If there are four doors instead of the usual three that the monty hall problem asks. Would you switch if the game host shows you a door without the prize in it. How can would you figure this out.
let's say you picked the door with the "goat". you have a 1/4 chance of doing this. in this case, switching gives you a 1/2 chance of winning the prize.
let's say you picked the door with the prize. you also have a 1/4 chance of doing this. in this case, swtiching gives you a 0 chance of winning the prize.
finally, let's say you picked a "empty" door. you have a 1/2 chance of doing this. in this case, switching gives you a 1/2 chance of winning the prize.
so you total chance of winning the prize is: (1/4)(1/2) + (1/4)(0) + (1/2)(1/2) = 3/8.
now let's analyze the "don't switch" strategy:
you pick the door with the "goat (booby prize)". you have a 1/4 chance of doing this, and in this case, you never win.
you pick the door with the prize. you have a 1/4 chance of doing this, and in this case, you always win.
you pick an empty door. you have a 1/2 chance of doing this, and in this case, you never win.
total chance of winning the prize is: (1/4)(0) + (1/4)(1) + (1/2)(0) = 1/4.
it's still better to switch.