4. Given that lines $P A$ and $P B$ are tangent to the circle $\odot O$ with radius 1 at points $A$ and $B$ respectively, and $|P O|=2$,
$$
\overrightarrow{P M}=2 \lambda \overrightarrow{P A}+(1-\lambda) \overrightarrow{P B} \text {. }
$$

If point $M$ is inside the circle $\odot O$ (excluding the boundary), then the range of the real number $\lambda$ is $\qquad$ .
4. $0<\lambda<\frac{2}{3}$.

As shown in Figure 3, extend $B O$ and $P A$ to intersect at point $N$.
It is easy to see that $P A=P B=\sqrt{3}, \angle A P B=60^{\circ}$.
Thus, $\angle P N B=30^{\circ}$.
Therefore, $\overrightarrow{P M}=2 \lambda \overrightarrow{P A}+(1-\lambda) \overrightarrow{P B}$
$=\lambda \overrightarrow{P N}+(1-\lambda) \overrightarrow{P B}$
$\Rightarrow M, N, B$ are collinear, $\frac{M B}{M N}=\frac{\lambda}{1-\lambda}$
$\Rightarrow \frac{M B}{N B}=\lambda \Rightarrow M B=3 \lambda$.
From $0<M B<2 \Rightarrow 0<\lambda<\frac{2}{3}$.