Recently I asked the question: How to reason about the Monty Hall problem with odds ratios? It was closed due to being deemed off-topic. I'd like to challenge that decision.
The two reasons that were provided for why it was deemed off-topic are 1) being controversial and 2) being about arithmetic, not statistics or machine learning.
From the initial comment:
This looks like you are merely asking the Monty Hall problem, which--because it is notorious for the disputes arising around it--doesn't work here on CV.
However, I am seeing many other questions about the Monty Hall problem that haven't been deemed off-topic and closed. So then, the established precedent seems to be that questions about Monty Hall problems are on-topic.
Tangentially, the Mathematics StackExchange has a
monty-hall tag indicating that over there, it is not only deemed to be on-topic but also popular and encouraged.
This I'm not sure of. Looking at the traditional forumla for Bayes' rule and the odds form, it doesn't seem obvious how to go from one to the other. In particular, the former has four terms (
P(B)) whereas the latter only has three (prior odds, likelihood ratio, and posterior odds).
In my experience on other Stack Exchange sites, the sense I get is that questions that are much simpler than this one are welcomed. However, I could be wrong about how simple this actually is, or about whether there is a higher bar here on CV, so I am happy to defer to the judgement of others here.
I think I figured out the answer to my original question. This fact seems tangential to the question of whether the original question should be closed as off-topic, however.
I don't think this is high quality enough to serve as an answer, but here's my thinking.
In performing a Bayesian update by utilizing odds ratios, first you determine what the prior odds are, then you figure out the likelihood ratio, and then to determine the posterior odds, you multiply the prior odds by the likelihood ratio.
Suppose you choose door number one and Monty opens door number three. The question we are trying to answer here is, given that Monty opened door number three which revealed a goat, what are the odds that door number one has the car? (Well, technically this question needs to be worded more precisely.)
The prior odds (before Monty opens the door) of the car being behind door number one are
1:2. It is equally likely to be behind each of the doors, and door one is one of three doors.
To determine the likelihood ratio, we have to ask ourselves two questions:
- If the car is behind door one, how likely is it that Monty would open door three?
- If the car isn't behind door one, how likely is it that Monty would open door three?
Let's explore each.
If the car is behind door one, how likely is it that Monty would open door three?
In this scenario, Monty will randomly choose a door to open amongst door two and door three, so there is a 50% chance here that he opens door three.
If the car isn't behind door one, how likely is it that Monty would open door three?
If the car isn't behind door one, that means it's either behind door two or door three.
There's a 50% chance that it's behind door two. If it is behind door two, there is a 100% chance that Monty opens door three.
There's a 50% chance that it's behind door three. If it is behind door three, there is a 0% chance that Monty opens door three.
0.5 * 1 + 0.5 * 0 = 0.5 + 0 = 0.5 tells us that, if the car isn't behind door one, there is a 50% chance that he opens door three.
With that, we can come up with our likelihood ratio of
50% : 50% or
Now that we have our prior odds of
1:2 and our likelihood ratio of
1:1, we can multiply them to get our posterior odds of
1:2 that door one contains the car.
Intuitively, the way I think about it is something like this:
We started out thinking that there's a 1/3 chance it's behind door one and a 2/3 chance that it's behind either door two or door three. Monty opened door three, revealing that there is a goat behind it. It's still the case that there is a 2/3 chance of the car being behind either door two or door three. But we now know that door three doesn't have the car, so it's a 2/3 chance of being behind door two and 0% chance of being behind door three.
Wikipedia also gives a similar explanation, but rather than focusing on the odds of
is behind door one-to-
isn't behind door one, it looks at the odds of
is behind door one-to-
is behind door two-to-
is behind door three.