Yes and no. In this particular doc the change you are seeing reflects the change of uncertainty of the estimation, but not the posterior.

For a similar model you *can* estimated the posterior uncertainty by increasing PPC, because for Bernoulli the variance (or uncertainty) is linked to the mean, but in general it is not the case.

Think of a Gaussian, where we have better intuition:

Say you have a data generation process y \sim \text{Normal}(\mu, \sigma^2), with \mu and \sigma being the model parameters. When both parameters are fixed, the uncertainty of y is fully explain by the property of Gaussian, and its uncertainty is quantify by a function of \sigma (e.g., if you quantify the uncertainty as variance to the mean then it is just \sigma^2).

Now, say you have \sigma fixed and \mu from some distribution (in PPC, it would means \mu \sim \pi_{posterior}(...)), then the distribution for y becomes a result of a convolution (see Colah’s blog on this where in the beginning he gives an very intuitive representation). Which to answer your initial question, the posterior uncertainty is in \pi_{posterior}(...), where as PPC uncertainty is the convolution of \pi_{posterior}(...) and a Gaussian with the fixed \sigma

All these is to show you that, increasing PPC sample *does not* necessary give you the ability to quantify the uncertainty of posterior for \mu, but it can let you *better estimate* of the uncertainty of y.