Problem 1639

1. Define the Problem and Notation:
- There are 256 players ranked from 1 to 256.
- The player with the higher rank wins with probability $\frac{3}{5}$.
- We need to determine the expected value of the rank of the winner.

2. Binary Representation and Initial Setup:
- Let $a_{n-1}$ be the player with rank $n$, where $n \in \{1, 2, \ldots, 256\}$.
- The binary representation of $n-1$ is $\overline{b_7b_6b_5b_4b_3b_2b_1b_0}_{(2)}$, where $b_i \in \{0, 1\}$ for $i \in \{0, 1, \ldots, 7\}$.

3. Round 1 Analysis:
- Players $\overline{A0}_{(2)}$ and $\overline{A1}_{(2)}$ play against each other, where $A = \overline{b_7b_6b_5b_4b_3b_2b_1}_{(2)}$.
- $\overline{A0}_{(2)} < \overline{A1}_{(2)}$.
- Probability of $\overline{A0}_{(2)}$ winning: $\frac{3}{5}$.
- Probability of $\overline{A1}_{(2)}$ winning: $\frac{2}{5}$.
- Probability to win in the first round: $p_1 = \frac{2^{b_0} \cdot 3^{1-b_0}}{5}$.

4. Round 2 Analysis:
- Players $\overline{B0x}_{(2)}$ and $\overline{B1y}_{(2)}$ play against each other, where $B = \overline{b_7b_6b_5b_4b_3b_2}_{(2)}$ and $x, y \in \{0, 1\}$.
- $\overline{B0x}_{(2)} < \overline{B1y}_{(2)}$.
- Probability of $\overline{B0x}_{(2)}$ winning: $\frac{3}{5}$.
- Probability of $\overline{B1y}_{(2)}$ winning: $\frac{2}{5}$.
- Probability to win in the first two matches: $P_2(n-1) = \frac{2^{b_0 + b_1} \cdot 3^{2 - (b_0 + b_1)}}{5^2}$.

5. Generalizing for 8 Rounds:
- Probability for player $a_{n-1}$ to be the winner:
$$P_8(n-1) = \frac{2^{b_0 + b_1 + \cdots + b_7} \cdot 3^{8 - (b_0 + b_1 + \cdots + b_7)}}{5^8}$$

6. Expected Value Calculation:
- Expected value for $n-1$:
$$n_{ex} - 1 = \sum_{m=0}^{255} m \cdot \frac{2^{b_0 + b_1 + \cdots + b_7} \cdot 3^{8 - (b_0 + b_1 + \cdots + b_7)}}{5^8}$$
- Simplifying, we get:
$$n_{ex} - 1 = \sum_{k=0}^8 S_k \cdot \frac{2^k \cdot 3^{8-k}}{5^8}$$
where $S_k$ is the sum of numbers $m$ for which $b_0 + b_1 + \cdots + b_7 = k$.

7. **Calculating $S_k$:**
- Number of numbers $m$ for which $b_0 + b_1 + \cdots + b_7 = k$ is $N_k = \binom{8}{k}$.
- Sum of these numbers:
$$S_k = 255 \cdot \frac{k \cdot \binom{8}{k}}{8} = 255 \cdot \binom{7}{k-1}$$

8. Final Summation:
- Substituting $S_k$:
$$n_{ex} - 1 = \sum_{k=1}^8 255 \cdot \binom{7}{k-1} \cdot \frac{2^k \cdot 3^{8-k}}{5^8}$$
- Simplifying further:
$$n_{ex} - 1 = 255 \cdot \frac{2}{5} \cdot \sum_{k=0}^7 \binom{7}{k} \cdot \left(\frac{2}{5}\right)^k \cdot \left(\frac{3}{5}\right)^{7-k}$$
- Using the binomial theorem:
$$\sum_{k=0}^7 \binom{7}{k} \left(\frac{2}{5}\right)^k \left(\frac{3}{5}\right)^{7-k} = \left(\frac{2}{5} + \frac{3}{5}\right)^7 = 1$$
- Therefore:
$$n_{ex} - 1 = 255 \cdot \frac{2}{5} = 102$$
- Hence, the expected value of the rank of the winner is:
$$n_{ex} = 102 + 1 = 103$$

The final answer is $\boxed{103}$