Merkle tree collision probability
Yes, the root has a higher probability of collision.

Take, for example, simple trees with two leaves. Label tree 1's leaves $A$ and $B$, and tree 2's leaves $X$ and $Y$.

The probability that the tree hashes collide at the root is the probability that $\langle h(A),h(B) \rangle \neq \langle h(X),h(Y)\rangle \land h(h(A)||h(B)) = h(h(X)||h(Y))$ plus the probability that $\langle h(A),h(B) \rangle = \langle h(X),h(Y)\rangle$. The probability of the second part is $p^{-2}$ and the probability of the first is $p^{-1}(1-p^{-2})$, for a total of $p^{-1} + p^{-2} - p^{-3}$, which is slightly greater than $p^{-1}$, the probability of a single collision.

That assumes all of your leaves have different values. If the leaves are all the same but one, the probability of collision at the root is proportional to the logarithm (base 2) of the number of leaves times the collision probability of the underlying hash function.