I am reading a pdf on pseudorandom function I found here https://www.cs.utexas.edu/~dwu4/courses/sp21/static/reductions.pdf
My problem/struggle is with the computation of the distinguisher's $B$ advantage.
According to the notes $b=0$ means that $B$ receives a sample from the function of interest, let's call it $F$, whereas $b=1$ means that they receive a sample from a truly random let's call it $f$.
Then the advantage is defined as:
$$
|\Pr[b'=1|b=0]-\Pr[b'=1|b=1]|
$$
The first probabillity $\Pr[b'=1|b=0]$ tells us that $B$ wrongly assumed that sample was from a truly random and the second one tells us that they correctly assumed that the sample was from a truly random.
Now it would make more sense to me if instead we computed the probabillity that they correctly assumed that the sample was from $F$ a.k.a $\Pr[b'=0|b=0]$ .
For example if we look at example 1 in the pdf : $G'(s)= G(s) ||s$ .
Now to my understanding : if $B$ receives $t=G(s)$ , they can think of it as $t= t_1||t_2$ and because they know the length of $s$ and $G(s)$ check if $t_2 = s \wedge G(s) = t_1$ .
So if $b=0$ then $\Pr[b'=0|b=0]=1$.
But in the paper instead $\Pr[b'=1|b=0]=1$ is said to be one. I am confused.