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?
Best Answer
Implicit in the problem statement is the assumption that the two colors red and black are on equal footing. The description is certainly invariant under the switching of the words "red" and "black" and informal notions of probability tends to consider such symmetries to preserve probabilities. Slightly more formally, consider a second participant in the experiment with the same role as you and who uses exactly the same strategy as you except switches the two colors. Given the information in the problem and an Occam's razor point of view, you would not expect them to perform better or worse than you. From this it is not hard to see that if there is a well defined probability of success with the mystery urn it has to be $1/2$.
On the other hand, consider a situation where you don't know how many red balls there are in urn $B$, but you do know that it has a million more black balls. Then the ratio $r/s$ is still unknown but it's clearly better to pick urn $B$ and black.
In other words, a reasonable restriction on the prior distribution of $r/s$ is that it should have all the symmetries that the problem description does. In this case that implies that the probability of success given urn $B$ is also $1/2$. In the absence of such a restriction, the choice of urn can matter.