A math problem, help!

R

Raydar

Veteran X
In a Nine Men's Morris tournament with 77 entrants, how many matches must be played to determine a champion? Two players participate in each match.

Whats the Answer?
 
I'm sure this thread will have alot of meaning to that one person who actually knows what a Nine's Men Morris tournament is.
 
64 requires 63

so to get down to 64 you need play 6 or 7 play in games

without thinking too much about it i'd say 69 or 70

edit: oops, that was stupid, i was thinking one game dealt with two people

i blame it on being out of school, its turned my brain to mush, n-1
 
Last edited:
its n-1.. so 76 in a single elim tournament

77 players, 1 champion.. which means 76 ppl have to lose.. so you have 76 matches
 
Slant_This said:
64 requires 63

so to get down to 64 you need play 6 or 7 play in games

without thinking too much about it i'd say 69 or 70
Click to expand...

64 players results in 63 matches because its n - 1 ;]

if you only had 70 matches, that would mean 7 ppl didnt lose (you can only lose in a match).

so you need 76 matches to get 1 person left standing
 
the way i thought of it was
64 requires 63 matches
that leaves 13 peeps
8 of those peeps play, thats 7 matches
4 of those peeps play thats 3 matches
that leaves 1 dude
1 dude plays winner of 64, winner of 8 plays winner of 4
then the winner of both those matches play eachother for the championship
so 63+7+3+1+1+1 = 76

damn i guess anvil is right
 
no need to draw anything out. You need 76 losers to leave one champion, and there's one loser per match, so 76 matches yeilds a champ
 
Christ, it's a two person game, as he said in the first fucking post. It doesn't matter what the game is. It could be Hide the fucking Salami and it wouldn't change a god damn thing as long as it was being played with two people.
 
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