Bayesian inference on correlation of signals

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Consider the following setup. There are two states $H$ and $L$ where $p=Pr(H)$ is the prior. A Bayesian agent observes two binary "signals" $s \in \{h,l\}$ on the state, which are determined as follows: with probability $\lambda$ a state of the world $ind$ occurs and two independent signals are drawn, but with probability $1-\lambda$ the state of the world is $cor$, only one signal is drawn and sent twice, so the signals are perfectly correlated (so $\lambda$ is the prior probability of state $ind$ and realization of $ind$ vs $cor$ and $H$ vs $L$ is independent). Each drawn signal is such that $\Pr(h \mid H)=Pr(l \mid L)=q.$

I am now interested in how Bayesian inference on both state $H$ vs $L$ and state $ind$ vs $cor$ looks like conditional on receiving one and two signals respectively. My understanding would be the following:

Considering updating on $ind$ vs $cor$, the first observed signal does not contain any information. When observing two different signals, it is clear that the state is $ind$. When observing two equal signals, inference should be as follows:

$$ \frac{\Pr \{ind \mid (h,h)\}}{\Pr \{cor \mid (h,h)\}} = \frac{\Pr\{(h,h) \mid ind\}}{\Pr\{(h,h) \mid cor\}} \times \frac{\Pr\{ind\}}{\Pr\{cor\}} =\frac{pq^2 + (1-p)(1-q)^2}{pq + (1-p)(1-q)} \times \frac{\lambda}{1-\lambda}$$

$$ \frac{\Pr \{ind \mid (l,l)\}}{\Pr \{cor \mid (l,l)\}} = \frac{\Pr\{(l,l) \mid ind\}}{\Pr\{(l,l) \mid cor\}} \times \frac{\Pr\{ind\}}{\Pr\{cor\}}= \frac{p(1-q)^2 + (1-p)q^2}{p(1-q) + (1-p)q} \times \frac{\lambda}{1-\lambda} $$

Inference on $A$ vs $B$ for the first signal again is standard. However, I am having trouble figuring out what happens upon observing the second signal, since the agent somehow jointly learns about $ind$ vs $cor$ and $A$ vs $B$. She should update from her one-signal posterior in the following fashion, right?

$$ \frac{Pr(H \mid h,h)}{Pr(L \mid h,h)} = \frac{Pr(h,h \mid h;H)}{Pr(h,h \mid h;L)} \frac{Pr(H\mid h)}{Pr(L \mid h)} = \frac{\lambda q + 1-\lambda}{\lambda (1-q) + 1-\lambda} \frac{Pr(H\mid h)}{Pr(L \mid h)} $$

I would be forever grateful if someone could point me to some resources dealing with these kind of updating problems or could maybe help me out!

Thanks!!