Game strategy- choose between two urns and two colors I found the following game in a (non-math) book:

We have two urns $A$ and $B$. $A$ contains black and red balls of equal amount. Whereas the ratio, $\frac{r}{s}$, of black and red balls of urn $B$ is unknown. You are free to choose an urn and then a color. Next the moderator draws a ball from the chosen urn. If it matches your color you win if not you loose.
Which urn should you choose?


I doubt that we can solve this game/find an optimal strategy without any further information.
If we choose urn $A$ then the probability to win, $P(A)$, can be found by law of total probability, i.e.
\begin{align*}
&P(A)=P(moderator~red\mid player~red)P(player~red)+P(moderator~black\mid player~black)P(player~black)\\
&=0.5\cdot P(player~red)+0.5\cdot P(player~black)=(P(player~red)+P(player~black))\cdot0.5=0.5
\end{align*}
If we choose urn $B$ and apply law of total probability, then
\begin{align*}
&P(B)=P(moderator~red\mid player~red)P(player~red)+P(moderator~black\mid player~black)P(player~black)\\
&=\frac{r}{r+s}\cdot P(player~red)+\frac{s}{r+s}\cdot P(player~black)=?
\end{align*}

However, the author states that it doesn't matter which urn we choose.
I suppose that he assumes that if the player chooses urn $B$ he decides with equal probability between both colors. This indeed yields $P(A)=P(B)=0.5$.
But why should we assume this? Or is there another reason or argument why both strategies should yield the same winning probability?
 A: Let $R_B=\frac{r}{b}$ be the ratio of red balls to black balls in urn B. It depends on the distribution of $ X=\frac{r}{r+b}$. Let's assume that $X=\frac{r}{r+b}= \frac{R_B}{1+R_B}$ is uniformly distributed on $[0,1]$.
After we choose an urn, if we  choose red with probability $p$ and black color with probability $1-p$, then
$$P(w|A) = P(w|A,r)P(r|A) + P(w|A,b)P(b|A) = \frac 12 p + \frac 12 (1-p) = \frac 12.$$
And for urn B we have,
$P(w|B) = P(w|B,r)P(r|B) + P(w|B,b)P(b|B)  = p\,P(w|B,r) + (1-p)\,P(w|B,b).$
Using the law of total probability and since $f_{X}(x)=1$ for urn B:
$P(w|B,r) = \int_{0}^{1} P(w|B,r, X=x) f_X(x)\,dx = \int_{0}^{1} P(w|B,r,X=x)\,  dx = \int_{0}^{1} x\,  dx = \frac 12.$
$P(w|B,b) = \int_{0}^{1} P(w|B,b, X=x) f_X(x)\,dx = \int_{0}^{1} P(w|B,b,X=x)\,  dx = \int_{0}^{1} 1-x \, dx = \frac 12.$
so,
$$P(w|B) = p\,\frac 12 +(1-p) \frac 12= \frac 12.$$
It seems it doesn't depend on $p$, i.e., if $X=\frac{r}{r+b}$ is uniformly distributed then both probabilities are the same and it doesn't matter which urn to choose. It only depends on the distribution of $X=\frac{r}{r+b}$.
UPDATED:
Case 2: Again let $f_X(x)$ be density distribution of $X=\frac{r}{r+b}$, but this time it is unknown. Then,
$P(w|B,r) = \int_{0}^{1} P(w|B,r, X=x) f_X(x)\,dx = \int_{0}^{1} x\,f_X(x)\,  dx $
$P(w|B,b) = \int_{0}^{1} P(w|B,b, X=x) f_X(x)\,dx = \int_{0}^{1} (1-x)\,f_X(x)\,  dx = 1 - \int_{0}^{1} x\,f_X(x)\,  dx = 1 - P(w|B,r).$
Now if $p=\frac 12$, (recall that $p$ is the probability of choosing the red color) then
$$P(w|B) = \frac 12 P(w|B,r) + \frac 12 (1-P(w|B,r)) = \frac 12. $$
which again shows it doesn't matter which urn to choose if $p=\frac 12$ but $X=\frac{r}{r+b}$ can have any distribution.
