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#### Sam’s Monty Hall Problem is a problem…

azryan

azryan
Total Posts:  14
Joined  30-10-2010

08 November 2010 15:06

sld,

1). Thanks for an answer, though I don’t think you should ‘tell’ people to “Take a course in probability and statistics.” just because they didn’t understand the logic of one simple question everyone on Earth seems to get wrong. If you didn’t mean it insultingly you should have just ‘suggested’ that ‘I might like to take a course like that’ because ‘you found it helpful and interesting’. Something along those lines at least. You came across like you were insulting me.

Again.. it’s the internet so I don’t really know for sure how you meant it and have to guess.

2). You wrote -“Your error is in assuming that Monty Hall doesn’t know what door has the goat and what door has the car.  He isn’t randomly choosing.-”

But isn’t it actually ‘your’ error in assuming it ‘isn’t’ random? Because I promise you I was NOT assuming whether he knew or not…since the ‘story’ doesn’t say. It’s not enough for me that it kinda, sorta, maybe, implies it.

Maybe that’s my problem with it? And that’d be an issue unrelated to the math of it exactly as I suspected?
If anything, I’d be more inclined to think he didn’t know and it was random.

I also mentioned Harris’s footnote that expands the number of doors just like you mentioned. He even used 1,000 doors rather than your 100. In that example it makes it ‘appear’ that Hall ‘must know’ which door the winner is behind based on the improbability of revealing all losers and ending up with only 2 doors.
BUT… I think that just confuses the issue of the ORIGINAL and only scenario actually in question. I don’t have a problem with that follow-up example. I don’t think it is logically the EXACT same problem though.

Also as I already wrote, the result of the 3 door gamble ‘could have been’ that Hall opens a ‘winning’ door that you didn’t pick and the game is over right there. Essentially that would be accepting your first attempted pick and they are just showing what’s behind doors you didn’t pick for whatever reason.

NOTHING in the story says that couldn’t have happened. It seems like you’re telling me (as well as Sam in his book) to accept a fact not in evidence (that Monty knows where the ‘winner’ is). That’s ‘faith’. Something Sam (and all his fans) is logically against.

Let me ask this then…say the ‘winning’ door is the one you first picked (but they didn’t reveal). Then, IF Hall knows that’s the right door (which you say he in fact does), isn’t it random which of the two ‘loser’ doors he opens?

Wouldn’t that make the result either a 1 out of 2 chance, OR…a 2 out of 3 chance, DEPENDING ON…if your first uncounted attempt actually had the ‘winner’?

Maybe the story just needs to be slightly re-worded? Or would you claim it doesn’t need to be changed at all?

[ Edited: 08 November 2010 15:22 by azryan]

azryan

azryan
Total Posts:  14
Joined  30-10-2010

08 November 2010 15:06

Some glitch caused dual posts.
—repost deleted—

Midwest Skeptic

Midwest Skeptic
Total Posts:  10
Joined  11-02-2011

11 February 2011 23:09

Imagine that you are going to play Monty Hall’s game with the 3 doors many times.  There are two basic strategies:

(1) Pick a door and then always stay with that door when given the chance to switch.
(2) Pick a door and then always switch doors when given the chance to switch.

These are the only two possible strategies.  It turns out that strategy (1) has a 1/3 chance of winning and strategy (2) has a 2/3 chance of winning.  Strategy (2) has a better chance because it takes advantage of information given to you in the middle of the game.  When Monty opens a door and shows you a goat - that is actually giving you information.

There are three doors so you would correctly expect that the chance of picking the right door at the start is 1/3.  When the game starts, you figure the chances like this:

Door 1 = 1/3 chance
Door 2 = 1/3 chance
Door 3 = 1/3 chance

A basic rule of probability is that all possible chances at any moment in the game have to add up to 1.

Suppose that you pick door 1 and then Monty open door 2 to reveal goat.  Now we can recalculate the chances.

If you never switched your door, you’d correctly expect to win 1/3 of the time.  So 1/3 is the correct chance number for door 1.  Also, we know door 2 has a goat, so the chance that door 2 has the good prize is zero.  Now we have:

Door 1 = 1/3 chance
Door 2 = 0 chance (goat)
Door 3 = ???

So what’s the chance on door 3?  All chances have to add to 1, so the chance has to be 2/3.

Door 1 = 1/3 chance
Door 2 = 0 chance (goat)
Door 3 = 2/3

If you play the price is right, always switch doors.  2/3 has been proven to be correct mathematically, experimentally, and through computer simulation.

Scruffy

Scruffy
Total Posts:  6
Joined  21-04-2011

12 May 2011 07:34

Midwest Skeptic:

Yes, that is the clearest explanation, and the one Sam probably should have put in his book. Although then the point might have been lost about how easily people can be so convinced that they absolutely know the right answer when they are completely wrong. And thank you to the previous posters for doing just that. Perhaps they would now like to explain to us why Albert Einstein’s theory of relativity must be wrong because there is obviously only one absolute ‘now’ that is common to the entire universe? I mean, obviously… right? The mystics have been saying it for millennia and stupid old Albert thinks he can disprove it with science and math?

