Bayes is going to win your car.
I am curently being taught Probability as a precursor to Econometrics. The more about probability I learn, the more counter-intuitive it seems to me. Let’s take the Monty Hall Problem we did in class yesterday. It goes something like this:
“You are a part of a game show (hosted by Monty Hall, thus the name) where you are shown three doors. Behind two of these doors is nothing, behind the third is a shiny new car. You, not knowing what’s behind each door are asked to choose one. Mr. Hall, then, opens one of the other two gates, following which he gives you a choice to switch from the one you initially chose to the one left unopened out of the ones you didn’t choose.” The question is should you switch. Quickly running it through your head would probably result in the conclusion that switching doesn’t really make a difference. You start out with each door having a one in three chance of winning you the car and while an un-car-possessing door being opened wouldn’t really make a difference, however, if you were to accomodate that information alongside your preliminary klnowledge of probability, you’d probably say, “yeah, so what? I now have a one in two chance of winning with either door so screw switching”. Or at least that’s how it ran in my head in the 30 seconds that the professor gives you between asking a question and answering it herself. Turns out, NOT. Switching actually, mathematically, increases your chances of winning. And while she demonstrated it using this (the Baye’s theorem) : I am just going to stick to the method a friend (who has nothing to do with economics or maths, academically) used while trying to answer this question when I quizzed him after class. So here we go: Say there are two contestants. You ,with your limited knowledge of probability , always stick to your first choice and Mr. Bayes, the other contestant, always switches. And the two of you take turns at playing this game. Both of you always initially choose door 1. Case 1: The car is behind door 1. Monty Hall opens door 2. You stick to door 1. Mr. Bayes switches to door 3. You win, he loses. 1-0 Case 2: The car is behind door 2. Monty Hall now has no choice but to open door 3. You stick to door 1. Mr. Bayes switches to door 2. You lose, he wins. 1-1 Case 3: The car is behind door 3. Monty Hall now has no choice but to open door 2. You stick to door 1. Mr. Bayes switches to door 3. You lose, he wins. 1-2. So out of the three cases, you win in one and lose in two while Bayes wins in 2 and loses in one. So, switching clearly is a better strtategy. {Mathematically: Let A1, A2, A3 be the events that the car is behind door 1,2 and 3 respectively. Say, you choose door 1. Let B be the event that door 3 is opened. Now we need to compare the probabilty of car being behind door 1 given that door is opened and it being behind door 2 given that door 3 is opened i.e P (A1|B) vs P(A2|B) Applying Baye’s Theorem: we know, P(B|A1)= 1/2 P(A1) = P( A2) = P(A3) = 1/3 P(B|A2)= 1 P(B|A3)= 0 1. P(A1|B) = [ P(B|A1) . P(A1)]/ [ P(B|A1) . P(A1) + P(B|A2) . P(A2) + P(B|A3) . P(A3)] = 1/3 2. P(A2|B) = [ P(B|A2) . P(A2)]/ [ P(B|A1) . P(A1) + P(B|A2) . P(A2) + P(B|A3) . P(A3)] =2/3 Therefore, when you choose door 1 and door 3 is opened, there is a higher chance that the car is behind door 2.}
