I know the basic Monty Hall problem and that I should decide to change the door after the first negative door is opened.
But let's say my gut feeling says, stay with the door you took and I want to give the door a 50% chance to be picked. Therefore after the first door is opened I toss a coin to decide, whether I switch to the other door or not. If the coin shows head I choose the other door, if it shows tails I stay with the door.
Since then both remaining doors have a chance of 50% to be picked, would that give me an equal chance of winning compared to the solution to always switch?
Or alternative: After the first door is opened there are only two doors left. A stranger comes into the room and doesn't know my vote. He picks one of the two doors that are not opened yet. Wouldn't that give him a 50% chance of winning? is that variant equal to me tossing a coin if I switch or not?
It is always better to switch. The original door wins with probability $\frac 13$ and the other wins with probability $\frac 23$. Your strategy would therefore win with probability $$\frac 12\times \frac 13+\frac 12\times \frac 23=\frac 12$$
One way to get intuition: suppose that after you chose you were always given the choice between sticking with your first choice or taking both of the other two. In that case, it's clear you should switch, no? But that's what is happening here! You already knew that at least one of the other two had no value, so exposing a worthless one provides no relevant information.