3 ms·
For the “The host knowledge and deliberate choice are irrelevant, if he opened a random door nothing would change.” crowd: If he’s going to open a random door
by kgwgk 6y ago
For the “The host knowledge and deliberate choice are irrelevant, if he opened a random door nothing would change.” crowd:
If he’s going to open a random door he may just leave it to you (because you don’t know anything).
The problem becomes:
0) you’re presented three doors, there is a car behind one of them
1) you pick one door
2) you choose one of the other doors and open it: there is a goat
Now, you have the choice between keeping your initial door and switching to the other unopened door.
- the_af 6y agoEven in this scenario, you should switch too. Failing to switch is equivalent to saying "I chose correctly the first time", which has 1/3 probability of being right.
- kgwgk 6y agoNo, in this sceneario it has a 1/2 probability of being right. P(I chose correctly the first time | all doors are closed) = 1/3 but P(I chose correctly the first time | I opened one of the other doors at random and there was a goat) = 1/2 Without loss of generality, we can say that I picked door A and opened door B findind a goat. P(car@A | goat@B) = P(car@A and goat@B) / P(goat@B) = P(goat@B | car@A) P(car@A) / P(goat@B) = 1 x 1/3 / 2/3 = 1/2 In case it's not clear where the formulas above come from: P(car@A and goat@B) = P(car@A) P(goat@B | car@A) is equal to P(goat@B and car@A) = P(goat@B) P(car@A | goat@B)
- the_af 6y agoHmmm. Makes sense. I iterated a simulation of 10000 cases, discarding every case where the second random door reveals a car (because this breaks the game; at this point there is nothing to guess anymore) and it gives a probability of winning of only 0.42 if you switch. So it's neither 1/2 nor 2/3. Curious!