$ and is zero otherwise. I will use the shorthand notation $ X(\omega_i)\equiv  X_i$


Derivations
The assumption of homogeneity implies that the $X_i$ are independent of one another and $P(X_i=1)=f$. Therefore, $P(X_i=1 \cap X_j = 1)=f^2, \;\; \forall i\neq j$. However, we don't care about the probability that any two particles will be $S_1$, we want to know the probability that any two touching particles are both $S_1$, assuming an arbitrary packing of identically sized spheres in 3-dimensions.
It turns out that the above situation has been studied in quite a bit of detail. Specifically, I will rely on the Karoly's Touching Pairs Theorem 1(i) to determine the number of touching pairs of particles. I will denote such a pair using indexed sets $\phi_k=\{\omega_i,\omega_j\}$, since the ordering of the particles in the pair is irrelevant and designates the same pairing.
Using our notation, the theorem states:
$$P_N<6N-0.926N^{2/3}$$
Therefore, we have a collection of $P_N$ pair sets $\{\phi_k:k \in \{1...P_N\}\}$
The assumption of homogeneity again implies that the composition of the pairs are independent of one another. However, we run the risk of double-counting the number of particles involved in the pairs, since some pairs may share a common particle.
To address this, I will calculate the average coordination number ($C_N$) and use it to calculate the correct number of unique particles among the parings.
$$C_N = 2\frac{P_N}{N}$$
For example, if we were dealing with a gas with weak interactions, then we may have every particle only being linked to one other particle. In this case, $P_N=\frac{1}{2}$ hence the average coordination number is $1$.
Therefore, we can think of each particle as being the center of a cluster with $C_N$ branches (rounding $C_N$ to nearest integer). The number of $(S_1,S_1)$ pairings  in the cluster with center particle being $\omega_i$ is given by $K_i$. 
The expected value is calculated conditionally: 


*

*$E[K_i|\omega_i=S_1] = C_Nf$ and

*$E[K_i|\omega_i=S_2] = 0$
Hence, $E[K_i]=P(\omega_i=S_1)E[K_i|\omega_i=S_1] = C_Nf^2$ and the number of $S_1$ particles paired with other $S_1$ particles is expected to be $NC_Nf^2$.
We now have what we need to get our estimate of what fraction of the $S_1$ particles are paired with other $S_1$ particles.

Main Results
$\tau_{N} < \frac{NC_Nf^2}{N_2} = \frac{f^22P_N}{N_2}=\frac{f^22P_N}{fN}=\frac{2fP_N}{N}=\frac{2f(6N-0.926N^{2/3})}{N}=2f\left(6-0.926N^{-1/3}\right)$
(Note that the coordination number of perfectly packed spheres is 12, so arbitrary packing of finite size are expected to generate less than optimal numbers of pairings)
What we want is $\tau_{\infty}$:
$\tau_{\infty} = \lim\limits_{N\to \infty} \tau_N < 12f$

Discussion
Essentially, an infinite and homogenous media of two types of spherical particles can only be bounded by the number of pairings for a perfectly packed media.
The derivation for the bound on $\tau_{\infty}$ implies that, at most, $12f\times 100\%$ percent of the $S_1$ particles are touching other particles.
Unfortunately, this bound is somewhat conservative, as it implies your mixture is not sufficiently diluted to guarantee anything less than 100% of the $S_2$ particles are paired (which is obviously false...but to what extent is it false?). You will need much more dilution.Therefore,  if we want to get $\tau_{\infty} < D$ then we need $f<\frac{D}{12}$!
In your case, if we want a very low pairing probability (say $D<.01$), then $S_1$ can only be $0.08\%$ of the mixture, for a dilution ratio of 1,250:1! Your apparatus may get quite  big (or conversely the effectiveness will be limited) by such a large bulk dilution ratio requirement.
