The tournament has 6 players, and each player plays against every other player exactly once. The total number of games played in a round-robin tournament with $n$ players is given by:

$\binom{n}{2} = \frac{n(n-1)}{2}$

For $n = 6$:

$\binom{6}{2} = \frac{6(6-1)}{2} = \frac{6 \times 5}{2} = 15 \text{ games}$

The number of wins listed so far is:

• Helen: 4 wins
• Ines: 3 wins
• Janet: 2 wins
• Kendra: 2 wins
• Lara: 2 wins

The sum of the wins is:

$4 + 3 + 2 + 2 + 2 = 13$

Since there are a total of 15 games and each game has exactly one winner, the total number of wins must also be 15. Thus, Monica must have won the remaining:

$15 - 13 = 2 \text{ games}$

Therefore, Monica won 2 games.

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Would you like more details or have any questions? Here are five related questions you might find interesting:

1. How many total games would there be if there were 8 players in the tournament?
2. What is the general formula for the number of games in a round-robin tournament with $n$ players?
3. If a player won all of their games, how many games would they have won in this 6-player tournament?
4. How would the number of games change if players played each other twice instead of once?
5. What is the difference between a round-robin tournament and a knockout tournament?

Tip: In a round-robin tournament, every player faces every other player, so you can use combinations to calculate the total number of games.