Is there an optimal scenario?

Let us find the probability $p$ of winning the game by deconditioning on the first picked card being the $k$ -th blue card. Then we have that $\begin{aligned} p &\stackrel{(i)}{=} \sum_{k=1}^n \frac{1}{n+m}\frac{\binom{n-1}{k-1}}{\binom{n+m-1}{k-1}} \\ &= \sum_{k=1}^n \frac{1}{n}\frac{\binom{n+m-k}{m}}{\binom{n+m}{m}} \\ &\stackrel{(ii)}{=} \frac{1}{n\binom{n+m}{m}} \sum_{i= 0}^{n-1} \binom{m+i}{m} \\ &\stackrel{(iii)}{=} \frac{1}{n\binom{n+m}{m}} \sum_{i= 0}^{n-1} \left[ \binom{m+i+1}{m+1} - \binom{m+i}{m+1} \right] \\ &\stackrel{(iv)}{=} \frac{\binom{n+m}{m+1}}{n\binom{n+m}{m}} = \displaystyle\frac{1}{m+1}. \end{aligned}$ Step $(i)$ comes from the fact that the probability to take exactly the $k$ -th blue card is $1/(m+n)$ and then, we need to take $k-1$ blue cards from the $n-1$ remaining. Step $(ii)$ is the change of variable $i=n-k$ . In $(iii)$ we used Pascal's rule . Finally, in step $(iv)$ we have a telescoping series . Notice that $p$ does not depend on the number of blue cards! Finally, the expected return is simply $E = p\,m + (1-p)\,(-1) = 0.$ Surprisingly, the expected return does not depend on $n,m$ . Therefore, the expected return is always zero!