You have to be careful with the truth table method. It's easy to get it wrong, as your events might not have equal probability.
For example, we'll label the three doors P, L, R, for the prize door, the leftmost goat door and the rightmost goat door. This gives possibilities
1. You choose P, Monty chooses L - switching loses
2. You choose P, Monty chooses R - switching loses
3. You choose L, Monty chooses R - switching wins
4. You choose R, Monty chooses L - switching wins
This seems to indicate that switching gives you a 50/50 likelihood of winning. In fact, this is the correct truth table to use if Monty doesn't know which door has the prize. In that case, these are all equally likely.
Instead, if he does know where the prize is and won't open that door, this truth table is misleading because the cases where switching loses are less likely. If you choose L, he'll definitely choose R, but if you choose P, he has two options to split the probability.
Similarly, if you model Monty as not knowing what's behind the doors, then the truth table you're thinking of doesn't work. Either way, the truth table on its own isn't convincing.
The thing I find more clear is using Bayes' rule, as it shows exactly how the model of Monty's behavior affects the probability that switching is beneficial.
Let H be the event that your door has a prize and G be the event that Monty shows you a goat door. Then P(H|G) = P(G|H) * P(H) / P(G). Obviously, P(H) = 1/3 and (assuming Monty must pick a door other than yours) P(G|H) = 1, so it just comes down to what P(G) is.
If he always shows a goat, then P(G) 1 and you should switch. If he doesn't know what's behind the doors, it's 2/3 and it doesn't matter whether you switch. If he knows but prefers to spoil the prize, it's 1/3 and you definitely have the prize door.