<s>[INST] Probability for repeated incidents . I had problems on understanding the following problem :

A die is continuously thrown until 4 is obtained.  What is the probability that the dice is thrown an even number of times?

Here , a solution states that probability is :
$$\frac{4}{6}\times \frac{2}{6}+\frac{4}{6}\times \frac{4}{6}\times \frac{4}{6}\times \frac{2}{6} + ...$$
$$=\frac{2}{5}$$ 
I didn't understand why such process was used and what's the the concept behind it?
Another problem :

(ii) If a biased coin is tossed two times  repeatedly , then what is the probability of occurring HH?

Can I do the problem (ii) similarly like (i) as :
$$\frac{1}{4}+\frac{3}{4}\times \frac{1}{4}+\frac{3}{4}\times \frac{3}{4}\times \frac{1}{4}+ ...$$
 [/INST] *

*See, that the probabilty of throwing $4$ in $k$-th throw ($k\geq 1$)is equal to:
$$p(k)=\frac{1}{6}\left(\frac{5}{6}\right)^{k-1}$$
So we have to sum up all the probabilities for even $k$:
$$P=\sum_{n=1}^{\infty}p(2n)=\sum_{n=1}^{\infty}\frac{5}{6}\left(\frac{5}{36}\right)^{2n-2}=\sum_{n=0}^{\infty}\frac{5}{36}\left(\frac{25}{36}\right)^{n}=\frac{5}{36}\frac{1}{1-\frac{25}{36}}=\frac{5}{11}$$

*Tossing a coin two times makes the probability of $HH$ equal to $p(H)^2$.
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