In a dance party initially there are $20$ girls and $22$ boys in the pool and infinitely many more girls and boys waiting outside. In each round, a participant is picked uniformly at random; if a girl is picked, then she invites a boy from the pool to dance and then both of them elave the party after the dance; while if a boy is picked, then he invites a girl and a boy from the waiting line and dance together. The three of them all stay after the dance. The party is over when there are only (two) boys left in the pool.

(a) What is the probability that the party never ends?

(b) Now the organizer of this party decides to reverse the rule, namely that if a girl is picked, then she invites a boy and a girl from the waiting line to dance and the three stay after the dance; while if a boy is picked, he invites a girl from the pool to dance and both leave after the dance. Still the party is over when there are only (two) boys left in the pool. What is the expected number of rounds until the party ends?
1. **Initial Setup and Problem Understanding:**
   - We start with 20 girls and 22 boys in the pool.
   - In each round, a participant is picked uniformly at random.
   - If a girl is picked, she invites a boy from the pool to dance, and both leave the party.
   - If a boy is picked, he invites a girl and a boy from the waiting line to dance, and all three stay after the dance.
   - The party ends when there are only two boys left in the pool.

2. **Invariant Analysis:**
   - The invariant here is the difference between the number of boys and girls in the pool, denoted as \( b - g \).
   - Initially, \( b = 22 \) and \( g = 20 \), so \( b - g = 2 \).

3. **Probability Calculation:**
   - We can treat this process as a random walk on a line starting at \( x = 20 \).
   - Each step either moves left (a girl and a boy leave) or right (a boy invites a girl and a boy from the waiting line).
   - The probability of moving left (a girl is chosen) is \( \frac{g}{b+g} = \frac{20}{42} = \frac{10}{21} \).
   - The probability of moving right (a boy is chosen) is \( \frac{b}{b+g} = \frac{22}{42} = \frac{11}{21} \).

4. **Expected Outcome:**
   - We need to find the probability that the party never ends, which means the random walk never reaches the state where there are only two boys left in the pool.
   - The probability of the party ending is equivalent to the probability of the random walk reaching \( x = 0 \) (i.e., no girls left in the pool).

5. **Telescoping Series and Catalan Numbers:**
   - The probability of a walk with \( s \) steps that makes it end up \( d \) steps to the right is given by:
     \[
     2^{-s} \cdot \frac{x+d+1}{x+1}
     \]
   - Let \( f_d(t) \) be the number of walks from \( (0,0) \) to \( (t-d,t) \) staying entirely below (or on) the line \( y = x + d \).
   - We are trying to find:
     \[
     1 - \frac{1}{21} \cdot 2^{-19} \sum_{t \ge 19} f_{19}(t) \cdot 4^{-t}
     \]
   - Using the definition of Catalan numbers and generating functions, we can simplify this to:
     \[
     2^d \sum_{t \ge d} f_d(t) \cdot 4^{-t} = 2
     \]

6. **Final Probability Calculation:**
   - For \( d = 19 \), the probability that the process stops after \( t \) steps is:
     \[
     f_{19}(t) \cdot 2^{-(t-19+t+1)}
     \]
   - Therefore, the probability that the party never ends is:
     \[
     1 - \frac{1}{21} \cdot 2^{18} \sum_{t \ge 19} f_{19}(t) \cdot 4^{-t} = \frac{20}{21}
     \]

The final answer is \( \boxed{ \frac{20}{21} } \)