In The Feynman Lectures on Physics Vol. III, Section 3-2 The two-slit interference pattern, the reason why "probability amplitudes [do not] interfere if we can know which alternative is chosen" is explained simply and quantitatively for the two-slit experiment, in terms of the laws for combining quantum mechanical amplitudes discussed in the book, and without resorting to philosophical notions, such as "collapse of the wave function."
(Please refer to Figure 3-3. I was not allowed to post it here, unfortunately.)
Let phi_1 = < x | 1 >< 1 | s > be the amplitude for an electron to go from the source to hole 1 and from hole 1 to position x on the backstop, and similarly
Let phi_2 = < x | 2 >< 2 | s > be the amplitude for an electron to go from the source to hole 2 and from hole 2 to position x on the backstop,
Let a = the amplitude for an electron at hole 1 to scatter a photon into detector D1, which we assume by symmetry also equals the amplitude for an electron at hole 2 to scatter a photon into detector D2, and
Let b = the amplitude for an electron at hole 1 to scatter a photon into detector D2, which we assume by symmetry also equals the amplitude for an electron at hole 2 to scatter a photon into detector D1.
Then the amplitude to detect an electron at position x coincident with a photon at D1 is a*phi_1 + b*phi_2, while the amplitude to detect an electron at position x coincident with a photon at D2 is a*phi_2 + b*phi_1. Since these two cases are distinguishable, their amplitudes must each be squared, to find their respective probabilities, before being added together to find the total probability of detecting an electron at position x, which is (See Eq. 3.10)
|a*phi_1 + b*phi_2|^2 + |a*phi_2 + b*phi_1|^2.
To figure out what this distribution in x looks like, you can imagine different cases of 'a' and 'b'. If, for example, you have arranged things carefully so that electrons at hole 1 are almost certain to scatter photons into detector D1 and not into D2 (and visa versa for electrons at hole 2), so that the magnitude of amplitude a is close to 1 while that of b is close to 0, then you get, for the total probability of detecting an electron at position x, something close to
|a*phi_1|^2 + |a*phi_2|^2
which is just the distribution |phi_1|^2 + |phi_2|^2, multiplied by a factor |a|^2. In this case, evidently, there is no interference between phi_1 and phi_2. On the other hand, if you have arranged things so that you are completely uncertain as to whether a photon detected at D1 was coincident with an electron at hole 1 or with one at hole 2 (and similarly for photons detected at D2), so that amplitudes a and b are approximately equal in magnitude, then the total probability of detecting an electron at position x is something close to
2*|a*phi_1 + a*phi_2|^2
which is just the distribution |phi_1 + phi_2|^2, multiplied by a factor 2|a|^2. And in this case there is interference between phi_1 and phi_2.