Question about Ekert protocol of quantum key distribution

recently I’ve self-studied the quantum computing from the book Quantum Computing for Everyone by Chris Bernhardt. Everything is so far so good until I came across with the Ekert protocol which Bernhardt use as example of Bell’s inequality application. Since the concept and notation that Bernhardt use to explain things in the book may differ from conventional textbooks, let me briefly summarise key concept before jumping to the question (for the sake of anyone without the book).

So, let’s start with Alice, Bob and Eve. Alice want to send a key to Bob, using quantum key distribution in form of Ekert protocol. Eve is the person who want to intercept the message. Since in the introduction Bernhardt use electron spin as example a lot in the book, he also use spin for proving Bell’s inequality and expressing Ekert protocol. Thus in this example of Ekert protocol he use a pair of entangled electron for sending qubits:

$frac{1}{sqrt2}|uparrow:rangle|uparrow:rangle:+ frac{1}{sqrt2}|downarrow:rangle|downarrow:rangle$

$|uparrow:rangle$ and $|downarrow:rangle$ are orthonormal basis vector for electron spin in a standard $0^{circ}$ position.

There’re also other 2 orthonormal basis (which Alice, Bob and Eve can use for measurement) which is $(|searrow:rangle,|nwarrow:rangle)$ for measuring electron spin in $120^{circ}$ and $(|swarrow:rangle,|nearrow:rangle)$ for measuring electron spin in $240^{circ}$

the bra-ket notation of all 6 basis can be expressed as followed

$|uparrow:rangle$ = $begin{bmatrix} 1 0 end{bmatrix}$,
$|downarrow:rangle$ = $begin{bmatrix} 0 1 end{bmatrix}$,
$|searrow:rangle$ = $begin{bmatrix} 1/2 -sqrt3/2 end{bmatrix}$,
$|nwarrow:rangle$ = $begin{bmatrix} sqrt3/2 1/2 end{bmatrix}$,
$|uparrow:rangle$ = $begin{bmatrix} -1/2 -sqrt3/2 end{bmatrix}$,
$|uparrow:rangle$ = $begin{bmatrix} sqrt3/2 -1/2 end{bmatrix}$,

In term of qubits, the first basis in each position ($|uparrow:rangle$, $|searrow:rangle$,$|swarrow:rangle$) is correspond to valve $0$ and the second basis ($|downarrow:rangle$, $|nwarrow:rangle$,$|nearrow:rangle$) correspond to valve $1$ in classical bit.

First, Alice will pick 1 out of 3 pairs of orthonormal basis randomly and use them to measure the spin of the entangled electron. Let’s denote each pair of orthonormal basis as followed

$X=(|uparrow:rangle,|downarrow:rangle)$ , $Y=(|searrow:rangle,|nwarrow:rangle)$ , $Z=(|swarrow:rangle,|nearrow:rangle)$

Suppose that Alice has picked up the basis and perform measurement for $3n$ time. Every times she will write down the sequence of basis she randomly choose and the corresponding bit of each measurement. In this way she will end up with 2 string sequence, each with length of $3n$:

$XYYZXYXZZYXZ…$
$101100101101…$

She then send the qubit through sequence channel to Bob. Bob will also pickup the 3 basis-pairs randomly for measurement of receiving qubit. He will write down the sequence of basis he chose and the result of his measurement (so he will end up with 2 sequence with $3n$ length just like Alice.

Alice and Bob will then broadcast the sequence of basis they chose publicly (the sequence $XYXZYX…$) and each will compare the other sequence with their. Since both randomly chose the 1 of 3 basis, there is $1/3$ chance that they will pick the same basis. Since both of their whole sequence have $3n$ length, when discarding the basis that they chose differently, there will be sequence with $n$ length left and the corresponding measured bit to this sequence will be identical between Bob and Alice (the qubit that they used is entangled). If they are sure that Eve is not listening to their communication, they can use this $n$ sequence as key for encryption and decryption.

Now both Alice and Bob have to test that whether Eve is intercepting their communication or not. They do this by examine the rest $2n$ sequence that they chose different basis. If no one make the measurement apart from Alice and Bob themselves, Bell’s inequality states that the proportion that their result of measurement will be identical is $1/4$. However, if Eve make measurement before both Alice and Bob, the proportion of their agreement in result will change to $3/8$ as stated in the book:

From the Bell inequality calculation, they know that if their states are entangled, in each place they should only agree 1/4 of the time. However, if Eve is measuring one of the qubits the proportion of times they agree changes. For example, if Eve measures a qubit before Alice and Bob have made their measurements, it is fairly straightforward to check all the possibilities to show that the proportion of times that Alice and Bob will agree increases to 3/8. This gives them a test for the presence of Eve. They calculate the proportion of agreement. If it is 1/4, they can conclude that nobody has interfered and use the key.

My issue is that I couldn’t derive the number $3/8$ mentioned in the text. I tried to list all possibility when Eve make measurement first and then sending the now unentangled qubit to Alice for measurement. Alice then send that qubit (which state is now altered due to her measurement) to Bob for final measurement. The number which I got is that Alice and Bob will agree for $1/6$ instead of $3/8$

Could anyone show how Eve’s interception cause Alice and Bob to agree in their measurement for $3/8$ when they chose different basis?