Monty Hall: Part Deux
Lately there's a new variant of the Monty Hall problem going around the internet:
You are playing the Monty Hall problem. However, you secretly know one of the goats is the former pet of an eccentric billionaire who lost it and is willing to pay an enormous amount for its return, way more than the car is worth. You really want that goat. The host is unaware of this. After you pick your door, as is traditional, the host opens one door, which he knows doesn't have the car. He reveals a goat, which you can tell is the ordinary goat and not the secretly valuable one. The host offers to let you switch doors. Should you?
If you're not familiar with the original problem, then congrats for being one of the lucky 10,000! (xkcd). Wikipedia is a good place to start (wiki)
If you want to solve the variant yourself, stop reading here, and come back after!
Solution / Spoilers
We'll define events:
- $G$: We select the "good goat" in the original door
- $B$: The "bad goat" is eliminated by the game show host
Solving this is just a straight-forward application of Bayes Rule: $$ P(G|B) = \frac{P(B|G)P(G)}{P(B)} $$ We have a uniform prior on the original selection, so: $$ P(G) = 1/3 $$ And, if we did select the "good goat" first then there is no choice except to eliminate the "bad" goat, thus: $$ P(B|G) = 1 $$ Determining $P(B)$ is more challenging (as always in Bayesian analysis). We'll work through cases of how the game could play out.
Case 1:
- Select: "good goat"
- Eliminate: "bad goat"
- Remaining: "car"
Case 2:
- Select: "bad goat"
- Eliminate: "good goat"
- Remaining: "car"
Case 3:
- Select: "car"
- Sub-case (a)
- Eliminate: "good goat"
- Remaining: "bad goat"
- Sub-case (b)
- Eliminate: "bad goat"
- Remaining: "good goat"
Here we can see that the only cases where $B$ occurs are case 1 and case 3(a). Now we have to be careful here because these cases do not occur with the same frequency. In case 1, the host has no choice and in case 3(a) they do. That is, case 1 occurs with probability $1/3$ and case 3(a) occurs with probability $(1/3)(1/2) = 1/6$
And thus: $$ P(B) = 1/3 + 1/6 = 1/2 $$ Putting it all together, we have: $$ P(G|B) = \frac{(1)(1/3)}{(1/2)} = 2/3 $$ So, there is a 2/3 chance that you've chosen correctly and you should not switch.
Post Hoc Intuition
We can obtain a solution by updating our prior based on the original puzzle. The solution to the Monty Hall problem is that there is a 2/3 chance that the car is behind the remaining door and the original has the goat. Since, the "bad goat" was eliminated, we must have selected the door with the "good goat" with probability 2/3.
I call this type of solution Post Hoc Intuition. Its useful to hone your understanding and makes you sound smart. However, its probably not an advised way of approaching most problem as its pretty easy to fool yourself. Famously, Richard Feynman did stuff like this where he solved problems the Hard Way (tm) in private and then only told others the "easy way" in public to sound smart.
A decent takeaway from Monty Hall problems is to proceed carefully and make fewer "intuitive" guesses.