In a Metropolis-Hastings algorithm for a MCMC one has the value $r$ as:
$$r(x, y) = \min \left(\frac{f(y)}{f(x)} \frac{q(x\mid y)}{q(y\mid x)} , 1\right) $$
where $f$ is the density and $q(x\mid y)$ the proposal from $y$ to $x$. Then, from a $X_i$ value one has $Y$ with probability $r$ and $X_i$ with probability $1 - r$.
Ok. But what I don't understand is why the quotient $\frac{q(x\mid y)}{q(y\mid x)}$ is not $\frac{q(y\mid x)}{q(x\mid y)}$. For me, it has more sense. Imagine that $f(y) > f(x)$ and $q(y \mid x) > q(x \mid y)$, then it seems that $Y$ is a value that we want our algorithm to visit, and indeed it should be easy to go from $x$ to $y$ since $q(y \mid x)$ has a big value.
However, if $q( y\mid x)$ is in the denominator, then the overall quotient with $f$ diminishes, perhaps below 1 and the probability of selecting $Y$ will be below 1. If $q(x\mid y)$ is in the numerator that probability would be 1, so a transition to our desired value of 1 will be very a sure thing.
Thanks.
The proposal distribution gives unwanted bias. As a silly example, take $f$ to be the uniform distribution on a finite set, and the proposal picks from some non-uniform distribution (independent of $X_i$). We want to sample from $f,$ so we need to compensate in some way for the bias introduced by $q.$ You said that if $q(y|x)>q(x|y)$ then we want our algorithm to visit $y$ more. That's not the case - we want our algorithm to visit $y$ with probability $f(y),$ with no effect from $q.$
The formal explanation is pretty quick; $r$ is chosen to satisfy the detailed balance equation $$f(x)p(x,y)=f(y)p(y,x)$$ where $p(x,y)$ is the transition probability from $x$ to $y,$ which is $$p(x,y)=q(y|x)r(x,y).$$ This gives $$f(x)q(y|x)r(x,y)=f(y)q(x|y)r(y,x)$$ and we just pick the largest possible values for $r(x,y)$ and $r(y,x).$
The detailed balance equation ensures that $f$ is a stationary distribution, so if the chain mixes then it converges to $f.$