Conditional probability greater than 1. Why?

Question

Let's say I roll a fair die independently many times. Let $X_i$ be the outcome of the $i$th roll. Assume that on the $k$th roll, I get a $1$ or $X_k = 1$ (Edit based on comments: assume this is the first $1$). What is the probability that I never rolled an even number before rolling this $1$?

I can solve this in two ways and get a gibberish answer if I condition. Let me demonstrate.

Gibberish with conditioning

$$ \begin{align} \mathbb{P}\{\overbrace{X_1 \notin \{2,4,6\}, \dots , X_{k-1} \notin \{2,4,6\}}^{A_{k-1}} \mid X_k = 1 \} &= \frac{\mathbb{P}\{X_k = 1 \mid A_{k-1}\} \mathbb{P}\{A_{k-1}\}}{\mathbb{P}\{X_k = 1\}} \\\\ &\stackrel{\text{ind}}{=} \frac{\mathbb{P}\{X_k = 1\} \mathbb{P}\{A_{k-1}\}}{\mathbb{P}\{X_k = 1\}} \\\\ &= \mathbb{P}\{A_{k-1}\} \\\\ &\stackrel{\text{ind}}{=} \prod_{i=1}^{k-1} \mathbb{P}\{X_i \notin \{2,4,6\}\} \\\\ &= \Big(\frac{1}{2}\Big)^{k-1} \end{align} $$

Now we just need to handle every possible $k$, so

$$ \sum_{k=1}^{\infty} \Big( \frac{1}{2} \Big)^{k-1} = \sum_{k=0}^{\infty} \Big( \frac{1}{2} \Big)^{k} \stackrel{\text{geometric series}}{=} \frac{1}{1 - \frac{1}{2}} = 2 $$

which is obviously gibberish.

Without conditioning

Note that if I don't condition, I'm okay:

$$ \begin{align} \mathbb{P}\{\overbrace{X_1 \notin \{2,4,6\}, \dots , X_{k-1} \notin \{2,4,6\}}^{A_{k-1}}, X_k = 1 \} &= \mathbb{P}\{X_k = 1 \mid A_{k-1}\} \mathbb{P}\{A_{k-1}\} \\\\ &\stackrel{\text{ind}}{=} \mathbb{P}\{X_k = 1\} \mathbb{P}\{A_{k-1}\} \\\\ &\stackrel{\text{ind}}{=} \prod_{i=1}^{k-1} \mathbb{P}\{X_k = 1\} \mathbb{P}\{X_i \notin \{2,4,6\}\} \\\\ &= \Big( \frac{1}{6} \Big) \Big(\frac{1}{2}\Big)^{k-1} \end{align} $$

and therefore

$$ \sum_{k=1}^{\infty} \Big( \frac{1}{6} \Big) \Big(\frac{1}{2}\Big)^{k-1} = \frac{1}{6} \cdot 2 = \frac{1}{3} $$

Answer 1

The since the first few rolls can't be even and can't be 1, the probability of each conditional is $1/3$, not $1/2.$

Answer 2

Synthesizing from the comments, my answer in notation is as follows. If two events $A$ and $B$ are disjoint, then $P(A \cap B) = P(A) + P(B)$. But in my problem, $A_{k-1}$ and $A_k$ are not disjoint events:

$$ \begin{align} A_{k-1} &= \{X_1 \notin \{1,2,3\}, \dots, X_{k-1} \notin \{1,2,3\}\} \\ A_{k} &= \{X_1 \notin \{1,2,3\}, \dots, X_{k-1} \notin \{1,2,3\}, X_k \notin \{1,2,3\}\} \end{align} $$

Therefore, my infinite series

$$ \sum_{k=1}^{\infty} \mathbb{P}\{A_{k-1}\} $$

does not adhere to basic rules of probability.