Why the variance of Maximum Likelihood Estimator(MLE) will be less than Cramer-Rao Lower Bound(CRLB)? Consider this example. Suppose we have three events to happen with probability $p_1=p_2=\frac{1}{2}\sin ^2\theta ,p_3=\cos ^2\theta $ respectively. And we suppose the true value $\theta _0=\frac{\pi}{2}$. Now if we do $n$ times experiments, we will only see event 1 and event 2 happen. Hence the MLE should be
$$\mathrm{aug} \underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2}$$
where $m_1$ is how many times event 1 happened and $m_2$ is how many times event 2 happened.
Obviously, the solution to the above optimization problem(also it's our MLE) is $\pi/2$ which is a constant and has nothing to do with $m_1,m_2$.
But let's see what is our fisher information, whose inverse should give us a lower bound of the variance of any unbiased estimator:
$$\begin{align}
&\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\cos ^2\theta \right] ^2}{\cos ^2\theta}
\\
&=2\frac{\left[ \partial _{\theta}\frac{1}{2}\sin ^2\theta \right] ^2}{\frac{1}{2}\sin ^2\theta}+\frac{\left[ \partial _{\theta}\cos ^2\theta \right] ^2}{\cos ^2\theta}
\\
&=4\cos ^2\theta +4\sin ^2\theta 
\\
&=4
\end{align}$$
Hence there's a conflict, where do I understand wrong?
Thanks in advance!
Edit
If I calculate MLE not based on the observed data, but instead, based on all the possible outcomes before we do experiments, it should be:
$$\mathrm{aug}\underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2}\left[ \cos ^2\theta \right] ^{m_3}$$
taking ln we will have
$$\left( m_1+m_2 \right) \left( 2\ln\sin \theta -\ln 2 \right) +2m_3\ln\cos \theta $$
taking derivative w.r.t. $\theta$ and set derivative to be zero we will get
$$2\left( m_1+m_2 \right) \frac{\cos \theta}{\sin \theta}=2m_3\frac{\sin \theta}{\cos \theta}$$
Hence if true value $\theta_0=\pi/2$, $m_3$ will always be $0$, and we will always have a conclusion that $\hat\theta=\pi/2$ which will have no variance at all. Hence the variance of $\hat\theta$ is $0$ which against the rule of Cramer-Rao bound.

Variance of your MLE estimator should be computed while still considering the results to be random, i.e.:
$$
\hat{\theta} = \mathrm{arg} \underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2} \left[ \cos ^2\theta \right] ^{m_3}
$$
$$
Var(\hat{\theta}) = Var \left( \mathrm{arg} \underset{\theta}{\max}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_1}\left[ \frac{1}{2}\sin ^2\theta \right] ^{m_2} \left[ \cos ^2\theta \right] ^{m_3} \right) = \dots
$$
where $m_3$ is the number of observation of the event 3. You can't just rule out $\left[ \cos ^2\theta \right] ^{m_3}$ just because you didn't observe the event 3 after the experiment. If you don't know the value of $\theta$, you must consider the possibility of event 3 happening.
Even if after doing $n$ experiments you would observe some events 3, i.e. $m_3 > 0$, the evaluation of the $\mathrm{arg}\max$ above would be a constant (if $m_1$, $m_2$, $m_3$ are some constant numbers). That would not mean the variance of the estimator is $0$ - we just evaluated the estimate accoring to results of experimenting, there is no randomness in it.
What is the variance of the estimator $\hat{\theta}$ of the random variable $\theta$? To compute it, you'd need some assumption about $\theta$ (e.g. which distribution does it come from?). However, what you know by the Cramer-Rao theorem, is that the variance will be higher or equal than $1/4$.
