ZWK
Footballguy
I'm getting an estimate of
88.4 wins, assuming that none of their games are against each other.
This is an instance of the type of question that asks "for n independent variables, each with the same distribution, what is the expected value of the largest of the n variables?" There is a nice answer to this question for uniform variables (it is n/(n+1), which would be 5/6 in this case). For normal variables it's messier - this post claims that the answer for n=5 is 1.16296 standard deviations above the mean. Here we are dealing with binomial variables, which are approximately normal for large sample sizes. Number of wins for a single team is approximately a normal variable with mean 81 and stdev 6.36396 (which is sqrt(np(1-p)) = .5 * sqrt(162)). For an individual team you have to correct for the fact that # of wins is an integer, but I think you don't need to worry about that for expected values so I'll just use the normal approximation without any corrections. 1.16296 times 6.36396 is 7.401032, so we should expect the team with the most wins to have about that many wins above average, which is about 88.401 wins.
The answer is different if the teams are sometimes playing each other. Focusing on the case with just two teams, for independent teams (who never play against each other) the expected value of the team with more wins is 1/sqrt(pi) standard deviations above the mean (according to the post linked above). For two teams that play all the games against each other, the mean of the team with more wins is sqrt(2)/sqrt(pi) standard deviations above the mean (since we're just dealing with a half-normal distribution, although it's possible that I've made a mistake by not correcting for continuity and therefore underweighting the case where they both finish 81-81). That gives an expected value of 84.59 wins for the better of two teams that never play each other, and 86.08 wins for the better of two teams that always play each other. My guess is that, with actual MLB schedules, this would bring the answer up to around 88.5 rather than 88.4 for the original question.
This is an instance of the type of question that asks "for n independent variables, each with the same distribution, what is the expected value of the largest of the n variables?" There is a nice answer to this question for uniform variables (it is n/(n+1), which would be 5/6 in this case). For normal variables it's messier - this post claims that the answer for n=5 is 1.16296 standard deviations above the mean. Here we are dealing with binomial variables, which are approximately normal for large sample sizes. Number of wins for a single team is approximately a normal variable with mean 81 and stdev 6.36396 (which is sqrt(np(1-p)) = .5 * sqrt(162)). For an individual team you have to correct for the fact that # of wins is an integer, but I think you don't need to worry about that for expected values so I'll just use the normal approximation without any corrections. 1.16296 times 6.36396 is 7.401032, so we should expect the team with the most wins to have about that many wins above average, which is about 88.401 wins.
The answer is different if the teams are sometimes playing each other. Focusing on the case with just two teams, for independent teams (who never play against each other) the expected value of the team with more wins is 1/sqrt(pi) standard deviations above the mean (according to the post linked above). For two teams that play all the games against each other, the mean of the team with more wins is sqrt(2)/sqrt(pi) standard deviations above the mean (since we're just dealing with a half-normal distribution, although it's possible that I've made a mistake by not correcting for continuity and therefore underweighting the case where they both finish 81-81). That gives an expected value of 84.59 wins for the better of two teams that never play each other, and 86.08 wins for the better of two teams that always play each other. My guess is that, with actual MLB schedules, this would bring the answer up to around 88.5 rather than 88.4 for the original question.