How Many Games?

Question

The rage at Chaos Middle School was a new “mind: game called Easy Money in which two players take the role of S&L Bank officials trying to outwit bank regulators and treasury agents so as to amass fortunes for themselves. Herbert, president of the Easy Money Club, proposed a tournament. Vinnie, a compulsive Easy Money player, recommended that all 160 members of the club play each other. The player who amassed the greatest number of wins would be crowned the champion. Herbert asked, how many games such a tournament would require? How many games would the 160 members of the Easy Money Club have to play if each member played all of the other member once?

Answer 1

This is a form of what's often called the handshake problem (because you could do it with handshakes instead of games, if you wanted to).  The best way to approach it is to consider it in the following way:

 

Person 1 (let's say it's Vinnie) goes through and plays a game with each other member - 159 games, Since there are 159 people other than himself that he could play a game with.  He's now completely done; there's no one else for him to play with, and if anyone else initiated a game with him it would be a repeat, so he sits out.

Person 2 now goes through and plays a game with each other remaining member - 158 more games.  She's now done as well, and sits out.

Person 3 goes through and plays 157 more games with the remaining members, and sits out himself.

 

It continues along these lines until the last two remaining members play one more game, which is the last.

 

The total number of games required is then just the sum of this series:  159 + 158 + 157 + 156 + ... + 3 + 2 + 1, a simple arithmetic series whose sum can be found by S_n = n(a₁ + a_n)/2.

 

n, the number of terms, is clearly 159.  a₁, the first term, is 1, and a_n is also 159.  The sum is then 159*80 = 12,720.