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May 25th, 2017, 06:05 PM   #1
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Find the expected value

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 uegame.

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?
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June 16th, 2017, 11:35 AM   #2
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Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics
Let's list out the possible outcomes, the gain from each, and the probability of each:

G: 1, 2/5
SGG: 1, 3/5 * 1/2 * 1/3 = 1/10
SGSG: 0, 3/5 * 1/2 * 2/3 * 1/2 = 1/10
SGSSG: -1, 3/5 * 1/2 * 2/3 * 1/2 = 1/10
SSGG: 0, 3/5 * 1/2 * 2/3 * 1/2 = 1/10
SSGSG: -1, 3/5 * 1/2 * 2/3 * 1/2 = 1/10
SSSGG: -1, 3/5 * 1/2 * 1/3 = 1/10

So the expected value is 1 * (2/5 + 1/10) - 1 * (1/10*3) = 0.2 > 0.
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