Bayesian updating when data is partially observed Consider two distinct coins (COIN_1 and COIN_2) whose respective prior probabilities are given by:
$Pr_1(HEAD_1)=\alpha_1$
$Pr_2(HEAD_2)=\alpha_2$
Suppose that those two coins are jointly tossed $n$ times but the outcome of each joint toss is observed if and only if "HEAD_1, HEAD_2" is realized.
Is it possible to compute $Pr_1(HEAD_1 | ``HEAD_1,HEAD_2" realized\ k\ times\ in\ n\ tosses)$?
 A: You have two unobserved events for tossing heads on the first and second coin, let's call the events $X$ and $Y$. What you observe is another event $Z = X \land Y$.
If you can assume that $X$ and $Y$ are independent, then by definition
$$
\Pr(Z) = \Pr(X, Y) = \Pr(X) \, \Pr(Y)
$$
In such a case, you can use a simple model using two latent variables for the probabilities $p_X$ and $p_Y$
$$\begin{align}
p_X &\sim \mathsf{Beta}(\alpha_X, \beta_X) \\
p_Y &\sim \mathsf{Beta}(\alpha_Y, \beta_Y) \\
Z   &\sim \mathsf{Bernoulli}(p_X p_Y)
\end{align}$$
If $\Pr(X)$ and $\Pr(Y)$ are very similar to each other, you wouldn't be able to differentiate between them. However, if they are distinct and you have reasonable priors, this could work. Below I provide a simple example, where informative priors on $p_X$ and $p_Y$ lead to not that bad estimates (parameters of beta distribution can be thought of as pseudocounts, so it is as if prior to seeing the data we observed single head and nine tails for $X$).
library("rstan")
set.seed(42)

x <- rbinom(500, size=1, p=0.14)
y <- rbinom(500, size=1, p=0.57)
z <- x * y

model <- "
data {
    int<lower=0> N;
    int z[N];
}
parameters {
    real p_x;
    real p_y;
}
model {
    p_x ~ beta(1, 9);
    p_y ~ beta(5, 5);
    z ~ bernoulli(p_x * p_y);
}
"

stan(model_code=model, data=list(z=z, N=length(z)))

##         mean se_mean   sd    2.5%     25%     50%     75%   97.5% n_eff Rhat
## p_x     0.18    0.00 0.05    0.10    0.14    0.17    0.20    0.30   818    1
## p_y     0.55    0.00 0.13    0.30    0.45    0.54    0.64    0.80   899    1
## lp__ -161.56    0.03 0.97 -164.31 -161.91 -161.27 -160.87 -160.61  1148    1


Without the assumption of independence, there is no simple relationship between marginal and joint distributions and your data gives you degraded information, so this wouldn't be that simple.
However if you can’t assume that $X$ and $Y$ are independent, but you can trust that the prior marginal probabilities are correct, then it’s even simpler. Notice that by the law of total probability
$$
\Pr(X) = \Pr(X, Y) + \Pr(X, \neg Y)
$$
So if you know $\Pr(X)$ and can estimate $\widehat\Pr(X, Y) = \tfrac{\#(X, Y)}{N}$, you can calculate $\Pr(X, \neg Y)$ by subtraction. The same logic applies to calculating $ \Pr(\neg X, Y)$. Next, observe that
$$
\Pr(X, Y) + \Pr(\neg X, Y) + \Pr(X, \neg Y) + \Pr(\neg X, \neg Y) = 1
$$
so you have all the missing pieces besides $\Pr(\neg X, \neg Y)$, than again, it can be obtained by simple algebra.
