There are many two kinds of black and white cards. When you have $ k$ cards in hand, consider the following procedure $ (\bf{A})$.

$ (\bf{A})$ You choose one card from $ k$ cards in hand with equal probability $ \frac {1}{k}$ and replace the card with different color one.

Answer the following questions.

(1) When you have 4 cards with 2 white and 2 black, first,  find the probabilty such that  you will first have 4 cards in same color after repeating the procedure $ \bf{(A)}$
$ n$ times.

(2) When you have 6 cards with 3 white and 3 black, first,  find the probabilty such that  you will first have 6 cards in same color after repeating the procedure $ \bf{(A)}$ 
$ n$ times.
### Part (1): Probability of having 4 cards of the same color when starting with 2 white and 2 black cards

1. **Define the state variable**:
   Let \( X_j \) be the number of white cards at step \( j \). Initially, \( X_0 = 2 \).

2. **Transition probabilities**:
   - If \( X_j = m \), the probability of choosing a white card and replacing it with a black card is \( \frac{m}{4} \).
   - The probability of choosing a black card and replacing it with a white card is \( \frac{4-m}{4} \).

3. **Recurrence relation**:
   The probability \( P(X_j = m) \) can be expressed in terms of the probabilities at the previous step:
   \[
   P(X_j = m) = P(X_{j-1} = m-1) \cdot \frac{4-(m-1)}{4} + P(X_{j-1} = m+1) \cdot \frac{m+1}{4}
   \]

4. **Boundary conditions**:
   - \( P(X_j = 0) \) and \( P(X_j = 4) \) are absorbing states, meaning once all cards are of the same color, they remain so.
   - Initially, \( P(X_0 = 2) = 1 \) and \( P(X_0 = 0) = P(X_0 = 4) = 0 \).

5. **Solving the recurrence**:
   We need to find the probability of reaching either \( X_j = 0 \) or \( X_j = 4 \) starting from \( X_0 = 2 \). This can be approached using a Markov chain or solving the recurrence relation directly.

6. **Symmetry argument**:
   Since the process is symmetric with respect to the number of white and black cards, the probability of ending with all white cards is the same as ending with all black cards. Therefore, the probability of having 4 cards of the same color is:
   \[
   P(\text{4 white or 4 black}) = 2 \cdot P(\text{4 white})
   \]

7. **Final probability**:
   Given the symmetry and equal initial conditions, the probability of ending with all white cards (or all black cards) is \( \frac{1}{2} \). Thus, the probability of having 4 cards of the same color is:
   \[
   \boxed{1}
   \]

### Part (2): Probability of having 6 cards of the same color when starting with 3 white and 3 black cards

1. **Define the state variable**:
   Let \( X_j \) be the number of white cards at step \( j \). Initially, \( X_0 = 3 \).

2. **Transition probabilities**:
   - If \( X_j = m \), the probability of choosing a white card and replacing it with a black card is \( \frac{m}{6} \).
   - The probability of choosing a black card and replacing it with a white card is \( \frac{6-m}{6} \).

3. **Recurrence relation**:
   The probability \( P(X_j = m) \) can be expressed in terms of the probabilities at the previous step:
   \[
   P(X_j = m) = P(X_{j-1} = m-1) \cdot \frac{6-(m-1)}{6} + P(X_{j-1} = m+1) \cdot \frac{m+1}{6}
   \]

4. **Boundary conditions**:
   - \( P(X_j = 0) \) and \( P(X_j = 6) \) are absorbing states.
   - Initially, \( P(X_0 = 3) = 1 \) and \( P(X_0 = 0) = P(X_0 = 6) = 0 \).

5. **Solving the recurrence**:
   We need to find the probability of reaching either \( X_j = 0 \) or \( X_j = 6 \) starting from \( X_0 = 3 \).

6. **Symmetry argument**:
   Since the process is symmetric with respect to the number of white and black cards, the probability of ending with all white cards is the same as ending with all black cards. Therefore, the probability of having 6 cards of the same color is:
   \[
   P(\text{6 white or 6 black}) = 2 \cdot P(\text{6 white})
   \]

7. **Final probability**:
   Given the symmetry and equal initial conditions, the probability of ending with all white cards (or all black cards) is \( \frac{1}{2} \). Thus, the probability of having 6 cards of the same color is:
   \[
   \boxed{1}
   \]