Freethought & Rationalism Archive

triplew00t · 07-23-2003, 02:55 PM

Anyone ever heard of Monty's dilemma? Assume you have three doors to choose from. One hides a car, the other two, nothing. You are asked to select a door. You pick door number 1. Monty Hall then opens door number two to reveal that there is nothing behind it. The car is either behind door number one or door number 3. You can keep door number 1 or switch. Which do you do? On the surface it appears that there is a 50/50 chance now, but actually your original door has a 1/3 chance of being the door. You anchored that probability when you chose it. Door number 3 has a 2/3 chance of being the car, as the probability that occured in the set of door number 2 and door number 3 (Monty's choices to eliminate) had a 2/3 chance of containing the car. When Monty eliminated door number 2, this probability, instead of being distributed equally between 1 and 3, was distributed only to door 3 when door 2 was actualized as 0.

This seems falacious to me on the surface until I thought of the probability acting as if door number 1 (your choice) is one set and door numbers 2 and 3 (Monty's choices) are another set. I did an experiment to see, using 10 cups with an object under one, and a chooser who did not know which. When he picked one cup, if he kept that cup after I eliminated 8 of the other cups that did not have the object under them, he had the object under his cup 1/10 times (actual test result 2/20). When he switched his choice to the cup that I did not eliminate, but not his original cup, the object was under his cup 9/10 times (actual test resul 19/20, though I imagine that the result would average down to 9/10 if more tests were done. With only 20 tests, the chance of statistical error of one plus or one minus is very high).

-Nero

jpbrooks · 07-23-2003, 03:33 PM

Quote:

Originally posted by triplew00t
Anyone ever heard of Monty's dilemma? Assume you have three doors to choose from. One hides a car, the other two, nothing. You are asked to select a door. You pick door number 1. Monty Hall then opens door number two to reveal that there is nothing behind it. The car is either behind door number one or door number 3. You can keep door number 1 or switch. Which do you do? On the surface it appears that there is a 50/50 chance now, but actually your original door has a 1/3 chance of being the door. You anchored that probability when you chose it. Door number 3 has a 2/3 chance of being the car, as the probability that occured in the set of door number 2 and door number 3 (Monty's choices to eliminate) had a 2/3 chance of containing the car. When Monty eliminated door number 2, this probability, instead of being distributed equally between 1 and 3, was distributed only to door 3 when door 2 was actualized as 0.

This seems falacious to me on the surface until I thought of the probability acting as if door number 1 (your choice) is one set and door numbers 2 and 3 (Monty's choices) are another set. I did an experiment to see, using 10 cups with an object under one, and a chooser who did not know which. When he picked one cup, if he kept that cup after I eliminated 8 of the other cups that did not have the object under them, he had the object under his cup 1/10 times (actual test result 2/20). When he switched his choice to the cup that I did not eliminate, but not his original cup, the object was under his cup 9/10 times (actual test resul 19/20, though I imagine that the result would average down to 9/10 if more tests were done. With only 20 tests, the chance of statistical error of one plus or one minus is very high).

-Nero

I still find the proposed "solutions" to the three door problem perplexing!

spacer1 · 07-23-2003, 03:44 PM

Quote:

On the surface it appears that there is a 50/50 chance now, but actually your original door has a 1/3 chance of being the door. You anchored that probability when you chose it. Door number 3 has a 2/3 chance of being the car, as the probability that occured in the set of door number 2 and door number 3 (Monty's choices to eliminate) had a 2/3 chance of containing the car.

If door 1 is "anchored" as being 1/3 probability from the outset, then why aren't doors 2 and 3 equally "anchored" to the same probability?
If the car isn't behind door 2, then there will be a 50/50 chance of being either door 1 or door 3.

triplew00t · 07-23-2003, 03:53 PM

Spacer,

I also thought the same at first, but after researching the problem, and doing the experiment, I found I was wrong. The probability operates in sets and does NOT fall equally. See Quarter 1, 2003 edition of Skeptic Magizine, or "There are two errors in the the title of this book*" by Robert M. Martin for more.

http://www.mste.uiuc.edu/reese/monty/solution.htm
http://illuminations.nctm.org/lessonplans/9-12/monty/
http://jwilson.coe.uga.edu/EMT668/EM...all/Monty.html
http://www.wps.k12.va.us/jhhs/math/l...tat/monty.html
http://www3.telus.net/timntoni/montyhall.htm

I found all these on a quick google search for 'monty's dilemma' There were over 12 pages of links just like these. All agree with my original explanation.

