You are playing with a friend:
You are tossing a (fair) coin. If it is a tail, you win. If not, then you are tossing 2 (fair) coins. If they are both tails, you win. If not, you are tossing 3 (fair) coins, if they are all tails you win. If not, you are tossing 4 (fair) coins and so on...
What is the probability for you to win??? Is it a rational number?
I came to this formula:
$1-\Pi_{n=1}^{\infty}(1-(\frac{1}{2})^n)$
The probability of losing the game is the limiting probability that a random binary $n \times n$ matrix is regular (over $GF(2)$): the probability that row $i$ is not in the span of the first $i-1$ rows is $1 - 2^{i-1}/2^n = 1-2^{n-i+1}$, so the probability that the matrix is regular is $(1-2^{-n}) (1-2^{1-n}) \cdots (1-2^{-1})$, which tends to the complement of your number.
Curiously, there is also a series representation of this number: $$ \prod_{n=1}^\infty (1-2^{-n}) = \frac{1}{2^2-1} - \frac{1}{(2^2-1)(2^3-1)} + \frac{1}{(2^2-1)(2^3-1)(2^4-1)} - \cdots. $$ This is a special case of a more general identity of the q-Pochhammer symbol.
Using this series representation, we can prove that the number isn't rational. Denoting it by $\alpha$, let us suppose that $\alpha = p/q$. Multiplying the right-hand side by $M =(2^2-1) (2^3-1) \cdots (2^n-1)$, we obtain that for some integer $z$, $$ M\frac{p}{q} = z \pm \frac{1}{2^{n+1}-1} \mp \frac{1}{(2^{n+1}-1)(2^{n+2}-1)} \pm \cdots. $$ On the right we have an alternating series whose terms decrease in absolute magnitude, and so $$ 0 < \left|M\frac{p}{q} - z\right| < \frac{1}{2^{n+1}-1}. $$ The central term is a multiple of $1/q$, and in particular, since it is positive, it is at least $1/q$. For large enough $n$, we obtain a contradiction.
1st try, Probability is $\frac{1}{2}$.
You fail and in the second try, you succeed with the probability $\frac{1}{2}\frac{1}{4}$.
You fail again and in the third try, you succeed with the probability $\frac{1}{2}\frac{3}{4}\frac{1}{8}$ and goes on to infinity.
Thus the required probability $\frac{1}{2}+\frac{1}{2}\frac{1}{4}+\frac{1}{2}\frac{3}{4}\frac{1}{8}+....$
Thus the required probability is $$\sum_{k=1}^{\infty}\frac{\left(\prod_{i=0}^{k-1}(2^i-1)\right)}{2^{\frac{k(k+1)}{2}}} = 0.711212$$
More in general: Let $a$ be the coin tail probability. Let $P_n$ be the probability of winning in $n$ or less tries. Then we have the recursion
$$P_{n} = P_{n-1} + (1-P_{n-1})a^n \tag1$$
with $P_0=0$. Letting $Q_n=1-P_n$ this turns into
$$Q_{n} = Q_{n-1}(1 - a^n) \tag2$$
with $Q_0=1$. Then $Q_1 = (1-a)$, $Q_2 = (1-a)(1-a^2)$ and
$$Q_{\infty} = \prod_{n=1}^\infty (1 - a^n) = (a;a)_\infty = \phi(a) \tag3$$
where $(a;a)_\infty$ is the q-Pochhammer_symbol (or alternatively, $\phi(a)$ is Euler's function).
The probability of winning the game is then
$$p=1-Q_{\infty}=1-\left(\frac12;\frac12\right)_\infty = 1-\phi\left(\frac12\right)=1-0.288788095=0.711211905$$
So, yes, your formula is right. I would bet that it's irrational, but don't ask me to prove that...
Let us assume you stop after $L$ times (you could let it go to infinity afterwards). That is, you got a tree with $L$ branches. The probability that you win could be written as
$$\Pr(\textsf{win})= \underbrace{p_1}_{\textsf{win }1^{st}\textsf{time}} + \Big(\underbrace{1-p_1}_{\textsf{not win }1^{st}\textsf{time}} \Big) \Big( \underbrace{p_2}_{\textsf{win }2^{nd}\textsf{time}} + (\underbrace{1-p_2}_{\textsf{not win }2^{nd}\textsf{time}})(\ldots) \Big)$$ where $p_k = \Pr(T_1\ldots T_k)$ is the probability of obtaining all tails at the $k^{th}$ toss. In other words, it is the probability that you win at the $k^{th}$ toss, given that you have not on the $(k-1)^{th}$ toss. It turns out, since all coins are "fair", that $$p_k = \frac{1}{2^k}$$ Hence using $p_ip_j = p_{i+j}$, we can observe
For $L = 1$: $$\Pr(\textsf{win}) = p_1 = \frac{1}{2}$$ For $L = 2 $: $$\Pr(\textsf{win}) = p_1 + p_2(1-p_1) = p_1+p_2 - p_1p_2 = p_1 + p_2 - p_3$$ For $L = 3 $: $$\Pr(\textsf{win}) =p_1 +p_2 -p_4 - p_5 + p_6 $$ $$\vdots$$
