There are multiple ways to solve this, but you can give each player the following score :

$\mathbf{S}(x) = \underset{\{i\ :\ \text{x won round i}\}}{\sum \frac{1}{p_i}}=\underset{\{i\ :\ \text{x won round i}\}}{\sum (n- i)}$

With $p_i = \frac{1}{n -i}$ being the probability of winning the round $i$

There are multiple ways to solve this, but you can give each player the following score :

$\mathbf{S}(x) = \underset{\{i\ :\ \text{x won round i}\}}{\sum \frac{1}{p_i}}=\underset{\{i\ :\ \text{x won round i}\}}{\sum (n- i)}$

With $p_i = \frac{1}{n -i}$ being the probability of winning the round $i$. So the earlier rounds count for more than the later ones.

If you want the effect of the round won to be smaller, you can take

$\mathbf{S}(x) = \underset{\{i\ :\ \text{x won round i}\}}{\sum 1 + \frac{1}{k}\frac{1}{p_i}}= \text{number of wins} + \underset{\{i\ :\ \text{x won round i}\}}{\sum \frac{1}{k}(n- i)}$

So you are still counting all the wins of the player, and then adding extra points for winning in the earlier rounds.