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LostInParadise's avatar

Can you use weighted averages to solve this problem?

Asked by LostInParadise (31639points) 2 months ago

Weighted averages are not usually officially taught in K-12, which is a little odd, because student grading is usually done using weighted averages. The following problem is easily solved using weighte averages.

A league has 11 teams. Eleven is used to simplify things since each team plays against 10 others. Each pair of teams plays the same number of games as every other pair. One team wins every game and the other 10 teams have an equal number wins and llosses against every other one of the 10 teams.

The one team has winning record of 100%. What is the winning percentage of the other teams?

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7 Answers

zenvelo's avatar

There is no need for using weighted averages. But the problem is not properly stated.

the other 10 teams have an equal number wins and losses against every other one of the 10 teams. That implies every team plays against ten teams, plus they all play against the 100% super team, which makes for a total of 12 teams, not 11. And since every team has an equal number of wins and losses, they all play to a record of 5 wins, 5 losses plus a loss to the super team. Every team ends with a record of 5–6.

Weighted averages have nothing to do with this.

LostInParadise's avatar

What I should have said is that each pair among the 10 other teams has an even record. There are just 11 teams.

For the weighted average, each team plays against 10 other teams. For any team except the grand winner, they have a winning rate of ½ for 9 of their opponents and 0 against the grand winner. Since the record against each other team is weighted by 1/10, the winning record for these teams is 9(½ * 1/10) + 1/10 * 0 = 9*(0.5 * 0.1) = 9 * 0.05 = 0.45

zenvelo's avatar

Again, the question makes no sense because it contradicts itself. I cannot have “an even record” against 9 opponents. I either win 5 and lose four, or I win four and lose 5.

And why complicate things by saying you have to have a weighted average?

LostInParadise's avatar

You could play 10 games against each of the nine opponents and win 5 of the 10. Note that the number of games against each opponent must be even. You were assuming 1 game against each opponent, which will not work.

filmfann's avatar

Everyone is making this complicated.
Since there are 10 imperfect teams with equal amounts of wins and losses against imperfect teams, so they played an even number of games against each other, so let’s say 2. The imperfect teams were 18–18 against each other, and 18–20 overall. They were 47.386%.

LostInParadise's avatar

Close. In your example, each of the imperfect teams wins 9 games against the other imperfect teams. They play a total of 20 games, so they win 9/20 or 45% of their games.

filmfann's avatar

@LostInParadise I see my mistake. They were 9–9 against imperfect teams, and 9–11 overall.

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