| Line 6: | Line 6: | ||
*4 winners of the 1st games play two games. | *4 winners of the 1st games play two games. | ||
*2 winners of the 2nd games play again for a third time. | *2 winners of the 2nd games play again for a third time. | ||
| − | *Average would be <math> \frac{( | + | |
| + | 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 | ||
Revision as of 17:04, 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)
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 $ \frac{(4*1 + 2*2 + 2*3)}{8} = \frac{14}{8} $
So:
- A=Number of Teams; n=maximum number of games played by a team
- $ Average = \frac{(\frac{A}{2}*1+\frac{A}{4}*2+...2*(n))}{A} $
- Pull out an A from the top and then...
- $ Average = \frac{2*(n+1)}{A}+\sum_{2}^{N} \frac{1}{2}^N $
