Monty Hall Problem and Bayes Theorem

It’s quite mind-twisting to think about this problem. But one can learn a good and clear way of thinking by solving this problem.

Citing the description of this problem from Wikipedia:

“Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?”

An intuitive answer is no need to switch. People naturally think it would be 50% to 50% once one goat door is open. However, take a lesson from the book “think fast, think slow”, we cannot let our gut feeling to obscure rigorous analysis. The very fact the host opens a door with the goat has changed the situation, hence, the probability distribution is not even anymore.

If you pick door 1, then the host could open either door 2 or door 3, there are 4 scenarios depicted as follows:

350px-Monty_tree_door1.svg

Apparently, if you switch, the chances of getting the car is 2/3, versus getting the goat is 1/3, hence you should definitely switch.

People say it’s quite troublesome to draw tree diagram like above, is there a more straightforward way to solve this problem swiftly?

Yes, we just need a thinking aid. Instead of rigidly assigning each door a probability of 1/3, we assume door1, which you pick is 1/3 while combining door2 and door 3 together, the probability should be 2/3. Now after the host opens one of them, he changes the situation, the 2/3 chance is transferred to the unopen door.

monty-hall-problem

It’s essentially important to note the distinctive difference of Monty Hall problem and another setting: if the host also doesn’t know which door has the goat behind it, so he randomly opened a door that is not picked by you int he first place. In this setting, half the time, the host also opens a door with the goat, and you get 50-50 chance to switch or to stay.

We should not be satisfied with just solving this problem, either by a rigorous deducing using tree diagram, or a shortcut. I would like to pause and ponder what on earth is “probability”.

It is tied to homogeneity. The chances of the car behind every single door are identical. Therefore, still, there are a lot of people argue that it should be 50-50 after removing one door out of the picture. However, the key thrust here is the host plays a role, altering the original homogeneity.

So I related this Monty Hall problem as a conditional probability problem. Then, the ensuing question I have is what’s the difference between this kind of condition probability versus the famous Bayes theorem?

theorem

Bayes theorem has wide applications to solve problems such as drug user tests.

“Suppose that a test for using a particular drug is 99% sensitive and 99% specific. That is, the test will produce 99% true positive results for drug users and 99% true negative results for non-drug users. Suppose that 0.5% of people are users of the drug. What is the probability that a randomly selected individual with a positive test is a drug user?”

If not equipped with Bayes formula, I would try to think in this way:

Suppose there is a pool of 1000 random people, so there are 5 people who are drug users. the chances this 5 people being selected for the drug test is 0.5%, and being positive is 0.5%*99%*1000 = 4.95. False positive should also be included as the conditional base, which is made up of not only 4.95, but also 995*1% = 9.95. Hence, the probability should be 4.95/(4.95+9.95) =  33.22%.

To apply Bayes, we need to assign events A and B correctly according to what is asked. For the above question, B is a positive test result, A is drug use. If we change the condition, asking “What is the probability that a randomly selected individual who uses the drug is tested positive?” Obviously, this is a no-brainer, as the answer is given, 99%. Would it also be correct if we plug in Bayes formula:

  • p(B) is the probability of drug user – 0.5%
  • P(A) is the probability of testing positive – (5*99% + 995*1%)/1000 = 1.49%
  • P(B|A) is the probability of drug user when testing is positive – 33.22%

P(A|B) = 33.22%*1.49%/0.5% = 99%

My purpose is not to twist mind by reverse plugging in the formula. I want to master the concept and thinking logic. Beyer’s formula sets a good footprint for us to follow a clear thinking map.

Now the question is, could we follow this thinking map to solve the Monty Hall problem since it is also of the type of conditional probability?

The answer is no. In Monty Hall problem, the host altered the probability scenario, while in the latter case, both event A and event B are stochastical or random, hence, we cannot apply Beyer’s way to solve Monty Hall.

 

 

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