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?
Best Answer
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.