Interference with a minus sign In Lectures on Physics, by Richard Feynman, pg 3-11, I found the following: 

In case of electrons, the interfering amplitudes for exchange interfere with a negative sign.

I was unable to understand this. On reading further, it seems that electrons have spins unlike alpha particles and that causes them them to interfere in this manner.
I am unable to understand the connection between the statement and the reason given. I would be grateful for any clarity on this matter. 
I didn't truly understand the + sign either, but it made more sense when I thought about it like either this or that happens to reach the final state, there are two ways to get there, so I need to add the probability amplitudes of both to get the final amplitude. (We always use + to indicate or in probability.)
 A: Strictly speaking, Feynman at this point did not give a reason for why the interference is so peculiar for electrons. He just gave a rule. The discussion on pg 3-12 about spins that followed immediately was not to give a reason for the minus sign. There the point was, if you can measure spins, you can distinguish electrons and, therefore, there will be no interference at all (the probabilities and not amplitudes will add up). 
Now to your question. My explanation below is not discussing the same experiment as on pg 3-11 of Feynman. But if you think about it carefully, you will see the connection. The interference occurs when we are talking about identical particles. Consider two identical particles. We describe their state by some probability amplitude. E.g., the square of this amplitude gives you probability of finding one electron at point A and another electron at point B. Imagine that you exchanged the two particles. In this process, something can happen to the amplitude, but only in such a way as to keep the square the same (if we exchange two identical particles, the probability of finding these two particles at points A and B will not change). Mathematically, this means that exchange operation can only affect the complex phase of the original probability amplitude (probability amplitude gets multiplied by $e^{i\varphi}$ with some $\varphi$). Now imagine, you exchange the two particles again. The amplitude again changes (by the same factor $e^{i\varphi}$). However, two exchanges is the same thing as not exchanging at all! So after two exchanges, you must have the original probability amplitude. In the end, this means that there could be two possibilities. The exchange operation keeps probability amplitude the same (trivial case) and the exchange operation multiplies the amplitude by a -1 (non-trivial). Now, this thinking will not lead any further, in a sense it cannot explain why electrons behave the way they do. Experimental evidence suggests that Nature realizes both cases, those particles that do not change the sign of amplitude upon exchange (bosons) and those that do change the sign of amplitude (fermions). Electrons belong to the second category. That's it. This is experimental fact, which we have to accept. At least this is the spirit, in which Feynman explains Quantum Mechanics. We cannot always understand all laws of nature, but we can still describe them (at least we hope to).
And yes, there is a connection to spins. Fermions can only have half-integer spins, bosons can only have integer spins. But this is too deep. Later on pg 4-3 Feynman talks about this relation (apologizing for not being able to explain any better) saying that Pauli derived it using quantum field theory and relativity.
A: This might be something related to particle statistics: Electrons have spin 1/2 and thus belong to the group of fermions. Exchanging fermions causes an additional minus sign for the state. Alpha particles are made up from two protons and two neutrons, all of them spin 1/2 particles. These spins can add up to either zero, one or two and these integer spins belong to the group of bosons. Boson-exchange comes without an additional minus-sign.
