Assume that the number of offspring for every man can be $0,1,\ldots, n$ with with probabilities $p_0,p_1,\ldots,p_n$ independently from each other, where $p_0+p_1+\cdots+p_n=1$ and $p_n\neq 0$. (This is the so-called Galton-Watson process.) Which positive integer $n$ and probabilities $p_0,p_1,\ldots,p_n$ will maximize the probability that the offspring of a given man go extinct in exactly the tenth generation?
1. **Define the Generating Function:**
   Let \( f(s) \) be the generating function for the offspring distribution, defined as:
   \[
   f(s) = \sum_{k=0}^{n} p_k s^k
   \]
   where \( p_k \) is the probability of having \( k \) offspring.

2. **Extinction Probability:**
   The probability of extinction in the \( k \)-th generation is given by the \( k \)-th iterate of the generating function evaluated at \( s = 0 \). Let \( q_k \) be the probability of extinction in exactly \( k \) generations. Then:
   \[
   q_k = f^{(k)}(0)
   \]
   where \( f^{(k)} \) denotes the \( k \)-th iterate of \( f \).

3. **Maximizing Extinction in the 10th Generation:**
   We need to maximize \( q_{10} = f^{(10)}(0) \). Given that only \( p_0 \) and \( p_1 \) are nonzero, the generating function simplifies to:
   \[
   f(s) = p_0 + p_1 s
   \]
   with \( p_0 + p_1 = 1 \).

4. **Iterating the Generating Function:**
   Since \( f(s) = p_0 + p_1 s \), iterating this function \( k \) times gives:
   \[
   f^{(k)}(s) = p_0 + p_1 f^{(k-1)}(s)
   \]
   Starting with \( f^{(0)}(s) = s \), we have:
   \[
   f^{(1)}(s) = p_0 + p_1 s
   \]
   \[
   f^{(2)}(s) = p_0 + p_1 (p_0 + p_1 s) = p_0 + p_1 p_0 + p_1^2 s
   \]
   Continuing this process, we find:
   \[
   f^{(k)}(s) = p_0 \sum_{i=0}^{k-1} p_1^i + p_1^k s
   \]
   Evaluating at \( s = 0 \):
   \[
   f^{(k)}(0) = p_0 \sum_{i=0}^{k-1} p_1^i = p_0 \frac{1 - p_1^k}{1 - p_1}
   \]

5. **Extinction in the 10th Generation:**
   For \( k = 10 \):
   \[
   q_{10} = f^{(10)}(0) = p_0 \frac{1 - p_1^{10}}{1 - p_1}
   \]
   To maximize \( q_{10} \), we need to maximize \( p_0 \frac{1 - p_1^{10}}{1 - p_1} \).

6. **Maximizing the Expression:**
   Since \( p_0 + p_1 = 1 \), we can write \( p_0 = 1 - p_1 \). Substituting this into the expression:
   \[
   q_{10} = (1 - p_1) \frac{1 - p_1^{10}}{1 - p_1} = 1 - p_1^{10}
   \]
   To maximize \( 1 - p_1^{10} \), we need to minimize \( p_1^{10} \). The minimum value of \( p_1^{10} \) occurs when \( p_1 \) is as small as possible. Since \( p_1 \) must be a probability, the smallest value is \( p_1 = 0 \). However, \( p_1 = 0 \) would imply \( p_0 = 1 \), which means no offspring are produced, leading to immediate extinction.

   Therefore, the optimal solution is to have \( p_0 \) and \( p_1 \) such that \( p_1 \) is very small but not zero, ensuring that extinction occurs in exactly the 10th generation with high probability.

The final answer is \( \boxed{ p_0 = 1 - p_1 } \) and \( p_1 \) is very small but not zero.