Why does the chances of particular Feynman diagram occurring reduces by 1% at each photon-electron interaction? I saw a youtube video regarding Quantum Electrodynamics which explained how one can eliminate the Feynman diagrams with complex photon-electron interactions or loops. The guy explained that each time a photon interacts with the electron the chances of that Feynman diagram reduces by 1%. Thus, the diagram involving least photon-electron interaction (or the "vertices" as he called them) is more likely to happen in an electron scattering event. My question is that - Is there any possible reason for this probability reducing?
 A: Each Feynman diagram represents a contribution to the complex-valued probability amplitude that the initial state in the diagram transitions into the final state in the diagram. (What happens “in between” cannot be described classically in terms of particle trajectories.)
Each vertex in a Feynman diagram for a QED process causes an extra factor of the dimensionless fine-structure constant
$$\alpha=\frac{1}{4\pi\epsilon_0}\frac{e^2}{\hbar c}\approx\frac{1}{137}$$
to appear in the diagram’s contribution to the calculated probability for that process, which is proportional to the squared magnitude of the amplitude. This is where the “1% reduction” comes from. Diagrams with more vertices thus contribute less to the probability than those with fewer.
Another way say it is that it is due the fact that the coupling constant between the electron-positron field and the photon field is $e$. In quantum field theory, the electric charge of any particle is a measure of how strongly its field couples to the electromagnetic (i.e., photon) field. Each vertex in a diagram contributes a factor of $e$ to the probability amplitude represented by that diagram, and thus a factor of $e^2$ to the probability itself.
