Answer:
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