-Nero

spacer1 · 07-23-2003, 04:05 PM

Nero,

Why does the probability operate in sets? How is door 2 in any way related to door 3, and how are either of them different to door 1? The only difference I can see is that you *chose* door 1, but I don't see how this effects the probabilities before the doors are opened.

spacer1 · 07-23-2003, 04:05 PM

Double post...nothing to see here.

triplew00t · 07-23-2003, 04:13 PM

The assumption is very important. The host of the experiment, or Monty, MUST be benevolent and MUST open the door (or in my case, expose the cups) that DO NOT contain the 'prize'. The host must KNOW which contains it, and expose all OTHERS that DO NOT. This is the only circumstance in which it is unequal as a 1/3 to 2/3 or a 1/10 to 9/10 probability distribution. If the host does not know which door contain what, and has a CHANCE of exposing the prize when eliminating the other door/cups, then the chance of all 3 doors are 1/3 at first. If we are lucky and Monty doesnt ACCIDENTILY open the door with the prize (assuming, for this example that he doesnt know which does), then the chances of each door being the prize after we know that door 2 (say) is empty are equally 50% or 1/2. This result is the same in my experiment, assuming neither I nor the chooser knows which cup contains what. There is an 8/10 chance that I ACCIDENTILY reveal the winning cup in that way, and IF I do not, then the chance of either of the two remaining cups being the correct one is 50% or 1/2.

The presumption that the host KNOWS which door/cup it is, and MAKES CERTIAN NOT to reveal the correct door/cup is very determining in the statistics. Since Monty always knew which door it was, and never revealed the real prize door accidentily (or purposefully) when opening one door, it IS correct to say that the probability sets are 1/3 to 2/3 at the end (or 1/10 to 9/10 in my case). It can only be said that the chance of either door/cup is equal if they only remain NOT exposed by the accident that the host, in his ignorance of the correct door/cup, did NOT reveal it in the elimination process unintintionally.

-Nero

triplew00t · 07-23-2003, 04:19 PM

Look at it this way. When there are ten cups, and my chooser selects one, if he did NOT select the correct cup, it is (by 9/10 chance) among the other 9. I then have to (in 9/10 of the time) AVOID exposing that one cup when I expose 8 of them. By doing this, I couple the probability to that remaining cup. Try doing the experiment yourself. It becomes VERY obvious when you can see it in front of you. I did not believe it until I did it either.

Nero

spacer1 · 07-23-2003, 04:21 PM

Nero,
I didn't realise this was part of the experiment:

Quote:

f you were the contestant, which of the following strategies would you choose, and why?

Strategy 1 (stick): Stick with the original door

Strategy 2 (flip): Flip a coin, stick if it shows heads, switch if it shows tails

Strategy 3 (switch): Switch to the other door

Therefore, forget what I said earlier.

tronvillain · 07-23-2003, 04:57 PM

I seem to recall that a few years ago we had quite a large thread on this, with a lot of Bayes' Theorem.

07-23-2003, 02:55 PM	#1
triplew00t Regular Member Join Date: May 2003 Location: North Carolina Posts: 141	Monty's Dilemma and Lets Make a Deal logic Anyone ever heard of Monty's dilemma? Assume you have three doors to choose from. One hides a car, the other two, nothing. You are asked to select a door. You pick door number 1. Monty Hall then opens door number two to reveal that there is nothing behind it. The car is either behind door number one or door number 3. You can keep door number 1 or switch. Which do you do? On the surface it appears that there is a 50/50 chance now, but actually your original door has a 1/3 chance of being the door. You anchored that probability when you chose it. Door number 3 has a 2/3 chance of being the car, as the probability that occured in the set of door number 2 and door number 3 (Monty's choices to eliminate) had a 2/3 chance of containing the car. When Monty eliminated door number 2, this probability, instead of being distributed equally between 1 and 3, was distributed only to door 3 when door 2 was actualized as 0. This seems falacious to me on the surface until I thought of the probability acting as if door number 1 (your choice) is one set and door numbers 2 and 3 (Monty's choices) are another set. I did an experiment to see, using 10 cups with an object under one, and a chooser who did not know which. When he picked one cup, if he kept that cup after I eliminated 8 of the other cups that did not have the object under them, he had the object under his cup 1/10 times (actual test result 2/20). When he switched his choice to the cup that I did not eliminate, but not his original cup, the object was under his cup 9/10 times (actual test resul 19/20, though I imagine that the result would average down to 9/10 if more tests were done. With only 20 tests, the chance of statistical error of one plus or one minus is very high). -Nero

