MHB Sorting $8$ Players in $24$ Hours: Is it Possible?

[mathmari]

Hey!

We have $8$ players and we want to sort them in $24$ hours.
There is one stadium. Each game lasts one hour.
In how many hours can we sort them?? (Wondering)

I thought that we could it as followed:

$$\boxed{P1} \ \boxed{P2} \ \boxed{P3} \ \boxed{P4} \ \boxed{P5} \ \boxed{P6} \ \boxed{P7} \ \boxed{P8} \\ \ \ \ \ \boxed{P1} \ \ \ \ \ \ \ \ \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P5} \ \ \ \ \ \ \ \ \ \ \ \boxed{P7} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P5} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P1}$$

So, the best player is $P1$.
These games took place in $7$ hours.

We know that $P5$ is the best player among the players $P5$, $P6$, $P7$ and $P8$.

$$\boxed{P2} \ \boxed{P3} \ \boxed{P4} \ \boxed{P5}\\ \ \ \ \ \ \ \ \boxed{P2} \ \ \ \boxed{P5} \\ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P2}$$

So, the second best player is $P2$.
These games took place in $3$ hours.

$$\boxed{P3} \ \boxed{P4} \ \boxed{P5}\\ \ \ \ \ \ \boxed{P3}\ \ \ \ \boxed{P5} \\ \ \ \ \ \ \ \ \ \ \ \ \boxed{P3}$$

So, the third best player is $P3$.
These games took place in $2$ hours.

$$\boxed{P4} \ \boxed{P5}\\ \ \ \ \ \boxed{P4}$$

So, the $4^{th}$ best player is $P4$ and the $5^{th}$ best player is $P5$.
This game took place in $1$ hour.

$$\boxed{P6} \ \boxed{P7} \ \boxed{P8}\\ \ \ \ \ \ \boxed{P6}\ \ \ \ \boxed{P8} \\ \ \ \ \ \ \ \ \ \ \ \ \boxed{P6}$$

So, the $6^{th}$ best player is $P6$.
These games took place in $2$ hours.

$$\boxed{P7} \ \boxed{P8}\\ \ \ \ \ \boxed{P7}$$

So, the $7^{th}$ best player is $P7$ and $8^{th}$ best player is $P8$.
This game took place in $1$ hour.

Therefore, we sorted the players $P1 \geq P2 \geq P3 \geq P4 \geq P5 \geq P6 \geq P7 \geq P8$ in $7+3+2+1+2+1=16$ hours.

Is it correct?? (Wondering)

 

[mathmari]

Is it maybe an application of Mergesort?? (Wondering)