How to solve infinite possible coin problem? I would need some help with this question:
You toss n coins, each showing heads with probability p, independently
of the other tosses. Each coin that shows tails is tossed again. Let X be the
total number of heads. What is the probability distribution?
The way I tried to solve it is by assuming I am interested in getting 2 heads. In my opinion that would express itself then to:
\begin{equation}
P(X=2) = \binom{\infty}{2} p^2 (1-p)^{\infty - 2}
\end{equation}
The reason I chose infinity as the number of trials is because there is a chance I will never get two heads. But this seems not to be the way to solve this. Any help is much appreciated!
 A: This question is old but unanswered, so this is for those who may be looking for an answer.
The way the event is defined in the question may lead to wrong interpretations. 
• Your interpretation of the event is : You toss the n coins infinitely many times until all coins show heads. 
• The actual event is : You toss the n coins. The coins that show tail are thrown again. Then, the heads are counted. It is not suggested that you repeat the process infinitely many times, you just do this once.
This results in three possible events given by Ω={TT,TH,H}, as shown in this event tree diagram. The subset of successful events is S={TH,H} as the success event is defined by head. The probability of success therefore is P(S)=p+(1-p)p.
The probability of failure is 1-P(S). Clearly, the number of final heads follows the Binomial distribution Bin(n,p+(1-p)p).
A: Assume each coin is tossed twice and counts as a success if and only if it turns up heads on at least one of the tosses. So $p=\frac{3}{4}$, and the distribution is binomial, with mean $np=\frac{3}{4}n$ and variance $np(1-p)=\frac{3}{16}n$. 
