world at finger tips

Answer is “n-1”.
Assumption: question is to find the minimum matches to be played
How:

for each match played one team is eliminated. thus in first round, n/2 matches are played.
in second round n/4 are played
.
.
process is continued still one team is left.
so total=n/2+n/4+n/8+….+1
=1+2+4+…+n/2
this is in geometric progression with a=1, r=2,  no. of terms =??(say k)

       for each match played one team is eliminated. thus after first round, n/2 teams are left.
       after 2nd round , n/4 teams are left
       .
       .
       after ‘k’th round n/(2^k) teams are left
       Now for n/(2^k)=1 team
       =>n=2^k
       =>k=log(n)
a=1,r=2,k=log n (base 2)

total matches = a(r^n – 1)/(r-1)
=2^(log n) -1
=n-1