In the NFL, a division consists of 4 teams, each of which plays each other team twice.
Assume that in any game, either team is equally likely to win (and there are no ties). What is the probability that, at the end of the season, the division has neither a perfect team with 6 wins nor a futile team with 6 losses?
I know there are 12 games total. How do I solve this? Thanks!
To solve this problem, we can use the concept of combinations.
First, let's calculate the total number of possible outcomes for the 12 games. In each game, there are two possible outcomes: either a win or a loss. Since there are 12 games, the total number of possible outcomes will be 2^12.
Next, let's calculate the number of outcomes where a team ends up with 6 wins or 6 losses. For a team to have 6 wins, it must win 6 out of the 12 games. This can be calculated using combinations. The number of possible combinations to choose 6 wins out of 12 games is given by 12C6, which is equal to 924. Similarly, the number of possible combinations to choose 6 losses out of 12 games is also 924.
Now, let's calculate the total number of outcomes where there is either a perfect team with 6 wins or a futile team with 6 losses. This can be done by adding the number of outcomes where there is a perfect team with 6 wins and the number of outcomes where there is a futile team with 6 losses. Therefore, the total number of outcomes where there is either a perfect team or a futile team is 924 + 924 = 1848.
Finally, we can calculate the probability of neither a perfect team nor a futile team by taking the ratio of the favorable outcomes (total possible outcomes without perfect or futile teams) and the total possible outcomes.
Probability = (Total possible outcomes - Favorable outcomes) / Total possible outcomes
= (2^12 - 1848) / 2^12
= (4096 - 1848) / 4096
= 2248 / 4096
= 0.5488
Therefore, the probability that, at the end of the season, the division does not have either a perfect team with 6 wins or a futile team with 6 losses is approximately 0.5488 or 54.88%.
To solve this problem, we can consider the possible outcomes of each game and calculate the probability of having neither a perfect team with 6 wins nor a futile team with 6 losses.
Since there are 12 games total in a season, we can think of each game as having two possible outcomes: a win or a loss. Therefore, there are 2^12 = 4096 possible outcomes for the entire season.
Next, let's think about the possible scenarios where there is a perfect team with 6 wins or a futile team with 6 losses. We can break down these scenarios as follows:
1. There is a perfect team with 6 wins: There are four teams in the division, so we have to choose one of them to be the perfect team. Additionally, for each team, we need to choose which six games they win out of the total 12 games. Therefore, the number of scenarios with a perfect team with 6 wins is 4 * (12 choose 6) = 4 * (12! / (6! * 6!)).
2. There is a futile team with 6 losses: Similar to the above case, we have to choose one team to be the futile team and select the six games they lose out of the total 12 games. Hence, the number of scenarios with a futile team with 6 losses is 4 * (12 choose 6).
Now, let's find the total number of scenarios where there is either a perfect team with 6 wins or a futile team with 6 losses. We can sum the two cases above:
Total number of scenarios = 4 * (12! / (6! * 6!)) + 4 * (12 choose 6)
Finally, we can calculate the probability of having neither a perfect team with 6 wins nor a futile team with 6 losses using the complement rule:
Probability = 1 - (Total number of scenarios / Total possible outcomes) = 1 - (Total number of scenarios / 4096).
Now, you can calculate the probability by plugging in the numbers into the formula.