Consider a biased coin with probability $p$ to show head. Then you toss it twice and consider the result $(H,T)$ as a "head" and the outcome $(T,H)$ as a "tail". If it does not occur neither "head" or "tail", you toss it twice again until you get a "head" or a "tail".
(a) Prove that this experiment corresponds to the unbiased version of the game of "tossing a coin until get head (or tail).
(b) Exhibit the probability mass function and the expectation of the number of tosses necessary to end the experiment.
MY ATTEMPT
(a) Based on the independence of the tosses, we have $$\textbf{P}(\{(H,T)\}) = p(1-p) = (1-p)p = \textbf{P}(\{(T,H)\})$$
(b) As far as I have noticed, such random variable has a geometric distribution:
$$X\sim G(p(1-p)) \Rightarrow f_{X}(x) = q(1-q)^{x-1}\,\,\text{for}\,\,x\geq 1$$
where $q = p(1-p)$. Thence we conclude that $\textbf{E}(X) = [p(1-p)]^{-1}$.
Am I reasoning correctly? I'd like to hear somebody else's opinion on the subject. Thanks!
For the second part:
It is geometric distribution with $p_1 = 2p(1-p)$ and $q_1 = p^2+(1-p)^2$ and is $Ge(p_1)$
Thus $P(X=x) = q_1^{x-1}p_1$, $x \ge 1$ where X is the number of trials(double tosses)
And the expectation is $E(\text{No of tosses}(2X)) = \frac{2}{p_1}$