Evan writes a random positive integer on a board: the integer $k$ has probability $2^{-k}$ of being written. He keeps writing integers in this way repeatedly until he writes an integer that he had written before. He then takes all the integers he has written besides his last, sorts them in the order he first drew them, and also sorts them in increasing order, forming two sequences. For example, if he wrote $5,8,2,3,6,10,2$ in that order then his two sequences would be $5,8,2,3,6,10$ and $2,3,5,6,8,10.$

Find the probability that for all $k \in \{ 1,4,34 \},$ that $k$ was written, and $k$ appears in the same position in both sequences.
1. **Understanding the Problem:**
   Evan writes random positive integers on a board with the integer \( k \) having a probability \( 2^{-k} \) of being written. He continues writing until he writes an integer that he has already written before. We need to find the probability that for all \( k \in \{1, 4, 34\} \), \( k \) was written, and \( k \) appears in the same position in both sequences (the sequence in the order they were first drawn and the sequence in increasing order).

2. **Sequence Analysis:**
   Let's denote the sequence of integers written by Evan as \( a_1, a_2, \ldots, a_n \) where \( a_n \) is the first repeated integer. We need to ensure that \( 1, 4, \) and \( 34 \) appear in the sequence and that they appear in the same position in both the original sequence and the sorted sequence.

3. **Probability Calculation:**
   The probability that a specific integer \( k \) is written at any given draw is \( 2^{-k} \). The probability that \( 1, 4, \) and \( 34 \) are written in that order before any repeats occur can be calculated as follows:
   
   - The probability of drawing \( 1 \) first is \( 2^{-1} \).
   - The probability of drawing \( 4 \) next (after drawing \( 1 \)) is \( 2^{-4} \).
   - The probability of drawing \( 34 \) next (after drawing \( 1 \) and \( 4 \)) is \( 2^{-34} \).

4. **Combining Probabilities:**
   Since the events are independent, the combined probability of drawing \( 1, 4, \) and \( 34 \) in that specific order is:
   \[
   2^{-1} \times 2^{-4} \times 2^{-34} = 2^{-(1+4+34)} = 2^{-39}
   \]

5. **Conclusion:**
   The probability that \( 1, 4, \) and \( 34 \) are written in the order \( 1, 4, 34 \) and appear in the same position in both sequences is \( 2^{-39} \).

\[
\boxed{2^{-39}}
\]