Hawking Radiation specific clarification [duplicate]

Question

If virtual particles on the edge of a black hole pop into existence randomly, wouldnt the likelyhood of a antiparticle generating on the inside of the black hole equal to a particle doing the same? So how is hawking radiation possible.

Answer 1

You are (classically) absolutely correct. We can think of the particle-antiparticle pair of having an equal probability for the two configurations (AP inside, P outside vs P inside, AP outside). However, Hawking radiation is not preferential--only one of the particles needs to escape for the black hole to lose mass and energy. It doesn't matter which particle gets the lucky break--only that one of the pair does.

Eventually, if this happens enough over a sufficiently long period of time, the black hole will have no mass left, a process called black hole evaporation.