Talk:Monty Hall problem/Arguments/Archive 8: Difference between revisions

Revision as of 02:31, 14 August 2012

Revision as of 02:39, 24 January 2013

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:The point is that people's intuition is what screws them up here. Appealing to a different intuition, even one that happens to produce the correct answer in this case, simply isn't helpful. A very good test of this is to ask the followup version - is it better to switch if the host blindly opens door 3 (without knowing where the car is) and happens to reveal a goat? Fill out the same table above. Think very carefully about the answers to #8 and #9. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 18:01, 7 January 2012 (UTC)

== Where all the confusion stems from -- explained simply. ==

The reason some people are having problems with this is that they are failing to see the difference between a host who ''knows'' where the car is (and will never choose that door), and a host who ''doesn't know'' where the car is (and randomly chooses a door). The two situations are '''completely different''', and yield different probabilities. ''One extra note: the contestant must also be aware of what the host knows and how the host will choose.''[[User:Hellbound Hound|Hellbound Hound]] ([[User talk:Hellbound Hound|talk]]) 05:40, 15 January 2012 (UTC)

:I agree the host knowledge is crucial. However, if the question is "'''should''' the contestant switch?" (which it generally is), what the contestant does or does not know is irrelevant. Specifically, if the host knows what is behind the doors and deliberately shows a goat (and picks between the remaining two doors randomly if the contestant's initial selection is the car) whether the contestant is aware of what's going or not, the contestant '''should''' switch and has a 2/3 chance of winning by doing so. Conversely, if the host opens a door without knowing what's behind the doors and happens to reveal a goat, the contestant's chance of winning by switching is 1/2 (whether the contestant realizes it or not).

:BTW - your conjecture that people arrive at the wrong answer because they are confused about whether the host is acting randomly or with knowledge of what's behind the doors has been experimentally tested (see the paper by Mueser and Granberg cited in the article). They found no significant difference between versions of the problem where this is explicitly clarified vs. those where it's not (e.g. the standard "vos Savant" version). The conclusion here is that it's not a misunderstanding of whether the host is opening a door randomly that confuses people - it's something else. Perhaps the definitive paper on the psychology of the problem is by Ruma Falk (also cited in the article). -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 06:57, 15 January 2012 (UTC)

== An alternative problem ==

At school, my teacher was talking about this problem, and I thought about something... If the candidate first selects a door, then the show host opens a door, the candidate switches door, he will have 66% chance of winning. What if someone else comes in after the host opens, and chooses between the two doors that are not opened yet, and then the original candidate switches doors. They both pick one of the 2 doors, but the other person who came in after the first door was opened, without knowing the candidate's first choice, has 50% chance (he chooses between 2 doors) and the original candidate has 66% chance, because he switches doors. This also works when throwing a coin after the host closed a door, as this makes you choose randomly between the two doors, not considering the other door (works exactly the same as someone else coming in after the host opened a door).

If you want to test this alternative problem, I wrote a script that simulates the 2 situations of switching doors (guy 1) or randomly picking one of the 2 unopened doors (guy 2): http://jsfiddle.net/tszp7/3/show/ [[User:Joeytje50|Joeytje50]] ([[User talk:Joeytje50|talk]]) 10:40, 26 January 2012 (UTC)

:Exactly. Let's say the first player initially chooses door 1 and the host then opens door 3 (after initially randomly placing the car behind one of the 3 doors, knowing what's behind the doors, and choosing which door to open randomly if the player's initial choice happens to be the one hiding the car). What these two problems distinguish is that "probability" is a function of information. The host knows where the car is, so (for the host) the car has a 100% chance of being behind one of door 1 or door 2 and a 0% chance of being behind the other. Given the set up as described, the first player (and the audience) knows enough to conclude the probability the car is behind door 2 is 2/3 - so if the player switches to door 2 (and we repeat this over and over, with the player always switching) the player will win the car about 2/3 of the time. Lacking the information about the initial set up, the second player can only choose randomly between two doors. Even if there is other information available to others allowing them to determine the probability between the two doors is not the same, the probability of a random choice between two alternatives being correct is 1/2. This is easy enough to prove. Say the actual probability the car is behind door 2 is p (for example, we started with 2 doors and the host picks a random number between 0 and 1 and hides the car behind door 2 if this number is p or less). If it's behind door 2 with probability p, the probability it is behind door 1 is 1-p. If you pick randomly between these doors you have a 50% chance of picking either door. So, your total chance of picking the door hiding the car is (p x 50%) + (1-p) x 50%, which is (p+1-p) x 50%, which is 1/2. What this means is that if you pick randomly over and over, you'll win about 1/2 the time overall regardless of what anyone might know about the probability the car is behind either door - even, for example, if there are only two doors and the host ''always'' puts the car behind door 2.

:All three of these are simultaneously true:

:1) the car has a 100% chance of being behind one of door 1 or door 2 and a 0% chance of being behind the other (this is the probability from the host's perspective)

:2) the car has a 1/3 chance of being behind door 1 and a 2/3 chance of being behind door 2 (these are the probabilities given what the first player, and the audience, know)

:3) picking randomly between door 1 and door 2 you have a 50% chance of winning the car, but if you do this repeatedly the audience will see that you win the car 1/3 of the time if you happen to pick door 1 (the first player's original choice) and 2/3 of the time if you happen to pick door 2 (the door the first player could switch to).

:The intermediate condition in the MHP, where the player knows ''something'' about the ___location of the car, is a little unusual. People are accustomed to the notion of knowing nothing about a situation (like the second player), even a situation where there is a 3rd party who knows with certainty (I flip a coin and look at without showing you - even though I now know whether it's heads or tails with 100% certainty you know nothing). Evaluating probabilities where you know only ''something'' is the ___domain of [[conditional probability]]. -- [[user:Rick Block|Rick Block]] <small>([[user talk:Rick Block|talk]])</small> 20:23, 26 January 2012 (UTC)