Does the coin flipping game terminate? Imagine I were to toss a coin, so that


*

*if it is tails, I will toss again.

*if it is heads, I stop the game.


It is clear that if I would to play the game forever the probability of me tossing tails over and over again tends to $0$, because $\lim\limits_{n \rightarrow \infty}{\frac{1}{2^{n}}} = 0$ and therefore the game terminates. My question is if it would still be correct to say that the game terminates after a finite number of tosses, for the reason that after a finite number of tosses the probability still gets smaller than every $\epsilon>0$ (if i get it right).
This is just confusing me for a while now and I would appreciate a detailed explanation.
Thx in advance!
 A: You are correct that when the game terminates it does so in a finite number of steps. Now, is it possible that the game does not terminate? Sure, there is no reason to assume that you will always end up getting a Head. 
However, the probability of the game not terminating is the probability of an infinite sequence of Tails, i.e.... zero.
Such is life with probability. There can be events that are possible (not impossible) but still have zero probability (for continuous random variables, many values are possible but usually all have zero probability).
Note that in practice you cannot perform this experiment: whatever resources you may have (millions of people tossing billions of coins for many many years) you are not guaranteed to ever get a Head. The above therefore relates to a thought experiment that is allowed to go on "forever".
A: How about the followings?
The probability that the game will terminate after N tosses = 1 - probability that the game doesn't terminate after N tosses = 
$$ 1 - \frac{1}{2^{N}}$$
If you define $\epsilon$ be the upper bound of the probability,
$$ 1 - \frac{1}{2^{N}} < \epsilon$$
$$ 2^{N} < \frac{1}{1 - \epsilon}$$
$$ N < -\frac{\ln(1 - \epsilon)}{\ln2}$$
So if there are N tosses and $ N < -\frac{\ln(1 - \epsilon)}{\ln2}$, the probability of the game terminating is less than $\epsilon$. As $N \rightarrow \infty, \epsilon \rightarrow 1.$
