Let me state the problem in terms of vases and water, I think it's clearer. We have three vases $A$, $B$ and $C$, with $A$ containing a certain amount of water, so let's say $A=1$ and $B=C=0$ empty. The vase $A$ is connected to the vase $B$ and $B$ is connected to $C$. What I want to do is to pour water from $A$ to $C$, but I can do that in steps and at each step I can pour at most a fraction $\varepsilon$ of the water in a vase inside the next, and no more. So, after the first steps I'll have (I will call $A_n$, $B_n$ and $C_n$ the quantity of water at step $n$):
$$ A_1 = 1- \varepsilon \quad B_1 = \varepsilon \quad C_1 = 0$$
$$ A_2 = 1- 2 \varepsilon + \varepsilon ^2 \quad B_2 = 2\varepsilon - 2\varepsilon ^2 \quad C_2 = \varepsilon ^2$$
$$ A_3 = 1- 3 \varepsilon + 3 \varepsilon ^2 -\varepsilon ^3 \quad B_3 = 3\varepsilon - 6 \varepsilon ^2 + 3\varepsilon ^3 \quad C_3 = 3 \varepsilon ^2 - 2 \varepsilon ^3$$
and it soon gets very complicated. Now, for two vases this is easy and after $N$ steps you are left in the first vase with $(1- \varepsilon) ^N$ water, and so there is $1 - (1- \varepsilon )^N \approx N \varepsilon$ in the second vase. In this case however, is there a formula or at least some known behavior?
P.S. I would actually be interested in the general case with $n$ vases, so to estimate how much water I can transport from the first vase to the last in a certain number of steps.
Suppose there are $M$ total vases connected to each other. The quantity of water in the vases $Q^{(1)},..., Q^{(M)}$ at time step $n$ obeys the recurrence relations
$$Q^{(1)}_n=(1-\epsilon)Q_{n-1}^{(1)}$$ $$Q^{(k)}_n=(1-\epsilon)Q_{n-1}^{(k)}+\epsilon Q_{n-1}^{(k-1)}~~,~~ 2\leq k\leq M-1$$ $$Q^{(M)}_n=Q_{n-1}^{(M)}+\epsilon Q^{(M-1)}_{n-1}$$
This is a linear system of equations and it can in principle be solved using matrix techniques; however the matrix does not have a complete eigenvector space and so its Jordan decomposition would be required to obtain results. Fortunately, this is not immediately necessary, as a solution can be constructed by solving for the quantity of water in the first vase and sequentially substituting in the other equations. We can ignore the last equation, since $Q^{(M)}_n=1-Q^{(1)}_n-...Q^{(M-1)}_n$.
Without delving too much into details, it can be shown that for $1\leq k\leq M-1$ the solution can be written as
$$Q_n^{(k)}=(1-\epsilon)^{n-k+1}\gamma_n^{(k)}$$ $$\gamma_n^{(k)}=\gamma_{n-1}^{(k)}+\epsilon\gamma_{n-1}^{(k-1)}$$
The latter set of recursive equations can be inductively shown to have an explicit solution:
$$\gamma_n^{(k)}=\epsilon^k{n\choose k}$$
Incidentally, the quantity of water in any of the first $M-1$ vases is the coefficient of the binomial expansion $[x^k](1-\epsilon)[(x\epsilon+(1-\epsilon))^n-1]$. Remarkably, the solution does not change its form as one increases the number of the vases considered: at any fixed time step the quantities $Q^{(1)}_n,..., Q^{(M-1)}_n$ are drawn from the coefficients of the binomial distribution and are padded by zeroes for $M-1>n$, as expected from the physics of the problem (the $k+1$-th vase stays empty until the $k$-th timestep).