# How many cricket matches have to be played?

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There are N teams in a cricket match. How many matches have to be played to know the winner and the runner-up?

Generalized solution for N teams.

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Detective Expert Asked on 3rd September 2015 in

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

bsm Curious Answered on 5th September 2015.

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on 8th September 2015.

n-1.
when each match is played a team gets eliminate so after 1st round n/2 teams remains
in next  n/4

and so on
till one team remains.
therefore
=n/2+n/4+…………+1
=1+2+4+………….._n/4+n/2
this is a GP sequence
a=1 r=2
and no. of terms = x(assumed)

after each match, one team is eliminated. thus after first round, n/2 teams are left.
after 2nd round , n/4 teams are left

and so on

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

akky8055 Starter Answered on 5th September 2015.

n = number of teams,
x = a variable

When n is in order of 2^x,

for example, n = 2^4 = 16, then number of matches will be n-1 = 16-1 = 15 (simple GP)

But when n is not in order of 2^x,

for example, n = 21,

then number of matches will be n-1 = 21-1 = 20???

Lets check this, n = 21 = 16 + 5 = 16 + 4 + 1 (all in order of 2^x like, 2^4+2^2+2^0)

when n is in order of 2^x, then i can use n-1 for finding the matches.

for 16 teams = 15 matches
for 4 teams = 3 matches
for 1 team = 0 match

winner of these 2 groups needs to play again, along with the remaining 1 team. (total 3 teams remaining, say N)….

hence, N = 3 = 2+1 (again in order of 2^x)

for 3 teams = 2 matches are required for deciding the winner, who is none but the winner of all 21 teams.

Total matches to decide a winner =15+3+2 = 20 matches. that is (n-1).

Genarilized solution will be n-1.

Kumaran Ilangovan Curious Answered on 6th September 2015.

Its just similar to 25 Horse and 5 track puzzle (First solve it)

Lets say N=10
Here we have,
12 Teams(Horses) and 2 (Teams can play a match)Track and find Top 2

MATCH #1 – 5 Matches
A1 – B1
A2 – B2
A3 – B3
A4 – B4
A5  – B5
A6  – B6

Lets say, all A’s are winner
So,
MATCH #2 – Next tournament(3 Matches)-
A1-A2
A3-A4
A5-A6
Now we have winner- A1,A3 and A5

MATCH #3 – Next Tournament – (3 Matches)
A1-A3
A3-A5
A1-A5
Now we’ll get 2 winners lets say – A1,A3(A1 is winner and A3 runnerup)

In MATCH #4 – (1 Matches)
We’ll take winners opponent in MATCH #2 – A1’s opponent is A4
A4 – A3
Lets say A4 win

In MATCH #5-(1 Matches)
We’ll take winners opponent in MATCH #1- A1’s Opponent B1
B1 – A4

Lets B1 wins So, B1 is 1st Runnerup.

A total of 14 Matches required to find out Winner and runner up in 12 Teams.

It is not a generalized solution but only a way

SherlockHolmes Expert Answered on 3rd September 2015.

After each match is played a team will get eliminated so after 1st round n/2 teams remain
in next  n/4 will remain in tournament and so on…………..till one team remains.

therefore
=n/2+n/4+…………+1
=1+2+4+…………..n/4+n/2
this is a Geometric Progression sequence with a=1 & r=2

after each match, one team will get eliminated. thus after first round, n/2 teams are left.
after 2nd round , n/4 teams are left

This implies 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

Prateek Jain Curious Answered on 7th September 2015.

Complete Copy of akky8055 answer below.

on 7th September 2015.

anujarora Default Answered on 7th September 2015.

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