I initially had like 83 with this.

That's a really nice solution.

Do you know how many points would the sqrt blocks solution score(applied on bigger P values).

Thanks in advance :)

Is there something I'm missing here?

Let's say $$$p=0.2$$$. Then $$$x=3$$$. Let's first count the average number of queries we spend if $$$n=2$$$ if we know there's at least 1 positive sample: we test the first sample(1 query) and if it's positive($$$\frac{5}{9}$$$ chance) we test the other one; $$$\frac{14}{9}$$$ queries on average. If we didn't know there was at least 1 positive sample, it's optimal to first check that(1 query) and then ask $$$\frac{14}{9}$$$ queries with a $$$\frac{9}{25}$$$ chance, for an average of $$$\frac{39}{25}$$$ queries.

Now let's move back to this solution. There is a $$$\frac{64}{125}$$$ chance that the test is negative on a block(1 query). If the test is positive, let's test just the first sample. There are 2 possible outcomes:

1) It's positive($$$\frac{1}{5}$$$ chance): we need to solve the other 2 samples without any knowledge on them, total: 1 query on entire block, 1 query on first sample, $$$\frac{39}{25}$$$ queries on the last 2 samples, $$$\frac{89}{25}$$$ queries in total

2) It's negative($$$\frac{36}{125}$$$ chance): we need to solve the other 2 samples knowing that one of them is positive, total: 1 query on entire block, 1 query on first sample, $$$\frac{14}{9}$$$ queries on the last 2 samples, $$$\frac{32}{9}$$$ queries in total

All in all, on a block $$$\frac{64}{125} \cdot 1 + \frac{1}{5} \cdot \frac{89}{25} + \frac{36}{125} \cdot \frac{32}{9} = \frac{281}{125}$$$ queries are spent in total, and with $$$333$$$ blocks the average number of queries should be $$$748.584$$$, which means the number of points shouldn't be higher than 93 according to the formula in the problem statement.