How does probability change the more times you perform a procedure? I have a question based on fair coins. Every round, two coins are flipped. If both are heads, we say "Success" and end the experiment. What is the probability of saying "Success" in any round i?
Initially, I thought it would be $$\displaystyle\dfrac{\displaystyle\frac{i!}{(i-2)!2!}}{2^i},$$ because of NCR and increasing probability, but I realize it doesn't work.
Right now, it seems as though a summation works as we're increasing chances of getting a "Success" the more rounds we have. We also need to stop once it reaches a "Success", but I don't get how we can stop and operation like that unless it simply reaches 100%. $$\sum_{n=1}^{i} \frac{1}{4}$$
Logically, it sounds right to me, but I'm unsure so I'd like to hear your thoughts on this. The $\displaystyle\frac{1}{4}$ is due to the [TT][TH][HT][HH] probability sample space.

The question continues to ask for an expression on the probability of getting a "Success" in terms on n and i. Here, n stands for the total number of rounds while i stands for the number of rounds before we stop running the procedure.
My solution for this part seems the same as the previous one, as I don't see what the question is asking differently. Any ideas?

And finally, the last part is asking what if we don’t stop after saying “Success", that is we repeat the procedure n times. How many times do you expect to say “Success”, express this in terms of n.
$$\sum_{i=1}^{n} \frac{1}{4}$$
This way, we can easily go over the limit and we'll find the total number of times that "Success" can be reached.
Thank you for reading this :)
 A: First question
The probability of having success in round 1 is $P(X=1)=\frac{1}{4}$. We agree.
The probability of having success in round 2 is $P(X=2)=\left(1- \frac{1}{4}\right)\cdot \frac{1}{4}=\frac{3}{4}\cdot \frac{1}{4}=\frac{3}{16}$
You have to take into acccount that the probability of having a success in round $2$ depends on wether you have success in round 1 or not. Here we need to have no success in round 1.
The probability of having success in round 3 is $P(X=3)=\left(1- \frac{1}{4}\right)^2\cdot \frac{1}{4}=\frac{3^2}{4^2}\cdot \frac{1}{4}=\frac{9}{64}$
Here we need to have no success in round 1 and round 2.
$...$
The probability of having success in round $i$ is $P(X=i)=\left(1- \frac{1}{4}\right)^{i-1}\cdot \frac{1}{4}=\left(\frac{3}{4}\right)^{i-1}\cdot \frac{1}{4}$

Second question
You need the closed form of $\frac{1}{4}\cdot \sum_{i=1}^n \left(\frac{3}{4}\right)^{i-1}$ For $j=i-1$ we get $\frac{1}{4}\cdot \sum_{j=0}^{n-1} \left(\frac{3}{4}\right)^{j}$. This is the partial sum of a geometric series.
Thus $\frac{1}{4}\cdot \sum_{j=0}^{n-1} \left(\frac{3}{4}\right)^{j}=\frac{1}{4}\cdot \frac{1-\left(\frac{3}{4}\right)^{n}}{\frac{1}{4}}=1-\left(\frac{3}{4}\right)^{n}$

Third question
I´m not sure if my understanding is right. But if we don´t stop at the first success the probability of getting $i$ successes and consequently $n-i$ failures is 
$P(Y=i)=\binom{n}{i}\cdot \left( \frac{1}{4} \right)^i \cdot \left( \frac{3}{4} \right)^{n-i}$.
This is the pdf of the binomial distribution. And the expexted value is $n\cdot p=\frac{1}{4}n$
