There is an extra layer of probability in the question which I think obscures the issue without adding anything. Suppose $X_1,...,X_{10000}$ are IID random variables (whose values are numbers which are not necessarily probabilities of anything). Suppose $Y_1,...,Y_{10000}$ are IID standard Gaussians. Let $X_i + Y_i=Z_i$ be observable, and let $a_1,...,a_{100}$ be the indices with the highest values among the $Z_i$. The question is how much larger $(Z_{a_1} + ... + Z_{a_{100}})/100$ is over $(X_{a_1} + ... + X_{a_{100}})/100$ (perhaps conditioned on the values observed). This problem seems fundamental, and I hope a statistician can say how this is typically addressed.
Even if the $X_i$ are constant, this is still interesting, and it shows that you need to know the values $100$ and $10000$, with a good approximation coming from the ratio $r=1/100$. If you simply know that you have noisy versions of the top $100$, you don't know how strong the bias is. (In concrete terms, you need to know how many experiments were not published to estimate the bias toward type I statistical errors in what is reported.)
The bias in the case that the $X_i$ are constant is an upper bound for the bias in general. The proof is simply that $\sum Y_{a_i}$ is at most the sum of the greatest $100$ values among the $Y_i$.
If we take the limit as the size of the sample goes to infinity, fixing the ratio between the sample and the population, then the bias can be computes in terms of the distribution for $X_i$. Let $T$ be the boundary of the top $r$ of the distribution of each $Z_i$. The bias is $1/r \int_{-\infty}^\infty \int_{T-x}^\infty y ~d\mu_Y d\mu_X$ which can be computed explicitly in some cases.
For example, if $X_i$ is distributed as $N(0,\sigma^2)$, then each $Z_i$ is distributed as $N(0,\sigma^2+1)$ so $T = \sqrt{\sigma^2+1}~ \Phi^{-1}(1-r)$, and the bias is $\frac{\exp(-T^2/(2\sigma^2+2)}{r~\sqrt{2\pi (\sigma^2 +1)}} = \frac{\exp(\Phi^{-1}(1-r)^2/2)}{r \sqrt{2 \pi(\sigma^2+1)}}$.
For comparison, the average value of $Z_{a_i}$ is $\frac{\exp(\Phi^{-1}(1-r)^2/2)\sqrt{\sigma^2+1}}{r \sqrt{2 \pi}},$ or $\sigma^2+1$ times as great, so $1/(\sigma^2+1)$ of the average observed value is the bias from the noise $Y_{a_i}$ and $\sigma^2/(\sigma^2+1)$ of the average observed value comes from the $X_{a_i}$.