Coin Toss Probability Question (Feller) I'm working out of Feller's "Introduction to Probability and its Application (Vol I.)" textbook and I'm stuck on a coin toss problem. I'll list the full problem and show where I'm having trouble.

A coin is tossed until for the first time the same result appears twice in succession. To every possible outcome requiring n tosses attribute probability 1/$2^{n-1}$. Describe the sample space. Find the probability of the following events: a.) the experiment ends before the sixth toss, b.) an even number of tosses is required. 

Alright so I'm not having any trouble describing the sample space and completing part a. This first part was solved by creating a possibility tree and adding up the probabilities (answer: 15/16). However, I'm stuck on part b and I don't understand how the 1/$2^{n-1}$ given in the problem is to be interpreted because if you toss the coin twice it makes it seem like HH, and TT each have a probability of 1/2 which is not the case. The sample space of two tosses would be {HH, HT, TH, TT} and each would have a probability of 1/4 and following this logic I arrived at 15/16 so I believe this is the correct thinking which makes the problem even more confusing. 
The answer to part b is 2/3 so I'm not sure if that will help. Thanks for any help. 
 A: I think you have interpreted the question slightly incorrectly. The probability that the game ends immediately after the $n$ tosses is $\dfrac1{2^{n-1}}$ and this the probability that has been given in the problem.
The sample space is $$\Omega = \{\underbrace{AA}_{1/2},\underbrace{A\bar{A}\bar{A}}_{1/4},\underbrace{A\bar{A}AA}_{1/8},\underbrace{A\bar{A}A\bar{A}\bar{A}}_{1/16},\underbrace{A\bar{A}A\bar{A}AA}_{1/32},\ldots\}$$ where $\bar{A}$ denotes the outcome which is not $A$.
Hence, the probability that the experiment ends before the $6^{th}$ toss is $$\dfrac12 + \dfrac1{2^2} + \dfrac1{2^3} + \dfrac1{2^4} = \dfrac{15}{16}$$
For the second part, the probability that the game ends after even number of tosses is 
\begin{align}
\sum_{n=2,4,6}^{\infty} \dfrac1{2^{n-1}} & = \dfrac1{2} + \dfrac1{2^3} + \dfrac1{2^5} + \dfrac1{2^7} + \cdots = \dfrac12 \left( 1 + \dfrac14 + \dfrac1{4^2} + \dfrac1{4^3} + \cdots\right) = \dfrac12 \dfrac1{\left(1- \dfrac14 \right)}\\
& = \dfrac12 \times \dfrac1{3/4} = \dfrac4{2 \times 3} = \dfrac23.
\end{align}
A: You can rest assured that you are correct the probability of HH and TT are both 1/4. An odd sequence would be HTT or THH.  While the other possibilities for three HTH THT HHH and TTT. So in two out of 6 equally likely cases the sequence ends in an odd number of tosses and consequently 1/3 is the probability of an odd number of tosses and 2/3 is the probability of an even number.
A: For part b) you can use the law of conditional probability:
The chance of the game ending on the 2nd toss is 1/2.  Call this event 1.  Then $P(E_1) = 1/2$.  But there is a 1/2 chance that the game continues to the 3rd toss, in which case there is a 1/2 chance that the game continues to the 4th toss, where there is a 1/2 chance of the game ending. Call this event 2.  Then $P(E_2) = 1/8$.  Continue this process to obtain event 3, event 4, etc and calculate their probabilities.
Since these events are disjoint, we see the probability of the game ending in an even number of tosses is $\sum_{n=1}^{\infty}\dfrac{1}{2^{2n-1}} = \frac{2}{3}$
