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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.

Controversial

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.

Furthermore, I am not seeing anything in the What topics can I ask about here? or What types of questions should I avoid asking? FAQs that would indicate that it is off-topic and should be closed.

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.

Arithmetic

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(A|B), P(B|A), P(A) and 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.


Update

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:

  1. If the car is behind door one, how likely is it that Monty would open door three?
  2. 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.

So then, 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 1:1.

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.

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  • $\begingroup$ Adam, concerning your update: great you found the solution, bingo! Before door three was opened, all doors had odds 1:2. Now, that only two doors are left, door 1 has odds 1:2 which means door 2 has odds 2:1 :-). I calculated the odds for door 2 directly with that odds form Bayes formula, which was more complicated. $\endgroup$
    – Ute
    Jul 19, 2023 at 9:10
  • $\begingroup$ Thanks, I appreciate it @Ute! $\endgroup$ Jul 19, 2023 at 17:39

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Questions that look for statistical insight related to working with odds, however simple they might seem, would likely be welcome here on CV if not already asked and answered. This isn't one of them (but possibly it could be turned into one with some constructive and thoughtful editing).

You misconstrue the reasons that were offered for closure. They have little to do with "simplicity" and everything to do with duplication and lack of topicality.

  1. Probabilities and odds are equivalent ways of expressing the same quantities. You convert from probabilities to odds and back again as explained, inter alia, at https://stats.stackexchange.com/a/133633/919 (which also covers odds ratios), so that's been asked and answered.

    A good analogy, IMHO, is between expressing temperatures in degrees Fahrenheit (F) and Celsius (C). Would an explanation of global warming on a (hypothetical) climate site merit a repeat just to re-express all temperatures in F rather than C? Perhaps if the purpose is to illustrate temperature conversions it would be appropriate on the Math site, but it wouldn't be suitable on a site for people interested in climate.

  2. You assert the Monty Hall problem has already appeared here on CV. (See https://stats.stackexchange.com/questions/373.) Why shouldn't that be sufficient reason to close a restatement of the question, especially one that asks nothing new?

Thus, you haven't established any reason why any new version of the Monty Hall question should exist.

In fact, wouldn't your argument, if it were accepted, imply anyone could re-ask all of the 100K+ questions related to probabilities that have already appeared here and request that they be re-explained with all probabilities expressed as odds?


BTW, it bears repeating that two illegitimate reasons for not closing a question are (a) "it was allowed before" and (b) "another site allows it." Sometimes off-topic questions remain for historical purposes (the site's policies and norms have changed a little over the years) or because they were never detected by moderators in the first place. And what other sites allow is their own business. If you think Math would accept your question, by all means post it there.

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  • $\begingroup$ It sounds like the crux is about how simple it is to, given an explanation of the Monty Hall problem using the traditional form of Bayes' theorem, deduce the explanation of the Monty Hall problem using the odds form of Bayes' theorem. You propose that it is analogous to converting between Fahrenheit and Celsius. I question whether that is true. To me it seems like it is actually somewhat difficult. Personally, I understand that 1:3 odds corresponds to a probability of 25% probability and 1:4 corresponds to a probability of 20%. However, I am having trouble applying that knowledge. $\endgroup$ Jul 17, 2023 at 23:44
  • $\begingroup$ In looking at this explanation, they took the traditional formula, rearranged it, and then explained what the terms are for that new formula. I find myself confused about whether you would do the same thing if you took an odds centric approach. Maybe you would figure out the likelihood ratio using a different thought process? Something more similar to this from Arbital? I find that explanation from Arital much easier to think about than ones using the traditional formula. $\endgroup$ Jul 17, 2023 at 23:49
  • $\begingroup$ In fact, wouldn't your argument, if it were accepted, imply anyone could re-ask all of the 100K+ questions related to probabilities that have already appeared here and request that they be re-explained with all probabilities expressed as odds? I don't think so. Given how popular the Monty Hall problem is, it seems more worthwhile to answer it from different angles than it would be to answer other questions from different angles. $\endgroup$ Jul 17, 2023 at 23:54
  • $\begingroup$ BTW, it bears repeating that two illegitimate reasons for not closing a question are (a) "it was allowed before" and (b) "another site allows it." Sometimes off-topic questions remain for historical purposes (the site's policies and norms have changed a little over the years) or because they were never detected by moderators in the first place. And what other sites allow is their own business. If you think Math would accept your question, by all means post it there. I agree that they aren't sufficient reasons, but they do seem like arguments that should carry some amount of weight. $\endgroup$ Jul 18, 2023 at 0:00
  • $\begingroup$ I appreciate your thoughtful consideration of the issues, Adam. $\endgroup$
    – whuber Mod
    Jul 18, 2023 at 16:45
  • $\begingroup$ Thank you. Same to you. $\endgroup$ Jul 18, 2023 at 17:39

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