Expected number of uniformly random points in unit square is in convex position.

If n points are uniformly generated in a unit square, following famous Erdős–Szekeres theorem. The probability of them be in convex position is $({\textstyle (}\genfrac{}{}{0ex}{}{2n-2}{n-1}{\textstyle )}/n!)}^{2$

My interest is to find the expected number of points be in convex position, and I am enumerating i points in the convex position and add all the possible is.

Here is what I did, $P(i)={({\textstyle (}\genfrac{}{}{0ex}{}{2i-2}{i-1}{\textstyle )}/i!)}^{2}$, the probability of having i points in convex position.

And the expected number of points be in convex position can be achieved via $\sum _{i=1}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{i}{\textstyle )}\ast P(i)$

And I was using WolframAlpha to get an idea of this number, for some reason the expected number is bigger than n, which is impossible

Can someone help me where did I do wrong?

If n points are uniformly generated in a unit square, following famous Erdős–Szekeres theorem. The probability of them be in convex position is $({\textstyle (}\genfrac{}{}{0ex}{}{2n-2}{n-1}{\textstyle )}/n!)}^{2$

My interest is to find the expected number of points be in convex position, and I am enumerating i points in the convex position and add all the possible is.

Here is what I did, $P(i)={({\textstyle (}\genfrac{}{}{0ex}{}{2i-2}{i-1}{\textstyle )}/i!)}^{2}$, the probability of having i points in convex position.

And the expected number of points be in convex position can be achieved via $\sum _{i=1}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{i}{\textstyle )}\ast P(i)$

And I was using WolframAlpha to get an idea of this number, for some reason the expected number is bigger than n, which is impossible

Can someone help me where did I do wrong?