A particle moves from $(0, 0)$ to $(n, n)$ directed by a fair coin. For each head it moves one step east and for each tail it moves one step north. At $(n, y), y < n$, it stays there if a head comes up and at $(x, n), x < n$, it stays there if a tail comes up. Let$ k$ be a fixed positive integer. Find the probability that the particle needs exactly $2n+k$ tosses to reach $(n, n).$
1. **Define the problem in terms of sequences:**
   - Each head (H) represents a step east.
   - Each tail (T) represents a step north.
   - To reach $(n, n)$, the particle needs exactly $n$ steps east and $n$ steps north.
   - The particle needs exactly $2n + k$ tosses to reach $(n, n)$.

2. **Determine the number of valid sequences:**
   - The total number of steps is $2n + k$.
   - We need $n$ heads (H) and $n + k$ tails (T) or $n + k$ heads (H) and $n$ tails (T).
   - The number of sequences with $n$ heads and $n + k$ tails is given by the binomial coefficient:
     \[
     \binom{2n + k}{n} = \frac{(2n + k)!}{n!(n + k)!}
     \]
   - Similarly, the number of sequences with $n + k$ heads and $n$ tails is:
     \[
     \binom{2n + k}{n + k} = \frac{(2n + k)!}{(n + k)!n!}
     \]
   - Since $\binom{2n + k}{n} = \binom{2n + k}{n + k}$, the total number of valid sequences is:
     \[
     2 \cdot \binom{2n + k}{n} = 2 \cdot \frac{(2n + k)!}{n!(n + k)!}
     \]

3. **Determine the total number of possible sequences:**
   - Each toss can result in either a head or a tail, so there are $2^{2n + k}$ possible sequences of $2n + k$ tosses.

4. **Calculate the probability:**
   - The probability that the particle reaches $(n, n)$ in exactly $2n + k$ tosses is the ratio of the number of valid sequences to the total number of possible sequences:
     \[
     P = \frac{2 \cdot \frac{(2n + k)!}{n!(n + k)!}}{2^{2n + k}} = \frac{2 \cdot (2n + k)!}{n!(n + k)! \cdot 2^{2n + k}}
     \]

5. **Bonus Question: Determine the value of $k$ that maximizes the probability:**
   - To maximize the probability, we need to maximize the expression:
     \[
     \frac{2 \cdot (2n + k)!}{n!(n + k)! \cdot 2^{2n + k}}
     \]
   - This is a combinatorial optimization problem. Generally, the probability decreases as $k$ increases because the denominator $2^{2n + k}$ grows exponentially while the numerator grows factorially. Therefore, the probability is maximized when $k = 0$.

The final answer is \( \boxed{ \frac{2 \cdot (2n + k)!}{n!(n + k)! \cdot 2^{2n + k}} } \)