Question on Simplest Oblivious Transfer Protocol In the paper The Simplest Protocol for Oblivious Transfer, in the scenario where the receiver chooses $c=1$, then $B$ is calculated as $Ag^b$.

Then $k_1$ is calculated as $H( (B/A)^a)$.

That means that the receiver can calculate the same encryption key, and decrypt message $M_1$ (because it's encrypted with $k_1$.

Does this imply that $H( (B/A)^a ) = H( Ag^b)$ ?

I don't understand how it works in this scenario (receiver choosing $c=1$)
As you have said, let's say the receiver chooses $c = 1$. Then his $B = Ag^b$. So In Alice's case, the $k_0 = H(B^a) = H((Ag^b)^a) = H(g^{a a} g^{a  b}) = H(g^{a a + a b})$. Similarly, $k_1 = H((\frac{B}{A})^a) = H(g^{ab})$. Receiver's own key $k_R$ is also, $H(A^b) = H(g^{ab})$. So you see, his key is similar to the senders key $k_1$. That is why, even though the sender is sending both encryptions, the receiver can only decrypt one of these. The one for which the keys are the same, i.e. message 1!