Say I have two coins and I toss them independently, what is the probability that I get two heads or two tails, assuming I discard all cases where I get HT or TH.
So let coin one be Bernoulli(p) and coin two be Bernoulli(q), then is it just: $\frac{pq}{pq + (1-p)(1-q)}$ ?
This does not seem right since it is too simple. Again I am in a game where I discard all HT and TH outcomes.
Yes. You have found a conditional probability.
If the first coin has probability $p$ of being heads and the second has probability $q$ of being heads and they are independent, then your expression $\dfrac{pq}{pq + (1-p)(1-q)}$ is the conditional probability that both are heads, given they are both head or both tails.
Conditional probability is as simple as $\mathbb{P}(A \mid B)=\dfrac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}=\dfrac{\mathbb{P}(A \cap B)}{\mathbb{P}(A \cap B) + \mathbb{P}(A^c \cap B)}$.