Z plays a game where independent flips of a coin are recorded until two heads in succession are encountered.
Z wins if 2 heads in succession occurs.
Z loses if after 5 flips, we have not encounter two heads in succession.
1) What is the probability that Z wins the game?
2) Suppose coin is fair. Z plays twice. What is the probability that both games have the same outcome?
The Attempt at a Solution
I get very confused by questions related to probability. :(
There are 5 flips at most so the sample space has 2^5 = 32 possible outcomes. When seen as 5 "slots", as long as two adjacent slots are filled with H (for "heads") then Z wins.
P(Z wins) = 1 - P(Z loses)
Let X = number of heads. Then, Z loses when X = 1. Z loses in some cases when X = 2,3.
X = 1 : There are 5 possible outcomes.
X = 2 : The two heads are either 1 slot apart (3 choose 1 = 3 outcomes) or 2 slots apart (2 choose 1 = 2 outcomes) or 3 slots apart (1 outcome)
X = 3 : The three heads need to be in slot 1,3,5 => 1 outcome.
So total outcome that results in a loss = 1 + 5 + 3 + 2 + 1 = 12.
Therefore, P(Z win) = 1 - 12/32 = 20/32 = 5/8.
Question : Is this method correct? Are there "cleaner" methods? I have no idea how to start part 2.