The answer is (A). We want to show that it is impossible for all the players, after 10 hands, to have zero points again.

Let $v$ be the number of hands won by the first player when everyone finds themselves again with 0 points, and $p$ the number of hands lost by him.

Then, by the rule given in the text on the assignment of points, we find
$$0 = 2k - p$$
By hypothesis we have $k + p = 10$, the total number of hands played; substituting into the previous equation we find $0 = 2k - (10 - k)$, that is $10 = 3k$.

The only solution to this equation is $k = \frac{10}{3}$, absurd since $k$, representing the number of hands won by the first player, should be an integer. The assumed situation is therefore impossible, and the probability that all players have 0 points is thus 0.