If a tournament consists of 64 teams playing sudden death games until 1 team is declared the winner. How many games, in total, are played during this tournament?
My approach: Teams: 64, 63, 62, ....., 1
Two teams play 1 game, therefore, all together 32 games. But where is the GP???
This is the old tennis ball question. How many tennis balls are needed if the loser takes the tennis ball home as a consolation prize.
Each match eliminates 1 team. Of the 64 teams, 63 need to be eliminated so that a winning team can be declared.
The actual answer for $N$ teams will be $(N-1)$ matches, where if a team is given a bye from one round to the next, that does not count as a match played.
Addendum
Responding to Simran's Comment/Question
$1 + 2 + 4 + 8 + \cdots + 2^{(n-1)} = 2^n - 1.$
As you said in your post, two teams play 1 game. Then, if 64 teams will play, only 32 teams will win that game. For the second game, only 16 teams will remain.
Notice the pattern that the number of teams is divided by 2 every round.
In this case, for only 1 team to win, we need to divide 64 by 64 or $2^{6}$. Since this calculation follows the pattern, then we need 6 matches for only 1 team to win.
Edit:
The geometric sequence in this problem is $a_{n} = \frac{1}{2}a_{n-1}$ or $a_{n} = a_{0}\left(\frac{1}{2}\right)^{r}$. It is given that $a_{0} = 64$ and we need to solve for $r$ such that $a_{n} = 1$.
A solution of this problem is like the following: \begin{align*}a_{n} &\;=\; a_{0}\left(\frac{1}{2}\right)^{r} \\ 1 &\;=\; (64)\left(\frac{1}{2^{r}}\right) \\ 2^{r} &\;=\; 64 \\ 2^{r} &\;=\; 2^{6} \\ r &\;=\; 6\end{align*}
Therefore, 6 games are needed.
