Infinite coin tosses

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Let $X = \{ 1, 2, \ldots, n, 2', 3', \ldots, n'\}$ be a set of $2n-1$ elements. Consider a coin with $2n-1$ sides, where each side of the coin corresponds to an element in $X$, and we get each side with equal probability. Consider throwing the coin infinitely many times. Each such experiment induces an outcome which is an infinite series: $s_1, s_2, s_3, \ldots$, where for all $i$, $s_i\in X$. For a finite set $A$, we denote by $A^{\omega}$ the set of infinite sequences of elements from $A$.

Consider the following sets $\alpha_1 = \{1\}, \alpha_2 = \{ 2, 2'\}$, $\alpha_3 = \{ 3, 3'\}$, ..., $\alpha_n = \{ n, n'\}$

Consider the following event (set of outcomes): $$B = \{ s= s_1, s_2, s_3, \ldots: \text{for every $j\in \{ 1, 2, \ldots, n\}$ there is an element in $\alpha_j$ that appears in $s$ infinitely often } \}$$

That is, the outcome $s = s_1, s_2, \ldots$ is in $B$ iff for all $j\in \{1, 2, \ldots, n\}$, there are infinitely many $i$'s with $s_i = m$, for some $m\in \alpha_j$.

Now consider the following sets $\beta_1 = \{1\}, \beta_2 = \{ 2'\}$, $\beta_3 = \{ 3'\}$, ..., $\beta_n = \{n'\}$ and consider following event $$ A = \{ s= s_1, s_2, s_3, \ldots: \text{for every $j\in \{ 1, 2, \ldots, n\}$ there is an element in $\beta_j$ that appears in $s$ infinitely often }\}$$

What is the probability of $A$ conditioned on $B$: $Pr[A|B]$? My goal is to bound $Pr[A|B]$ from below by a positive fraction. My guess is that $Pr[A|B] \geq \frac{n}{2n-1}$ since this is the fraction of numbers in that we require to appear infinitely often, but I'm not sure. How can I prove that formally?

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The probability that all of them occur infinitely often is $1$. Suppose some number $k$ appears only finitely many times. That is, for some $N$ it never occurs after toss $N$. The probability that $k$ does not occur on any given toss is $\frac{d-1}d$, and the probability that it doesn't occur on the $m$ tosses starting with toss $N$ is $\left(\frac{d-1}d\right)^m\to0$ as $m\to\infty$.

The conditional probability is also $1$.