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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?

Enter your answer as a fraction in simplest form.

I have approached this question using a combination of casework and complementary counting.

I set up a table for a team winning all games and another table for a team winning absolutely no games.

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    $\begingroup$ One way would be to list the $2^{12}=4096$ equally possible outcomes and count those with no complete wins or losses $\endgroup$ – Henry Nov 15 '16 at 11:05
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You need to consider the important outcomes and their probabilities

  • The probability that Team A wins all $6$ matches is $\dfrac{1}{2^6}$
  • The probability that Team D loses all $6$ matches is $\dfrac{1}{2^6}$
  • The probability that Team A wins all $6$ matches and Team D loses all $6$ is $\dfrac{1}{2^{10}}$
  • The probability that a team wins all $6$ matches is $\displaystyle 4 \times \dfrac{1}{2^6}$
  • The probability that a team loses all $6$ matches is $\displaystyle 4 \times \dfrac{1}{2^6}$
  • The probability that a team wins all $6$ matches and another team loses all $6$ is $\displaystyle 12 \times \dfrac{1}{2^{10}}$
  • The probability that a team wins all $6$ matches but no team loses all $6$ is $\displaystyle 4 \times \dfrac{1}{2^6} - \displaystyle 12 \times \dfrac{1}{2^{10}}$
  • The probability that no team wins all $6$ matches but a team loses all $6$ is $\displaystyle 4 \times \dfrac{1}{2^6} - \displaystyle 12 \times \dfrac{1}{2^{10}}$
  • The probability that no team wins all $6$ matches and no team loses all $6$ is $$1- \left(4 \times \dfrac{1}{2^6} - 12 \times \dfrac{1}{2^{10}}\right)- \left(4 \times \dfrac{1}{2^6} - 12 \times \dfrac{1}{2^{10}}\right) - 12 \times \dfrac{1}{2^{10}} =\dfrac{227}{256}$$
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