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May 25th, 2017, 06:05 PM  #1 
Member Joined: Apr 2017 From: PA Posts: 34 Thanks: 0  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? 
June 16th, 2017, 11:35 AM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 849 Thanks: 307 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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