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Error in the Monty Hall example
 Posted: 17 June 2011 09:33 AM [ Ignore ]
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I think ‘The Moral Landscape’ is brilliant, but there is an error in that book on page 86: the Monty Hall Problem. The error is not to see that the assessment of probability of an event depends upon the current information that the observer has.

For instance, seeing the three doors shut, the observer clearly assigns (all else being equal) one-third probability to each door.  However, if he chooses door #1, and before it is opened then door #2 is opened to reveal the goat, then at that point his information changes, and he should then, on his then current information, assign one-half probability to each remaining closed door.

To say that before door #2 was opened he chose door #1 with 1/3 probability, therefore it must stay fixed at 1/3 even after door #2 is open, and so the balance if you like of 2/3 must all go to door #3, is to fail to appreciate that assignment of probability depends on the information held by the observer.  And in that example, that information changes on the opening of door #2, so the observer’s probability of door #1 should also then change.

In a footnote this effect is exaggerated to 1000 doors.  The claim is that the probability that he assigned originally and correctly to door #1 was 1/1000, which must therefore stay fixed even after 998 doors have been opened, so that all that extra information must then go to the other remaining closed door alone.  This is clearly faulty.

In itself this does not greatly impact the book, except that this example is used to make strong comments about those naive people who fail to reason correctly and think that the probability should be 1/2 for each door after door #2 is opened.  Unfortunately, they are actually correct, which rather spoils the force of the argument.

However, as I say, the book as a whole is a tour de force, and this small blip should not diminish it at all.

—Mike

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 Posted: 17 June 2011 02:22 PM [ Ignore ]   [ # 1 ]
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Greetings, Mr. Finch.

This forum community has migrated to SH’s other site, Project Reason. Really. They told us to.

We just did a thread on the Monty problem. I forget where. You’re welcome to drop in and look for it.

http://www.project-reason.org/forum/

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Delude responsibly.

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 Posted: 18 June 2011 06:03 AM [ Ignore ]   [ # 2 ]
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Thanks for the information.

—Mike

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 Posted: 15 August 2012 11:05 AM [ Ignore ]   [ # 3 ]
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Mike Finch - 17 June 2011 09:33 AM

.......

—Mike

You make some great points and convinced me for a moment.. err, more than a moment.. A few months until I decided to put your argument to some people..  who said the right things for me to think about.  But extremely eloquently done..  I wouldn’t have figured it out without your post

You’re right i’m sure, that probability is based on the information you have..

But, it isn’t 50/50..

And i’m not claiming to fully understand this

but you missed something.  Here’s what you missed.

Let’s use your example, he picks a door1 in round1 and in round2, door#2 is opened

round1, door1.  Probability of it being a goat 2/3   Probability of it being a car is 1/3

So, most likely it’s a goat, and a car is one of the other two doors.

Now it’s round 2.

Monty Hall is not an unbias guy, he knows whether the player chose a goat or a car in round1. And Monty will never reveal a car when he opens a door.

Suppose the player chose a car in round1 door1.  Monty when opening other doors, will choose between them, doesn’t matter which he chooses, both are goats.

But, Suppose the player chose a goat in round1 door1 (this is most likely).
Monty would then look at the other two doors, one is a car the other a goat, and he’ll reveal the goat.

So, if the player wants to get the car, it doesn’t have to be a 1/3 probability like in the first round!

I don’t fully understand it since as you say, it’s based on what’s available.. But there is something in the fact that it’s not an unbias choice.. that monty makes.

And at round1, it is most likely you chose a goat in round1..

If you were right that the probabily is 50/50 it’d mean the probability of the door you chose being a goat, goes down from 2/3 to 50/50/ and I can’t see how that can be either.

Another example illustrating this, Sam gave I think in a speech.

Imagine that there are 100 doors.  Behind one is a car, and behind the other 99 are goats..

The player chooses a door (probabiliy of car is 1/100)

Monty looks at the remaining 99 doors..
If you chose a car, he reveals any 98 of them, all 99 are goats. he leaves the 99th,  a goat.

If you chose a goat(most likely!)  he reveals 98 that are goats. (since he cannot reveal a car). The one left is definitely a car. (if you chose a goat(which you knew in the first round at least, was most likely).

If your thinking is correct about 50/50 then it looks even crazier. the idea that the probability that you picked a car is now 50/50 just because he revealed all those goats (which you knew were there anyway).

Now the chances of the remaining door being a car, are equal to the chances of you picking a goat in that first round (99/100).
That one remaining door is not a random door. It’s a door that if you picked a goat, was particularly left there because it contains the car!  And you know that that’s information for you to take into consideration when you calculate the probability!
i.e. not 50/50!

Many thanks for your post.. I never understood the monty hall problem.. Then I saw your post and thought Ah, there was a misake in it no wonder I didn’t get it.  But when putting your argument to people.. they told me a few things. I thought and thought and now i’m there.  Look forward to your response if you get an email notification and care to give a response!

So, the reason why it’s not 50/50.. Even though there are 2 choices.. Is because it’s like a weighted coin.. is not 50/50.
Of the two options, one of the options is more than likely to be a car. To a probability equal to you choosing a goat in the first round.  There’s a bias.  It’s as 50/50 as a weighted coin.

[ Edited: 15 August 2012 11:23 AM by qarop]
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 Posted: 15 August 2012 12:24 PM [ Ignore ]   [ # 4 ]
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The problem really is that there are two Monty Hall problems! And unless the person who tells it gets it clear, then there is room for endless confusion:

#1 is when after you have picked a door (but it remains closed), then Monty opens one of the two remaining, and it is guaranteed that he will open a door with a goat (because he knows what is behind each door).

#2 is when Monty opens one of the unchosen doors, but he does NOT know that it has a goat (but it happens to be a goat).

If it is #1, then the odds change as Sam Harris says, because the info that Monty knows now comes into the picture. Unfortunately, Sam described #2 in his book (it is a long time since I read it).  Or at least, when I read it, I assumed from his description that the scenario was #2, in which case the odds would remain as they were before Monty opened the door, as I described in my original post.

—Mike

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 Posted: 15 August 2012 12:35 PM [ Ignore ]   [ # 5 ]
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Ah, yes you’re absolutely right.  From the description you give(from your recollection from sam’s book), it looks like Monty doesn’t know. In which case there’s no bias.

Supposing there are 100 doors, and he reveals 98.
Monty as he reveals the doors is near the end, going to be wondering if he’ll run into the car. By unlikely chance, he doesn’t. Once he has revealed all 98, he is as unsure whether it’s the remaining one, as he is whether it’s the chosen one. It is indeed 50/50.

You’re right.. I considered the possibility with monty not knowing whether a car or goat was chosen, and not knowing what’s behind the doors, and also found it to be 50/50 and confirmed it with some mathematical people I asked.

A monty that doesn’t know, is more like.. a guest,  one may as well get rid of monty.  when the description is given like that.

and most of the time Monty would then reveal the car, which may ruin the program causing him to get sacked.

Wikipedia has a “Extended description of the standard version” which gives these assumptions.

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