Monty Hall Problem - Probabilities of the final choice Is the probability of swapping to the only other available door the same as the probability of the result of a coin flip to determine your last decision? For example, Heads you stay on the current door, Tails you swap doors. As opposed to just swapping without flipping.

No, because the door you did not choose has a higher probability than the one you did choose originally.

You start off with 3 choices and make a selection. Thus, said selection has a $\frac{1}{3}$ chance of being the "prize door".

After they eliminate a door, you're then left with your choice and another door. The difference is that your choice still holds a chance of $\frac{1}{3}$ of being the correct door, while the other has a chance of $\frac{2}{3}$.