Let there be N balls all together and Np of them are blue. The probability of selecting a blue ball is thus p, and selecting a yellow (1-p).
Given your setup the probability of selecting the correct color ball is $1/2p + 1/2(1-p) = 1/2, ~~ \forall p ~\ni~ 0 \leq p \leq 1$
At the extremes of p you have 50% chance of either making a certainly correct choice or a certainly incorrect choice.
Your original logic is correct.