There have been several questions (and discussions) on whether an NBA team can finish a season without a loss. I know there is a very very small possibility but by approximately how much?
Help me out Math wizzes!|||It really depends on how good the team is compared to other teams, and the probability of the team winning any 1 game.
Assuming all teams are pretty much equal, then the probability of winning any one game is 0.5 (50%). Therefore, probability (P) of winning all 82 games is:
P = (0.5)^(82) = 0.000000000000000000000000206795 or
P = 1/4,835,703,278,458,516,698,824,704
Even if the team is really good, and probability of winning any one game is 0.8 (80%), then probability of winning all 82 games is:
P = (0.8)^(82) = 0.000000011307821214581659709333
P = 1/88,434,366
So even with a very good probability of winning any one game, the probability of winning all 82 games is less than 1 in 88 million.|||There are too many unknown conditions to allow a reasonable answer. If the probability of victory in each game were 1/2, then the probability of a sweep season would be 1/2^82, but that simply is not a reasonable condition. Your question is one of applied mathematics. We cannot answer it without data.
If you look back at records from the early days of organized sports, you will be able to find many scores and season records that would be considered unattainable today. That is because there was much greater disparity in talent among the competing teams.|||Never will happen. There are far too many games in the NBA to win them all. Considering the best anyone has ever done is win 72 games in a season (the 1995-96 Bulls), I think this is impossible. It doesn't matter how good you are, eventually you will have a bad night when someone else has a good night.|||2^-82 or 1/(2^-82) or 2 x 10^-83
%26lt; 0.000000000000000000000000000000000001% to put it in further perspective|||well
it really depends on how good the team is
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