[ Edited: 12 May 2011 07:38 by Scruffy]

Lanskey

Lanskey
Total Posts:  1
Joined  13-11-2011

13 November 2011 15:19

Another way to explain it:

The answer reduces down to one simple point:  Only when you have chosen correctly in the first round can a losing alternative be offered in the second round.  And nearly everyone will agree that based on random odds, the chances of a correct pick in the first round is 1 out of 3.  So switching your choice is wrong only 1 out of 3 times making it the best choice.

To understand this better, consider what your first choice will force.  There are only two possible outcomes:

1) Your first pick was the correct door (1/3 of the time) -  it is then necessary to eliminate one of two wrong doors leaving a wrong door (because you already picked the winning door leaving only 2 wrong doors)

2)Your first pick was an incorrect door (2/3 of the time) - it is then necessary to eliminate a wrong door and leave the winning door (because you have already picked one of the wrong doors and only a wrong door will be eliminated.)

Where people get confused is in regards to the odds in the second round by assuming that everything is still random as in the first round.  This is not the case.  In round two, the other door is determined by systematically eliminating a wrong door out of the two remaining doors.  This increases the winning odds of whatever door is left.

Essentially, each door has a 1/3 chance of winning but by always eliminating a wrong door out of the remaining two, you are combining the probabilities of a winning pick from the other two doors into one single remaining choice.

Lance Armstrong

Lance Armstrong
Total Posts:  1
Joined  05-05-2012

05 May 2012 11:05

I agree that the version Harris uses in his book does not make all of the necessary details clear, but a quick google of the problem will illuminate its nature if you have made different assumptions by what he meant with his concise description.

Here is what Harris wrote:

“Imagine that you are a contestant on a game show and presented with three closed doors: behind one sits a new car; the other two conceal goats.  Pick the correct door, and the car is yours.

The game proceeds this way: Assume that you have chosen Door #1.  Your host then opens Door #2, revealing a goat.  He now gives you a chance to switch your bet from Door #1 to the remaining Door #3.  Should you switch?”

Critical to the problem working is the assumption that Monty will, in all cases, open a losing door that is not the door you have already picked.  This is the “standard version”, and Harris does not make that explicit.  However, it’s a famous problem, and we all have access to the internet.  You can accuse Harris of being less than perfectly clear about the nature of the problem, but that’s as far as it goes.

Regarding the problem itself, this explanation may make the solution seem more intuitive:
http://heartheretic.blogspot.com/2012/05/making-monty-hall-intuitive.html

monty video

monty video
Total Posts:  1
Joined  27-12-2012

27 December 2012 10:58

The basic point is that Sam made a mistake in not specifying that Monty always chooses to reveal a goat, and that the mistake should be fixed in future editions of the book..

AntonDrake

AntonDrake
Total Posts:  4
Joined  09-04-2013

14 April 2013 18:08

I agree with the general mathematical solution to the Monty Hall problem that Sam gives, however it’s possible there’s a bit more to the scenario.

In a specifically 3-door game, it seems like we’d have to take into account some meta-game considerations. For instance, if Monty Hall knew that you knew the solution to the Monty Hall problem, and didn’t want you to win, he could basically change the outcome by offering this option… If we’re assuming that Monty doesn’t want you to win, for one thing, he has no incentive to now offer you a fresh chance to win when you’ve already made a losing choice, and conversely he has a very good reason for offering you a chance to make a mistake and move you off of your original 33% EV , when he knows which door wins and you do not. Again this effect is stronger if Monty is aware that you know the mathematical answer to the Monty Hall problem, and allows him to trivially outplay you.

In this case it seems like the best bet in a 3-door game is just make your random pick, take your 33% EV and roll the dice, win or lose.

[ Edited: 14 April 2013 18:12 by AntonDrake]

AntonDrake

AntonDrake
Total Posts:  4
Joined  09-04-2013

14 April 2013 18:15

monty video - 27 December 2012 10:58 AM

The basic point is that Sam made a mistake in not specifying that Monty always chooses to reveal a goat, and that the mistake should be fixed in future editions of the book..

Yes, if Monty Hall has randomly opened a door, and you know that he has randomly opened a door, then it is unequivocally correct to change your pick to the other door.

PalmerEldritch

PalmerEldritch
Total Posts:  1
Joined  03-05-2013

03 May 2013 05:32

AntonDrake - 14 April 2013 06:15 PM
monty video - 27 December 2012 10:58 AM

The basic point is that Sam made a mistake in not specifying that Monty always chooses to reveal a goat, and that the mistake should be fixed in future editions of the book..

Yes, if Monty Hall has randomly opened a door, and you know that he has randomly opened a door, then it is unequivocally correct to change your pick to the other door.

If Monty has randomly opened a door (that happens to contain a goat), then it makes no difference whether you stay or switch

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