Expected value problem, balls in a box 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.
I'm not sure how to go about doing this problem. Do I only have to find the expected value for the conditions given, or do I have to figure out all possible outcomes and there probabilities to show this?
 A: The question does not ask me to find the value of the game, merely to show that it's favorable. Being very lazy, I will do the minimum required.
As usual with such problems, it's convenient to consider a space of 5! equally likely outcomes, namely, all possible orderings of the 5 balls. (This means that I regard all balls as distinguishable, and I pretend that I keep drawing "for fun" after the game is officially over.)
Plainly, in any play of the game, I win a dollar, lose a dollar or break even. I just have to show that there are more winning than losing outcomes.
Clearly, any losing outcome ends with me drawing a gold ball on the fifth turn. Hence, the reversal of any losing outcome is a winning outcome, with me drawing gold on the first turn. This already shows that there are at least as many winning as losing outcomes. 
To show that the game is favorable, all I have to do is find a winning outcome which is not the reversal of a losing outcome, e.g., an outcome of the form GSSSG whose reversal is also a winning outcome.
A: There are ${5\choose 2}$ ways to put the balls in a random order.
For each one, decide how many dollars you will win or lose if the balls come out in that order.
A: You are given a specific strategy and asked to find its expected value.  The problem is small enough that drawing the tree is not difficult.  If your first draw is gold (what probability) you win a dollar and quit, so the tree stops.  If your first draw is silver, you lose a dollar and draw again.  Find the probability and value of each leaf on the tree and add them up.
A: I would approach it with a decision tree -- it doesn't get very big, because some branches terminate early.
On the first turn, there's a probability of 0.4 that you draw a gold ball and go home $1 richer.
There's a 0.6 chance of getting to the second turn by drawing silver on the first, in which case there are now 2 silver and 2 gold and you are 1 down. Keep going, drawing it as a tree diagram, writing the probabilities on each branch and the number of gold and silver balls in each node, along with your winnings/losses so far if you get to that node. I took a few short-cuts (doubling up on nodes and not bothering to draw them when the outcome was inevitable), and ended up with just 10 nodes: quite manageable. And I got the answer that the probability of winning 1 is 0.5, the probability of breaking even is 0.2 and the probability of losing 1 is 0.3, giving an expected value of 0.20 (but check my calculations!)
A: Building on @Michael's solution:
These are the permutations where you win 1 (of course you'd stop drawing as soon as you are winning):
GGSSS
GSGSS
GSSGS
GSSSG
SGGSS
These get you 0:
SGSGS
SSGGS
And these get you -1:
SGSSG
SSGSG
SSSGG
Adding all these up give you an expectation of 2/10.
