Difference between revisions of "Talk:Single Elimination Forever MA181Fall2008bell" - Rhea

Latest revision as of 18:04, 18 October 2008

The solution is very simple, and doesn't require that you use series, just some inductive logic and limits.--John Mason 16:58, 17 October 2008 (UTC)

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 The solution is very simple, and doesn't require that you use series, just some inductive logic and limits.--[[User:Jmason|John Mason]] 16:58, 17 October 2008 (UTC)
-If you had 8 teams:
-*All 8 play at least 1 game.
-*4 winners of the 1st games play two games.
-*2 winners of the 2nd games play again for a third time.
-So:
-*4 teams played only 1 game
-*2 teams played 2 games
-*2 teams played 3 games
-*Average would be <math> \frac{(4*1 + 2*2 + 2*3)}{8} = \frac{14}{8}</math>
-So:
-*A=Number of Teams; n=maximum number of games played by a team
-*<math> Average = \frac{(\frac{A}{2}*1+\frac{A}{4}*2+...2*(n))}{A} </math>
-*Pull out an A from the top and then...
-*<math> Average = \frac{2*(n+1)}{A}+\sum_{2}^{N} \frac{1}{2}^N