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07-23-2003, 03:53 PM	#4
triplew00t Regular Member Join Date: May 2003 Location: North Carolina Posts: 141	Spacer, I also thought the same at first, but after researching the problem, and doing the experiment, I found I was wrong. The probability operates in sets and does NOT fall equally. See Quarter 1, 2003 edition of Skeptic Magizine, or "There are two errors in the the title of this book*" by Robert M. Martin for more. http://www.mste.uiuc.edu/reese/monty/solution.htm http://illuminations.nctm.org/lessonplans/9-12/monty/ http://jwilson.coe.uga.edu/EMT668/EM...all/Monty.html http://www.wps.k12.va.us/jhhs/math/l...tat/monty.html http://www3.telus.net/timntoni/montyhall.htm I found all these on a quick google search for 'monty's dilemma' There were over 12 pages of links just like these. All agree with my original explanation. -Nero

07-23-2003, 04:05 PM	#5
spacer1 Senior Member Join Date: Jun 2003 Location: Australia Posts: 564	Nero, Why does the probability operate in sets? How is door 2 in any way related to door 3, and how are either of them different to door 1? The only difference I can see is that you chose door 1, but I don't see how this effects the probabilities before the doors are opened.

07-23-2003, 04:13 PM	#7
triplew00t Regular Member Join Date: May 2003 Location: North Carolina Posts: 141	The assumption is very important. The host of the experiment, or Monty, MUST be benevolent and MUST open the door (or in my case, expose the cups) that DO NOT contain the 'prize'. The host must KNOW which contains it, and expose all OTHERS that DO NOT. This is the only circumstance in which it is unequal as a 1/3 to 2/3 or a 1/10 to 9/10 probability distribution. If the host does not know which door contain what, and has a CHANCE of exposing the prize when eliminating the other door/cups, then the chance of all 3 doors are 1/3 at first. If we are lucky and Monty doesnt ACCIDENTILY open the door with the prize (assuming, for this example that he doesnt know which does), then the chances of each door being the prize after we know that door 2 (say) is empty are equally 50% or 1/2. This result is the same in my experiment, assuming neither I nor the chooser knows which cup contains what. There is an 8/10 chance that I ACCIDENTILY reveal the winning cup in that way, and IF I do not, then the chance of either of the two remaining cups being the correct one is 50% or 1/2. The presumption that the host KNOWS which door/cup it is, and MAKES CERTIAN NOT to reveal the correct door/cup is very determining in the statistics. Since Monty always knew which door it was, and never revealed the real prize door accidentily (or purposefully) when opening one door, it IS correct to say that the probability sets are 1/3 to 2/3 at the end (or 1/10 to 9/10 in my case). It can only be said that the chance of either door/cup is equal if they only remain NOT exposed by the accident that the host, in his ignorance of the correct door/cup, did NOT reveal it in the elimination process unintintionally. -Nero

07-23-2003, 04:19 PM	#8
triplew00t Regular Member Join Date: May 2003 Location: North Carolina Posts: 141	Look at it this way. When there are ten cups, and my chooser selects one, if he did NOT select the correct cup, it is (by 9/10 chance) among the other 9. I then have to (in 9/10 of the time) AVOID exposing that one cup when I expose 8 of them. By doing this, I couple the probability to that remaining cup. Try doing the experiment yourself. It becomes VERY obvious when you can see it in front of you. I did not believe it until I did it either. Nero

07-23-2003, 04:57 PM	#10
tronvillain Veteran Member Join Date: Oct 2000 Location: Alberta, Canada Posts: 5,658	I seem to recall that a few years ago we had quite a large thread on this, with a lot of Bayes' Theorem.

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