Six teams play a single round-robin tournament, in which each team plays every other team exactly once, there are no ties, and for each team, the probability of winning any given game is 1/2. What is the probability that at least two different teams finish the tournament with the same number of wins? Express your answer as a common fraction.

In a single round-robin tournament with 6 teams, each team plays every other team exactly once. The total number of matches played in such a tournament is given by the formula for combinations:

\[
\binom{n}{2} = \binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15
\]

Each match results in either a win or a loss, and since there are no ties, there are \(15\) total games played. Each of the 6 teams will play \(5\) games (one against each other team), leading to a cumulative total of \(30\) game outcomes (wins + losses).

Let \(X_i\) represent the number of wins for team \(i\), where \(i = 1, 2, \ldots, 6\). The number of wins by all teams must satisfy:

\[
X_1 + X_2 + X_3 + X_4 + X_5 + X_6 = 15
\]

The number of possible wins \(X_i\) for each team can range from \(0\) to \(5\). Each team can thus have a certain number of wins, which leads us to determining the possible distributions of wins among the teams.

To find the probability that at least two teams have the same number of wins, we can use the principle of complementary counting. First, we calculate the total number of distinct distributions of wins that allow each team to potentially have a unique number of wins.

If all teams could hypothetically have different win totals, the possible unique win totals for 6 teams would be \(0, 1, 2, 3, 4, 5\). Assigning these unique totals would count each configuration exactly once, which is essential for our calculation.

The total number of configurations where all teams have distinct numbers of wins can be determined as follows:
1. **Assign each unique winning combination \(0\) to \(5\)**:
- We have \(6\) distinct values (the possible number of wins).
- The only configuration that permits all teams to have distinct numbers of wins is to assign the unique values to each team:

The number of ways to arrange these distinct outcomes among 6 teams is simply \(1\) (each configuration corresponds uniquely).

Thus, the total configuration of win distributions is \(6!\) (for \(6\) teams with distinct outcomes) which represents the permutations of the unique win counts. Only one is valid in terms of sum:

\[
\text{Valid Distinct Outcomes} = \{0, 1, 2, 3, 4, 5\}
\]

Thus, the favorable (distinct) cases yield \(1\) possibility.

2. Next, we calculate the total possible outcomes from which we will derive the probability of not having duplicates. Given that every match outcome is independent and equally likely, from the 15 matches where each can produce either a win or a loss, the total possible outcomes from these matches is:

\[
2^{15}
\]

3. **Calculating the complementary probabilities**:

Therefore, the probability that all teams finish with distinct win numbers (at least one valid outcome) is given by the ratio of valid distinct distributions to total possible configurations (\(2^{15}\)):

\[
P(\text{All Distinct}) = \frac{1}{2^{15}}
\]

Finally, we must find the probability that at least two teams have the same win count:

\[
P(\text{At least 2 teams have the same}) = 1 - P(\text{All Distinct}) = 1 - \frac{1}{2^{15}} = \frac{2^{15} - 1}{2^{15}}
\]

Lastly, computing \(2^{15} = 32768\):

\[
P(\text{At least 2 teams similar}) = \frac{32768 - 1}{32768} = \frac{32767}{32768}
\]

Thus, the final probability that at least two different teams finish the tournament with the same number of wins is:

\[
\boxed{\frac{32767}{32768}}
\]