Bayes’ Rule and the Monty Hall Problem
http://en.wikipedia.org/wiki/Monty_Hall_problem
Estimating probabilities can be a real problem for humans. Consider the Monty Hall Problem, which has the given situation: there are three doors, with a prize randomly put behind one of them. You choose one door that you hope has the prize behind it. A door you have not chosen is open, which has nothing in it (the door with the car will not be opened in case you chose a door with nothing). Should you switch your guess to the other unopened door?
Probability wise, you should always switch your guess! However, this observation seems very unintuitive. We can use Bayes’ Rule to verify this decision, and also identify where human reasoning goes faulty.
Let B be the event that there is nothing behind the to-be opened door.
Let A be the event that the car is behind your chosen door.
We must calculate P(A|B) and P(NOT A|B) given these two values.
In order to calculate Bayes’ Rule: we must calculate P(B|A), P(A), P(B|NOT A), P(NOT A), and P(B).
P(B) = Probability there is nothing behind the to-be opened door = 1 by problem constraint.
P(B|A) = Probability there is nothing behind the to-be opened door, given that the car is behind your chosen door = 1
P(A) = Probability that the car is behind your chosen door = 1/3
P(B|NOT A) = The probability there is nothing behind the opened door given that the car is behind an unchosen door = 1. Although there is ½ chance that one of the two unchosen doors has the car behind it, the door with nothing behind it is always chosen.
P(NOT A) = Probability that the car is not behind your chosen door = 2/3
Here we insert a check that P(B) is indeed definitely 1 through calculation:
P(B|A) * P(A) = 1 * 1/3 = 1/3
P(B|NOT A)*P(NOT A) = 1 * 2/3 = 2/3
P(B) = P(B|A) * P(A) + P(B|NOT A)*P(NOT A)
= 1/3 + 2/3 = 1
By Bayes’ Rule:
P(A|B) = P(B|A) * P(A) / P(B) = 1/3
P(NOT A|B) = P(B|NOT A) * P(NOT A) / P(B) = 2/3
Since the probability that there is not a car behind your chosen door is higher than the probability that there is a car behind your door given that there is nothing behind the to-be opened door, you are better off switching doors!
We can see that the key part in the decision of changing are the probabilities of A and NOT A. It is important to view the probabilities from the starting state of the problem as opposed to after the door is open. If we view the problem after the door is open, it seems that there’s a 50/50 chance that the car is behind your door, since there is only one car distributed among the two unopened doors. However, if we view the probabilities from the starting state, there is a 1/3 chance the car is behind your door, and a 2/3 chance the car is behind the two other doors. After one of these two other doors is open. It is like the 2/3 chance is focused into that other unopened door. Since the unchosen door now has a 2/3 chance of having the car, you should switch!
The Monty Hall problem just goes to show how important it is to calculate your probabilities instead of placing and keeping bets based on human intuition.
– kfc35