Does NTRU decrypt correctly now?
The likelihood of a decryption failure can be made arbitrarily small. IEEE P1363.1 says in appendix A.4.10:

For ternary polynomials with $d$ $+1$s and the same number of $-1$s, the chance of a decryption failure is given by [B30]:
$$\operatorname{Prob}_{(q, d, N)}(\text{Decryption fails}) = P_{(d, N)} \left( \frac{q - 2}{6} \right)$$
where
$$P_{(d, N)}(c) = N \times \operatorname{erfc} \left( \frac{c}{\sigma\sqrt{2N}} \right)$$
and
$$ \sigma(d, N) = \sqrt{\frac{8d}{3N}}$$

where $\operatorname{erfc}(x)$ is the Gauss error function.
As a practical example, for the EES1087EP2 parameter set where $N=1087$, $q=2048$, and $d=120$, the failure probability is $5.62·10^{-78}$, which is a bit less than $2^{-256}$. Those parameters have been chosen for a $256$-bit security level in general, and the failure probability is also smaller than $2^{-256}$. So exploiting a decryption failure requires just as much work as breaking other parts of the system.
Ps. The [B30] reference is to the paper &quot;Hybrid
Lattice reduction and Meet in the Middle Resistant Parameter Selection for NTRUEncrypt&quot; by P. Hirschhorn, J. Hoffstein, N. Howgrave-Graham, J. Pipher, J. H. Silverman and W. Whyte.