Background
So let us first familiarise ourselves with the following trick:
Let us assume that the problem $P_1$ that there are $4$ red balls and $6$ blue balls in a box. The probability of it being red is
$$ p_1(R) = .4 $$
And the probability of it being blue is:
$$ p_1(B) = .6 $$
Now consider the problem $P_{1+2}$ there is another box and one takes the previous ball from $P_1$ and puts it in this box. This box has $3$ red balls and $4$ blues balls. Now, the probability of it being red is:
$$ p_{1+2}(R) = \frac{3+ .4}{7 +1}$$
and the probability of it being blue is:
$$ p_{1+2}(B) = \frac{4+ .6}{7 +1}$$
Note: the $1$ in the denominator comes from $1=.6 + .4$
Hence, for a problem $P_{1+2+3+\dots+n}$
$$p_{1+ 2 + 3 + \dots + n} (R) = \frac{\text{# of red balls in P(1+2+..+n)} + p_{1+2 +\dots+ n-1}}{\text{# of balls in P(1+2+..+n)} +1 }$$
One can nest this using:
$$p_{1+ 2 + 3 + \dots + n-1} (R) = \frac{\text{# of red balls in P(1+2+..+n-1)} + p_{1+2 +\dots+ n-2}}{\text{# of balls in P(1+2+..+n-1)} +1 }$$
Question
My question is can one construct a problem $\tilde P$ where the probabilities associated it being red $\tilde p(R) = I $ and the probabilities associated with it being blue are $\tilde p(B) = 1 -I$ where I is an irrational number.
Note: since we are using starting with rational numbers it would take an infinite series to produce an irrational number?
Where $\tilde P$ one construct an expression to reach the probabilities related to $\tilde P = P_{1+2 + 3 + \dots}$. Does the trick method have a faster rate of convergence (as a measure of the time taken) as opposed methods used by the usual method?
The usual method would include any other method which will always work (for example one can merely always guess the probability is $I$ and get the answer right this time) as opposed to a continued fraction (/trick) method? Does this method beat it in the finite case as well? Does one have a method to compute things like $\tilde p(R)^2 = \text{rational fraction}$ and then squareroot it? In which case what happens if the irrational probability is $\frac{1}{\pi}$
In 1987, the author C. Badea proved the following criterion of irrationality:
For an infinite convergent series $\sum_{k=1}^{\infty}{\frac{b_k}{a_k}}$ of positive rational numbers to have an irrational sum, $(a_k)$ and $(b_k)$, $k \geq 1$, must be two sequences of positive integers such that...
$$a_{k+1} \gt \frac{b_{k+1}}{b_k}a_k^2 - \frac{b_{k+1}}{b_k}a_k + 1$$
In your example, you have...
$$P\left[R_n\right] = \sum_{k=1}^{2^{n-1}}{P\left[R_n\vert E_k\right]P\left[E_k\right]}$$
Where $n$ is the number of urns and the $E_k$s are permutation paths. E.g.,...
$$P\left[R_3\right] = P\left[R_3\vert E_1\right]P\left[E_1\right] + P\left[R_3\vert E_2\right]P\left[E_2\right] + \ldots + P\left[R_3\vert E_4\right]P\left[E_4\right]$$
$$= P[R_3|B_1B_2]P[B_1B_2] + P[R_3|B_1R_2]P[B_1R_2] + \ldots + P[R_3|R_1R_2]P[R_1R_2]$$
This can obviously be reduced to a single integer ratio $\frac{b_k}{a_k}$, i.e., it would be possible to choose marble ratios s.t. the first relation is satisfied when $n$ is taken to approach infinity.