If you mean if you started with two closed doors and a car randomly assigned to be behind one of them and a goat behind the other, then thats's right: on your first guess, you would have a $50\%$ chance of getting the car. If you mean if you started with 3 closed doors with the car randomly assigned to one and goats assigned to the other two, then you selected a door, and a goat was revealed behind one of the two other doors, then if you keep your selection, you have a $33.\overline{3}\%$ chance of winning, if you switch, you have a $66.6\%$ chance of winning, and if you flip a coin to decide which remaining door to choose, you have a 50% chance of winning. This 50% can be calculated from the probabilities $P(\text{heads})P(\text{car behind original door}) + P(\text{tails})P(\text{car behind other door}) = .5(.\overline{33}) + .5(\overline{.66}) = .5$ Note that you can do *better* than this $50\%$ by always switching.

In case it's helpful, a useful trick for understanding this intuitively is to imagine there are $1000000$ doors. There is a car behind one door and goats behind the other $999999$. You pick a door and $999998$ other doors are revealed to have goats behind them. You only had a $1$ out of $1000000$ chance ot getting the car on your first try, and if you didn't pick the car on your first try it *must* be behind the other remaining closed door.
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