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]

a) P{W>0} = 0.5

b) P{W<0} = 0.5

c) E[W] = 0

Step-by-step explanation:

Solution:

a) P(W > 0) is 0.5 because there is a 0.5 chance the player will win on their first gamble and stop with a net profit.

b) P(W = 0) is 0.25 because there is a 0.5 chance the player will lose on their first gamble, then also a 0.5 chance they will win on their second gamble and stop with a profit of 0.

All other combinations of gambles result in a net loss, so P(W < 0) = 1−P(W > 0)−P(W = 0)

c) The expected value can be obtained by summing the products of each profit and the probability of that profit. In this case, you have 1∗0.5+0∗0.25+(−1)∗0.125+(−2)∗0.0625, etc. So E(W) is the sum to infinity:

                               = (1 - n) / 2^( n + 1 )

                               = 0

answer: 0

step-by-step explanation: the positive, like 5, + its opposite, like -5 would be 0

5+(-5)=0, or 5-5=0