Question about H-gate on entangled qubits
Assuming that the initial quantum state is $\left| 00 \right\rangle$, the effect of the circuit is
$$
\frac{1}{\sqrt{2}}
\begin{bmatrix}
1 &amp; 0 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 1 \\
1 &amp; 0 &amp; -1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; -1
\end{bmatrix} \cdot
\begin{bmatrix}
1 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; 0
\end{bmatrix} \cdot
\frac{1}{\sqrt{2}}
\begin{bmatrix}
1 &amp; 0 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 1 \\
1 &amp; 0 &amp; -1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; -1
\end{bmatrix} \cdot
\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} =
\begin{bmatrix} 1/2\\1/2\\1/2\\-1/2 \end{bmatrix},
$$
where the rows and columns are indexed by $\left|q_0q_1\right\rangle = \left|00\right\rangle,\left|01\right\rangle,\left|10\right\rangle,\left|11\right\rangle$. When you measure the two qubits at the end, you should get each of the four possibilities with probability $(\pm 1/2)^2 = 1/4$, which is exactly what you have observed.

If, however, you apply the CNOT gate with input roles reversed, you get
$$
\frac{1}{\sqrt{2}}
\begin{bmatrix}
1 &amp; 0 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 1 \\
1 &amp; 0 &amp; -1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; -1
\end{bmatrix} \cdot
\begin{bmatrix}
1 &amp; 0 &amp; 0 &amp; 0 \\
0 &amp; 0 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 0
\end{bmatrix} \cdot
\frac{1}{\sqrt{2}}
\begin{bmatrix}
1 &amp; 0 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; 1 \\
1 &amp; 0 &amp; -1 &amp; 0 \\
0 &amp; 1 &amp; 0 &amp; -1
\end{bmatrix} \cdot
\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} =
\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix},
$$
so in this case you will always measure $\left|00\right\rangle$.