Modelling Y=min(c,X) for different c

Question

Assume I have a random variable $X \sim Poisson(\lambda)$ which models the potential nr of people entering some room. Now consider this room has a capacity $c$ so that whenever $X > c$ we observe $c$, so basically we consider the random variable $Y = min(c, X)$.

Now if we consider that $\lambda$ is fairly small, say $\lambda = 5$ and we consider that $c = 5$ then if I point estimate by taking the mean of $Y$ I will get some value $y \approx4$.
However, if I take the original room and put in a bunch of walls to make sub-rooms where each sub-room now has a capacity of $c=1$ and I divide up the intensity between these new rooms so that $\lambda = 1$ for each room and then point estimate by the mean again I get values $y_i \approx0.6, i = 1,...,5$. If I sum these different means I will get $y \approx3$

So by putting in those walls in my room and uniformly divided up the intensity between them I have essentially "lost" 1 whole expected person entering. I've been trying to wrap my head around it but haven't been able to come up with a satisfying answer of

1. How this is bad modelling (which I assume it is)
2. How to model this kind of situation better

Any ideas?

Also, a snippet of code in R to illustrate:

f <- function(lambda, C) {
  lambda * ppois(C-2,lambda) + C * ppois(C-1, lambda, lower.tail=FALSE)
}

censored1 = f(1, 1)
censored2 = f(5, 5)

list("manySmallRooms" = censored1*5, "oneBiggerRoom" = censored2)

$manySmallRooms
[1] 3.160603

$oneBiggerRoom
[1] 4.122663

Answer 1

The two are not equivalent. Think of it this way:

Case A: There is one big queue to get into the single room, where the number of arrivals is $\mathrm{Poisson}(5)$.

Case B: There are five separate queues (one for each sub-room), each of which is $\mathrm{Poisson}(1)$.

While it is true that the distribution of the total number of arrivals is the same in both cases, suppose that there are exactly two arrivals. How many of them can get in a room?

In Case A the capacity of the room is 5, so both people can get in.

In Case B if the two people happen to queue for the same room, only one of them can get in.