If the host opens a door truly at random... - you choose door 1, goat -- host chooses door 1, goat --- switch: 1/2 chance of being right --- stay: you're wr

If the host opens a door truly at random... - you choose door 1, goat -- host chooses door 1, goat --- switch: 1/2 chance of being right --- stay: you're wrong -- host chooses door 2, goat --- switch: you are right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 2, goat -- host chooses door 1, goat --- switch: you are right --- stay: you're wrong -- host chooses door 2, goat --- switch: 1/2 chance of being right --- stay: you are wrong -- host chooses door 3, car --- switch: you are right --- stay: you are wrong - you choose door 3, car -- host chooses door 1, goat --- switch: you are wrong --- stay: you are right -- host chooses door 2, goat --- switch: you are wrong --- stay: you are right -- host chooses door 3, car --- switch: you are wrong --- stay: you are right so there are 6 times out of 9 where the host shows you a goat total, and you are right if you switch 3 out of those 6 times. if you stay each of those 6 times, you're right 2 out of the 6 times. there are 4 times the host shows you a goat that you didn't choose, and if you switch you are right 2 out of those times. but in this scenario, you are actually right 2 out of those 4 times if you stay, too. this seems opposite of what i expected. I guess if the host is choosing at random, and you conditionalize on "being shown a goat AND a door you didn't choose" then you're reducing that particular sample space down to be more heavy in the "you got it right originally" scenario, because the "you got it wrong originally" scenarios rule out half the goat doors from the hosts choices we're considering.