Show that, if you draw until you are ahead by 1 dollar or until there are no more gold balls, this is a favorable game. A box contains two gold balls and three silver balls. You are allowed to choose
successively balls from the box at random. You win 1 dollar each time you
draw a gold ball and lose 1 dollar each time you draw a silver ball. After a
draw, the ball is not replaced. Show that, if you draw until you are ahead by
1 dollar or until there are no more gold balls, this is a favorable game.
for this, I thought about summing up 1 (2/5)-1 (3/2). Then I took a scenario of if I already picked a gold ball. So by then I would sum up 1 (1/4)-(3/4). Then, I took anther scenario that we already took a silver ball, so we would have 1 (1/2)-1 (1/2). Then, I took another secenario if I took out 2 silver balls to sum up 1 (2/3)-(1/3). How would to find the correct expected value?
 A: You might want to list all the possible scenarios that could arise:

G
  SGG
  SGSG
  $\vdots$
  (G=Gold, S=Silver)

There should be seven possible sequences. For each one, calculate the payoff and the probability of drawing balls in that order. For instance, for the sequence SGSG, the payoff is $-1+1-1+1=0$, and the probability is $\frac35\frac24\frac23\frac12=\frac1{10}$.
Finally, sum up the payoffs scaled by the corresponding probabilities.
A: With probability $\frac25$, you draw a gold ball first, are up by a dollar, and quit.
Otherwise, you can never be up by a dollar unless you’ve drawn all the gold balls, so you draw until you’ve drawn both gold balls. The expected winnings will be $s-1$, where $s$ is the expected number of silver balls at the end of a random one of the $24$ arrangements of the letters in $GgSs$. There are no silver balls at the end of arrangements ending in $G$ or $g$ (half of them, or $12$), in case of which your winnings are $-\$1$; there is one silver ball at the end of arrangements ending in $GS$, $Gs$, $gS$, or $gs$ (two arrangments for each ending, so $8$ or one-third of them), where you win $\$0$; and there are no silver balls at the end of arrangements ending in $Gg$ or $Gg$ ($2+2$ or one-sixth of them), where you win $\$1$. So the expected value in this “otherwise” situation, where you drew a silver ball first and must continue until drawing all the gold balls, is $\frac12(-1)+\frac13(0)+\frac16(1)=-\frac13$ dollars.
The expected winnings of the game are therefore $\frac25\cdot1-\frac35\cdot\frac13=\frac15$ dollars.
