4 ms·
No, 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
by kgwgk 6y ago
No, 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!