@wolfies comment, and my attempt at a response to it revealed an important problem with my other answer, which I will discuss later.
Specific Case (n=16)
There is a fairly efficient way to code up the full distribution by using the "trick" of using base 2 (binary) numbers in the calculation. It only requires 4 lines of R code to get the full distribution of Y=∑ni=1Zi where Pr(Zi=1)=pi. Basically, there are a total of 2n choices of the vector z=(z1,…,zn) that the binary variables Zi could take. Now suppose we number each distinct choice from 1 up to 2n. This on its own is nothing special, but now suppose that we represent the "choice number" using base 2 arithmetic. Now take n=3 so I can write down all the choices so there are 23=8 choices. Then 1,2,3,4,5,6,7,8 in "ordinary numbers" becomes 1,10,11,100,101,110,111,1000 in "binary numbers". Now suppose we write these as four digit numbers, then we have 0001,0010,0011,0100,0101,0110,0111,1000. Now look at the last 3 digits of each number - 001 can be thought of as (Z1=0,Z2=0,Z3=1)⟹Y=1, etc. Counting in binary form provides an efficient way to organise the summation. Fortunately, there is an R function which can do this binary conversion for us, called intToBits(x) and we convert the raw binary form into a numeric via as.numeric(intToBits(x)), then we will get a vector with 32 elements, each element being the digit of the base 2 version of our number (read from right to left, not left to right). Using this trick combined with some other R vectorisations, we can calculate the probability that y=9 in 4 lines of R code:
exact_calc <- function(y,p){
n <- length(p)
z <- t(matrix(as.numeric(intToBits(1:2^n)),ncol=2^n))[,1:n] #don't need columns n+1,...,32 as these are always 0
pz <- z%*%log(p/(1-p))+sum(log(1-p))
ydist <- rowsum(exp(pz),rowSums(z))
return(ydist[y+1])
}
Plugging in the uniform case p(1)i=i17 and the sqrt root case p(2)i=i√17 gives a full distribution for y as:
y012345678910111213141516Pr(Y=y|pi=i17)0.00000.00000.00030.00260.01390.04910.11810.19830.23530.19830.11810.04910.01390.00260.00030.00000.0000Pr(Y=y|pi=i√17)0.05580.17840.26520.24300.15360.07100.02480.00670.00140.00020.00000.00000.00000.00000.00000.00000.0000
So for the specific problem of y successes in 16 trials, the exact calculations are straight-forward. This also works for a number of probabilities up to about n=20 - beyond that you are likely to start to run into memory problems, and different computing tricks are needed.
Note that by applying my suggested "beta distribution" we get parameter estimates of α=β=1.3206 and this gives a probability estimate that is nearly uniform in y, giving an approximate value of pr(y=9)=0.06799≈117. This seems strange given that a density of a beta distribution with α=β=1.3206 closely approximates the histogram of the pi values. What went wrong?
General Case
I will now discuss the more general case, and why my simple beta approximation failed. Basically, by writing (y|n,p)∼Binom(n,p) and then mixing over p with another distribution p∼f(θ) is actually making an important assumption - that we can approximate the actual probability with a single binomial probability - the only problem that remains is which value of p to use. One way to see this is to use the mixing density which is discrete uniform over the actual pi. So we replace the beta distribution p∼Beta(a,b) with a discrete density of p∼∑16i=1wiδ(p−pi). Then using the mixing approximation can be expressed in words as choose a pi value with probability wi, and assume all bernoulli trials have this probability. Clearly, for such an approximation to work well, most of the pi values should be similar to each other. This basically means that for @wolfies uniform distribution of values, pi=i17 results in a woefully bad approximation when using the beta mixing distribution. This also explains why the approximation is much better for pi=i√17 - they are less spread out.
The mixing then uses the observed pi to average over all possible choices of a single p. Now because "mixing" is like a weighted average, it cannot possibly do any better than using the single best p. So if the pi are sufficiently spread out, there can be no single p that could provide a good approximation to all pi.
One thing I did say in my other answer was that it may be better to use a mixture of beta distributions over a restricted range - but this still won't help here because this is still mixing over a single p. What makes more sense is split the interval (0,1) up into pieces and have a binomial within each piece. For example, we could choose (0,0.1,0.2,…,0.9,1) as our splits and fit nine binomials within each 0.1 range of probability. Basically, within each split, we would fit a simple approximation, such as using a binomial with probability equal to the average of the pi in that range. If we make the intervals small enough, the approximation becomes arbitrarily good. But note that all this does is leave us with having to deal with a sum of indpendent binomial trials with different probabilities, instead of Bernoulli trials. However, the previous part to this answer showed that we can do the exact calculations provided that the number of binomials is sufficiently small, say 10-15 or so.
To extend the bernoulli-based answer to a binomial-based one, we simply "re-interpret" what the Zi variables are. We simply state that Zi=I(Xi>0) - this reduces to the original bernoulli-based Zi but now says which binomials the successes are coming from. So the case (Z1=0,Z2=0,Z3=1) now means that all the "successes" come from the third binomial, and none from the first two.
Note that this is still "exponential" in that the number of calculations is something like kg where g is the number of binomials, and k is the group size - so you have Y≈∑gj=1Xj where Xj∼Bin(k,pj). But this is better than the 2gk that you'd be dealing with by using bernoulli random variables. For example, suppose we split the n=16 probabilities into g=4 groups with k=4 probabilities in each group. This gives 44=256 calculations, compared to 216=65536
By choosing g=10 groups, and noting that the limit was about n=20 which is about 107 cells, we can effectively use this method to increase the maximum n to n=50.
If we make a cruder approximation, by lowering g, we will increase the "feasible" size for n. g=5 means that you can have an effective n of about 125. Beyond this the normal approximation should be extremely accurate.