In a chemistry experiment, a tube contains 100 particles, 68 on the right and 32 on the left. Each second, if there are $a$ particles on the left side of the tube, some number $n$ of these particles move to the right side, where $n \in \{0,1,\dots,a\}$ is chosen uniformly at random. In a similar manner, some number of the particles from the right side of the tube move to the left, at the same time. The experiment ends at the moment when all particles are on the same side of the tube. The probability that all particles end on the left side is $\tfrac{a}{b}$ for relatively prime positive integers $a$ and $b$.  Compute $100a+b$.

[i]Proposed by Alvin Zou[/i]
1. **Define the problem in terms of probabilities:**
   Let \( L(t) \) be the number of particles on the left side of the tube at time \( t \). Initially, \( L(0) = 32 \) and \( R(0) = 68 \), where \( R(t) \) is the number of particles on the right side of the tube at time \( t \). The total number of particles is \( N = 100 \).

2. **Transition probabilities:**
   Each second, a random number of particles \( n \) from the left side move to the right, and a random number \( m \) from the right side move to the left. The probability of choosing \( n \) particles from the left side is \( \frac{1}{L(t) + 1} \) and the probability of choosing \( m \) particles from the right side is \( \frac{1}{R(t) + 1} \).

3. **Symmetry argument:**
   The problem is symmetric in terms of the left and right sides. The probability of all particles ending up on the left side is the same as the probability of all particles ending up on the right side. Since the total probability must sum to 1, the probability of all particles ending up on one side is \( \frac{1}{2} \).

4. **Expected outcome:**
   Given the symmetry, the probability that all particles end up on the left side is \( \frac{1}{2} \).

5. **Express the probability in the form \( \frac{a}{b} \):**
   The probability that all particles end up on the left side is \( \frac{1}{2} \). Here, \( a = 1 \) and \( b = 2 \).

6. **Compute \( 100a + b \):**
   \[
   100a + b = 100 \cdot 1 + 2 = 102
   \]

The final answer is \( \boxed{102} \).