So given the circuit and $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ above I would like to see what the probabilities of each possible measurement are.
To do this I tried expanding the circuit as follows:
after first Hadammard and CNOT we get $(\alpha|0\rangle+ \beta|1\rangle) \otimes \frac{1}{\sqrt{2}}(|00\rangle \otimes |11\rangle)$
Then we apply the second CNOT to get: $\frac{1}{\sqrt{2}}(CNOT \otimes I)(\alpha|000\rangle + \alpha|011\rangle + \beta|100\rangle + \beta|111\rangle) = \frac{1}{\sqrt{2}}(\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle)$
Now apply the final Hadamard to get: $(H\otimes H \otimes I)\frac{1}{\sqrt{2}}(\alpha|000\rangle + \alpha|011\rangle + \beta|110\rangle + \beta|101\rangle) = \frac{1}{\sqrt{2}}(\alpha|+00\rangle + \alpha|+11\rangle + \beta|-10\rangle + \beta|-01\rangle) = \frac{1}{2}(\alpha|000\rangle + \alpha|100\rangle + \alpha|011\rangle + \alpha |111\rangle + \beta|010\rangle - \beta|110\rangle + \beta|001\rangle - \beta|101\rangle)$
So now we have $Pr[000]=Pr[011]=Pr[111] = \frac{1}{4}\alpha^2, Pr[010]=Pr[001]=Pr[110]=Pr[101] = \frac{1}{4}\beta^2$
So now if we measure $00$ on the first two qubits:
$Pr[00] = Pr[000] + Pr[001] = \frac{1}{4}$.
And by similar logic $Pr[11]=PR[01]=Pr[10]=\frac{1}{4}$
But then when I implement this circuit in IBM quantum I get that probability of $|10\rangle,|01\rangle$ is $0$ so I'm not sure what I did wrong.