I was lying in bed last night, wondering how the Monty Hall trick works, and finally figured it out by twisting around my perception of the problem.
What is the Monty Hall problem? Consider three doors. Behind one is a car, and behind two others are goats. You are asked to choose one door – if it’s the one with the car, you win. otherwise, you lose.
you choose one car. Then Monty (the gameshow host) opens one of the other doors to show a goat behind it. You are given the chance to change your choice of door to the other closed one – should you?
The answer is Yes. the /intuitive/ answer is that there is a 50% chance either way, but mathematically, there is actually a 66% chance.
Took me a good few minutes to figure out how to verify it. I kept thinking of it as “there is a 33% chance of me picking the right one. door opens. I now have… 50%?”. Couldn’t seem to make the leap for some reason.
That was a result of wrong perception – you need to think of it from the point of view of what is /not/ the right door.
- choose one door. the chance of the car being behind one of the other doors is 66%.
- one of the other doors opens. the chance is still 66%!.
- You now have two closed doors. the door you /did not originally choose/ has a higher chance than the one you did choose, so you should switch.