jhon13

05-23-2012, 11:20 AM

Studying for a stats final, and came across this problem that I'm not quite sure what to do, any help is appreciated.

Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win. Find:

a) P(W > 0)

b) P(W < 0)

c) E(W)

For part (a), since the gambler only needs net win greater than 0, there is only one way that will happen is when he wins the first time. Hence, there P(W>0) = 1/2

For part (b), since the gambler can lose infinite times, and have just one win. Thus, the gambler can lose once and win once which gives probability of 1/4 (net winning = -1). But the gambler can lose twice and win once (net winnings = -2), which gives a probability of 1/8, and so on and so forth. So I would have, 1/4 + 1/8 + 1/16 + .... = 1/2. However, the textbook says it is 1/4. That is where I am confused.

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Then for part (c), how do I account for the infinite combination of negative winnings?

Gambles are independent, and each one results in the player being equally likely to win or lose 1 unit. Let W denote the net winnings of a gambler whose strategy is to stop gambling immediately after his first win. Find:

a) P(W > 0)

b) P(W < 0)

c) E(W)

For part (a), since the gambler only needs net win greater than 0, there is only one way that will happen is when he wins the first time. Hence, there P(W>0) = 1/2

For part (b), since the gambler can lose infinite times, and have just one win. Thus, the gambler can lose once and win once which gives probability of 1/4 (net winning = -1). But the gambler can lose twice and win once (net winnings = -2), which gives a probability of 1/8, and so on and so forth. So I would have, 1/4 + 1/8 + 1/16 + .... = 1/2. However, the textbook says it is 1/4. That is where I am confused.

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Then for part (c), how do I account for the infinite combination of negative winnings?