Ok I had some more thoughts last night and have it reduced to the solution of an intractable differential equation, but one that can probably put some pretty reasonable decent bounds on to get a good bench mark estimate, and I think its simple enough to be posted here.
Part 1:
I want to introduce a new variable L. This is the number of times a ball hits the bottom of the ramp. This is not quite the same as y, it could be smaller (since due to z some balls don’t make it all the way down) and it could be larger than y since a certain number of balls will hit a filled basket and so have multiple runs down the ramp.
If we have L shots at a bucket and there are n buckets to choose from then number of times a bucket gets hit will follow a Poisson distribution with an average of L/n balls hitting the bucket, so defining
P(k,L)=Probability bucket gets hit k times given L shots on goal = ({L/n})^k e^{-L/n}/k!
Now what I am particularly interested in is the probability that a bucket gets full (i.e has x or more hits) lets call it q(L)
q(L)=\sum_{k=x}^{\infty}{{(L/n)}^k\ e^{-L/n})/k!}
or more easily calculated one minus the probability that you had fewer than x hits
q(L)=1-\sum_{k=1}^{x-1}{{(L/n)}^k\ e^{-L/n})/k!}
Part 2:
So all we need to do is calculate L, the number of times a ball reaches the bottom. To make things more complicated I’m going to introduce a new notation L(t) is the number of times a ball has reached the bottom after t of the original y balls have completed their journey. L(y) is going to be the final number of hits on bottom.
To calculate L(t), lets look at what happens to ball t+1 as it starts its journey down the ramp:
With probability z it gets knocked off course and gets zero hits
With probability (1-z) it makes it to the bottom and so counts as one hit on bottom.
Then if it did hit bottom there is also probability of about q(L(t)) that it hits a filled bucket and gets knocked back up to the top.
If we let E(L(t)) be the expected number of hits on bottom from this one ball
E(t) = z*0 + (1-z) + (1-z)*q(L(t))*E(t)
The first term is for getting knocked off the ramp. The second term is for getting the single hit on the bottom while the third term is for reaching the bottom and then getting kicked up to the top at which point by definition it has an average of E(t) shots on goal.
Doing some algebra we get,
E(t)=\frac{(1-z)}{1-(1-z)q(L(t))}
Now we have so many balls rolling down and so many buckets to fill that we are firmly in the area where the law of large number kicks in and we can more or less assume relatively speakdng that the expected number of balls, is the actual number of balls. So that the difference between the number of hits down stream after t original balls vs after t+1 original balls is E(t).
In other words L(t+1) - L(t) = E(t)
now again since we have so many balls rolling down the discreteness of L isn’t much of an issue and we can just assume that this difference equation actually represents a differential equation, or in other words
\frac{dL(t)}{dt}\approx\frac{L(t+1)-L(t)}{1}=E(t)=\frac{(1-z)}{1-(1-z)q(L(t))}
Part 3:
From here we would ideally start with the boundary condition that L(0)=0, solve the differential equation for L(y), and then plug that value in for L in the Poisson in part 1 and get your answer.
But differential equation is non linear and very intractable given that q(L(t)) itself is a complicated sum so we won’t be able to solve it explicitly
However it is a continuous monotonic function of q(L(t)), and q(L(t)) is itself continuous, monotonic and bounded between 0 and 1, so it is probably possible to solve it numerically. Or more simply just make reasonable ball park assumption about how large q(t) is likely to get. to bound your solution. Unfortunately differential equations is not my forte so someone else will have to chime in to carry this further.