Which of the following is most likely to be a change address?

reading paper: ‘even s. A randomized protocol for signing contracts [J]. ACM SIGACT news, 1983’, there has been a question why this scheme cannot be directly extended from 1-out-of-2 OT to 1-out-of-n OT. The OT extension in the later paper adopts very complex methods to expand.

Brief description of the original scheme:
Alice is the sender and Bob is the receiver. Alice has messages M0, M1; Bob has choice bit: $b in {0, 1}$；

-Step 1: Alice generates random numbers x0, x1 and sends them to Bob;

-Step 2: Bob generates a random number k and calculates $c_b = x_b + ENC_{PKA}(k)$, and send $c_b$ to Alice;

-Step 3: Alice calculates $k0 = DEC_{ska}(c_b – x0)，k1 = DEC_{ska}(c_b – x1)$, and then calculates $e0 = M0 bigoplus k0$, $e1 = M1 bigoplus k1$, and sends e0, e1 to Bob;

-Step 4: Bob calculate the result: $Mb = eb bigoplus k$.

I understand that this protocol should be directly extended to 1-out-of-N OT protocol by:
Alice is the sender and Bob is the receiver. Alice has messages M1,M2,…, Mn; Bob has choice number $b in [1, n]$;

-Step 1: Alice generates random number x1, …, xn, and send all of them to Bob;

-Step 2: Bob generates a random number k and calculates $c_b = x_b + ENC_{PKA}(k)$，and send $c_b$ to Alice;

-Step 3: Alice calculates: $k1 = DEC_{ska}(c_b – x1),…, kn = DEC_{ska}(c_b – xn)$, and then calculates $e1 = M1 bigoplus k1, …, en = Mn bigoplus kn$, and sends e1,…en to Bob;

-Step 4: Bob calculate the result: $Mb = eb bigoplus k$.

My question is, why do the subsequent papers not do this, but usually use very complex methods to achieve 1-out-of-N OT? Is it for safety or efficiency?