Adding my own explanation, because I think it clicks better for me (especially when I write it down):
- Pick a door. You have a 66% chance of picking a wrong door, and a 33% of picking the right door.
- Monty excludes a door with 100% certainty
- IF you picked a wrong door, then there’s a 100% chance the remaining door is correct (so the contingent probability is
p(switch|picked wrong) = 100%)
, so the total chance of the remaining door being correct isp(switch|picked wrong)* p(picked wrong) = 66%
. - IF you picked the right door, then Monty’s reveal gives you no new information, because both the other doors were wrong, so
p(switch|picked right) = 50%
, which means thatp(switch|picked right) * p(picked right) = 50% * 33% = 17%
. p(don't switch|picked wrong) * p(picked wrong) = 50% * 66% = 33%
(because of the remaining doors including the one you picked, you have no more information)p(don't switch|picked right) * p(picked right) = 50% * 33% = 17%
(because both of the unpicked doors are wrong, Monty didn’t give you more information)
So there’s a strong benefit of switching (66% to 33%) if you picked wrong, and even odds of switching if you picked right (17% in both cases).
Please feel free to correct me if I’m wrong here.
I work in the risk assessment space, so they are kind of critical to be aware of, for me :)