In a basketball tournament, there are teams named 1 through 8, such that a lower number team is better than a higher numbered team. (1 is best, 2 is second best... 8 is worst) Also, a better team will always win over a worse team. (There are no upsets)

?-\__
?-/  |
     |--\
?-\__|  |
?-/     |
        |-WINNER
?-\__   |
?-/  |  |
     |--/
?-\__|
?-/

Here is the grid for the tournament

If the better team always wins (there are no upsets) and if the pairing is completely random, what is the easiest way to figure the probability that team 2 doesn't win second place?

(In reply to So why not do it the hard way? by Jer)

If you insist:
Divide 8 initial placesinto 2 halves A & B
For teams 1 and two to be in half A there are 4*3 possibilities,same to be in B;
ERGO- 24 possibilities to be in the same half/
The remainig 6 teams ars dispersed in 6! ways

Allexisting arrangments : 8!

24*6!/8!=24/56=3/7

There so many ways to skin a cat....
ady

Posted by Ady TZIDON on 2004-06-15 12:05:18