Best game strategy in draws from a bin A bin has 2 white balls and 3 black balls. You play a game as follows: you draw balls one at a time without replacement. Every time you draw a white ball , you win a dollar, but every time you draw a black ball , you loose a dollar . You can stop the game at any time.Devise a strategy for playing this game which results in an expected profit.
According to my reasoning the best strategy is not to play the game at all : the expected value at every extraction remains the same and it's always negative $$E[X]=\left(\frac{2}{5}\right)*1 +\left(\frac{3}{5}\right)*(-1) =-\frac{1}{5}$$ So since I can treat it as a sum of expectations of random variables ,the best strategy is not to play…so the best expected profit is zero dollars right?

There may be a more elegant and/or general way to do this, but here's a brute-force approach.
Consider the following strategy: Draw balls until you've drawn more white balls than black, or if that's no longer possible, until you've drawn both white balls.
Under this strategy, only the following outcomes are possible:
$\begin{align}\\[1ex]
\text{W:} \quad & \text{win \$1} && p=\dfrac{2}{5}\\[1ex]
\text{BWW:} \quad & \text{win \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{10}\\[6ex]
\text{BWBW:} \quad & \text{break even} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}=\dfrac{1}{10}\\[1ex]
\text{BBWW:} \quad & \text{break even} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}=\dfrac{1}{10}\\[6ex]
\text{BWBBW:} \quad & \text{lose \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}\cdot1=\dfrac{1}{10}\\[1ex]
\text{BBWBW:} \quad & \text{lose \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{1}{2}\cdot1=\dfrac{1}{10}\\[1ex]
\text{BBBWW:} \quad & \text{lose \$1} && p= \dfrac{3}{5}\cdot\dfrac{1}{2}\cdot\dfrac{1}{3}\cdot1\cdot1=\dfrac{1}{10}\\[1ex]
\\
\end{align}$
Then your expected winnings are
$\left( \dfrac{2}{5} + \dfrac{1}{10} \right) * (1) + \left( \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10}\right) * (-1) = \boxed{\dfrac{1}{5}}$
