The Monty Hall problem

I've just rediscovered the so-called Monty Hall problem. Deliciously simple, yet rather counter intuitive.

The scenario is such: you are given the opportunity to select one closed door of three, behind one of which there is a prize. The other two doors hide 'goats' (or some other such 'non-prize'), or nothing at all. Once you have made your selection, Monty Hall will open one of the remaining doors, revealing that it does not contain the prize. He then asks you if you would like to switch your selection to the other unopened door, or stay with your original choice. Here is the problem: Does it matter if you switch?

The answer is yes but it's not obvious why. Intuitively nothing has changed and the probably is still a third.

The key realization is that the when the host opens one of the remaining doors, the host is forced to remove one of the wrong doors from the mix. If you picked a wrong door, which you'll most likely do, a strategy of always switching will land you on the right door in that case.

graph LR;
start((Start))-- 1/3 -->11(Picked the correct door)
start-- 1/3 -->12(Picked the wrong door)
start-- 1/3 -->13(Picked the wrong door)
11-- switch -->2w(Switched to the wrong door)
12-- switch -->2c(Switched to the correct door)
13-- switch -->2c