There is a 33% chance to pick a winning door and a 66% chance to pick a losing door.
If you pick one of the losing doors, the other losing door is revealed and switching gives you 100% success. That combines to 100% in 66% of cases.
If you pick the winning door one of the losing doors is revealed, switching gives you a 0% chance of success. That combines to 0% in 33% of cases.
Always switching gives you 66% chance of success overall.
Always staying is betting on having picked the correct door when you only had a 33% chance of picking correctly.
For all the pedants out there: the remaining 1% is the chance of summoning the ghost of Monty Hall who then drags you to probability hell. Probably.
Imagine you chose a door, then you are immediately offered to switch your choice to both of the two other doors. If one of the two is the correct one, it will go there automatically and you win.
It should be obvious that switching and betting on two doors is better. But this scenario is actually effectively the same as the original one where a failure door is revealed before you can choose to switch.
First choice had a 33% success chance, second will be 50% let’s switch
after switching the probability becomes 66%. we talked about this one in my theory of probability class, it’s very counterintuitive!
Help a noob out, is that because it’s still 3 but one failure is revealed?
You have Doors “Win” “Lose” and “Lose”
There is a 33% chance to pick a winning door and a 66% chance to pick a losing door.
If you pick one of the losing doors, the other losing door is revealed and switching gives you 100% success. That combines to 100% in 66% of cases.
If you pick the winning door one of the losing doors is revealed, switching gives you a 0% chance of success. That combines to 0% in 33% of cases.
Always switching gives you 66% chance of success overall. Always staying is betting on having picked the correct door when you only had a 33% chance of picking correctly.
For all the pedants out there: the remaining 1% is the chance of summoning the ghost of Monty Hall who then drags you to probability hell. Probably.
Imagine you chose a door, then you are immediately offered to switch your choice to both of the two other doors. If one of the two is the correct one, it will go there automatically and you win.
It should be obvious that switching and betting on two doors is better. But this scenario is actually effectively the same as the original one where a failure door is revealed before you can choose to switch.