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I've been scratching my head over it, but as a mere high school student I can't seem to figure it out.

As a big school project I've been doing research after the Monty Hall Problem. I'm trying to figure out whether or not the number of doors in the problem influences the choice of the player (to switch or not to switch doors).

I've calculated that by avarage players chooce to switch 16.3% of the time. I did this by looking at the control groups mentioned in this paper. I took the average of the two control groups: $$\frac{17 \times 35 + 16 \times 67}{35+67} \approx 16.3\%$$

Now my results return (in a test with 20 doors in stead of 3) a 29.3% ratio. (58 test subjects, 17 switchers, 41 non-switchers)

However, I don't know how to prove this result whether this result is significant or not, and I was hoping someone could explain me how.

ilanman
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Master_Grid
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  • Your setup is somewhat similar to an [A/B test](https://en.wikipedia.org/wiki/A/B_testing) (the binomial variant, specifically). However if I understand it right you collected data on "option B" (20 doors) yourself, but got "option A" data (3 doors) from a previous study? So you can formally apply a binomial test, but note that the two populations are not necessarily comparable. (For maximum quality control, you would want [randomized assignment](https://en.wikipedia.org/wiki/Random_assignment).) – GeoMatt22 Jan 08 '17 at 23:33
  • With that caveat, [this post](http://stats.stackexchange.com/questions/178854/a-b-tests-z-test-vs-t-test-vs-chi-square-vs-fisher-exact-test) may be helpful. – GeoMatt22 Jan 08 '17 at 23:37
  • It seems that all the work and variations of the problem got momentum from Marilyn Vos Savant's column in Parade magazine many years ago. The key mistake that people make is to think that by Monte opening a door that does not contain the prize adds no information. So the most common wrong answer given by many including Paul Erdos was that the probability of either door having the prize is now 1/2 so switching doesn't matter. But in fact 2/3 rd of the time switching will win. So the strategy to switch is best. – Michael R. Chernick Jan 09 '17 at 01:35
  • If you increase the number of doors to n and Monte shows you a door that does not contain the prize then the remaining n-2 doors all have the same probability which is greater than yours. so it is best to switch to one of the remaining doors. This advantage decreases as n get large. – Michael R. Chernick Jan 09 '17 at 01:46
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    @MichaelChernick I believe the question is about the impact of $n$ on the "biases and heuristics of intuitive reasoning", i.e. the *psychology* rather than the probability. – GeoMatt22 Jan 09 '17 at 03:25
  • @geoMatt22 I realized that. I was not commenting on the statistical analysis of the psychological testing in the linked paper. I was just giving a little background on the Monte Hall problem and its extension to more that 3 doors. – Michael R. Chernick Jan 09 '17 at 03:33
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    @Michael Chernick wrote "2/3 rd of the time switching will win." Au contraire, mon frère. You have made the "mistake" of assuming that the host always opens a door. What if the host is malevolent, only opening a door when the original guess is correct? – Mark L. Stone Jan 09 '17 at 03:59
  • @ Mark L. Stone That argument is what some mathematicians used to overcome the embarrassment of their criticism of Marilyn and also avoid admitting a mistake. The problem as posed assumes that the door is opened and is never the prize since if the prize door was opened there would be no reason for the contestant to be given the opportunity to switch. Also I doubt that most of those that made the mistake of saying the probability of 1/2 to switch never considered the issue of a malevolent host. It is worth pointing out that in the actual game Monty never gave the contestant a switch option. – Michael R. Chernick Jan 10 '17 at 14:43

1 Answers1

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Your question cannot really be answered as you wrote it. You cannot know if the number of doors impact switching rate because there are only two sets of doors, either the regular amount or your amount. If you tested some people with 3, 7, 21 and 50, for example, then you could test the association between doors and rate.

What you can do is test if the two groups have a different rate. The test you would do is Fisher's Exact Test. In an ideal world, you would have randomized two groups to make sure. It is possible that there were other properties to the groups that made for different results. For example, you subjects may have spoken to each other or watched a movie with the Monte Hall problem in it. There are movies with them.

The most you can answer is whether the two groups are different. Wikipedia has a good article on how to calculate the Fisher Exact Test. Your test could act as partial confirmation that the number of doors matter, but you should state your restrictions.

You chose an excellent problem.

GeoMatt22
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Dave Harris
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