If your matrices are drawn from standard-normal iid entries, the probability of being positive-definite is approximately $p_N\approx 3^{-n^2/4}$, so for example if $N=5$, the chance is 1/1000, and goes down quite fast after that. You can find an extended discussion of this question [here][1].

You can somewhat intuit this answer by accepting that the eigenvalue distribution of your matrix will be approximately [Wigner semicircle][2], which is symmetric about zero. If the eigenvalues were all independent, you'd have a $(1/2)^N$ chance of positive-definiteness by this logic. In reality you get $N^2$ behavior, both due to correlations between eigenvalues and the laws governing large deviations of eigenvalues, specifically the smallest and largest. Specifically, random eigenvalues are very much akin to charged particles, and do not like to be close to each other, hence they repel each-other (strangely enough with the same potential force as charged particles, $\propto 1/r^2$, where $r$ is the distance between adjacent eigenvalues). 


  [1]: https://mathoverflow.net/questions/118481/the-probability-for-a-symmetric-matrix-to-be-positive-definite
  [2]: https://en.wikipedia.org/wiki/Wigner_semicircle_distribution