Mythbusters- Monty Hall 3 Door Pick

[aslan]

On the Monty Hall game show there is a three door pick-a-door for a prize segment; it goes like this: The person picks a door. Immediately another door is opened which does not have a prize behind it. Now you are asked it you want to change your choice of door, with only the two remaining doors unopened. Part of the Mythbusters test was to see if people would stick with their original choice, or change. The claim is that most people will stick to their original choice. In fact, in the test, 100% decided to stick with their first choice, confirming that part of the test.

The part that confuses me is the second part. Mythbusters states that it is statistically wiser to switch doors. How can that be? What have I missed? It seems like with two doors remaining, you have a 50/50 chance of winning no matter which door you pick. Why is it that they state that statistically you should always switch doors for a better chance?

Are they saying that there is a 1/3 chance in the initial selection and that once one door is opened, it makes leaves the remaining door with a 2/3 chance and the original door still has a 1/3 chance? This is very difficult to see.

Besides simulation, is there a good way to prove it? It would be useful in creating gambling propositions, if it holds true.

[Dodo]

Besides simulation, is there a good way to prove it? It would be useful in creating gambling propositions, if it holds true.

Ask Jeebus or the Pope? What would Jeebus do?

(Less obtusely, there is no reason to be an independent thinker... always rely on someone else's authority... it has worked for you all of your life so far... why change?)

[zengrifter]

On the Monty Hall game show there is a three door pick-a-door for a prize segment; it goes like this: The person picks a door. Immediately another door is opened which does not have a prize behind it. Now you are asked it you want to change your choice of door, with only the two remaining doors unopened. Part of the Mythbusters test was to see if people would stick with their original choice, or change. The claim is that most people will stick to their original choice. In fact, in the test, 100% decided to stick with their first choice, confirming that part of the test.

The part that confuses me is the second part. Mythbusters states that it is statistically wiser to switch doors. How can that be? What have I missed? It seems like with two doors remaining, you have a 50/50 chance of winning no matter which door you pick. Why is it that they state that statistically you should always switch doors for a better chance?

Ask Jeebus or the Pope? What would Jeebus do?

Actually this question even threw some top math people, as it also threw me ...

Jeebus would switch, the reason being that He would know that most of the time people's first pick would be wrong because they made their pick from a larger set. But once it comes down to only two, picking anew is a 50-50 proposition.

Got it, Einstein?

[Dodo]

I would not, but Jeebus probably would, try to encourage players to buy deals from the banker (moneychanger) and give a (common) reluctant-to-switch player a better proposition than keeping his or her original choice--(partial) insurance for a "reasonable" portion of the deal's payoff.

And the last choice is the offer to switch cases... a definite must if one has not taking a deal before then.

"Las Vegas Casinos with 'Deal or No Deal' Slot Machines": http://traveltips.usatoday.com/las-v...es-103385.html

[sagefr0g]

On the Monty Hall game show there is a three door pick-a-door for a prize segment; it goes like this: The person picks a door. Immediately another door is opened which does not have a prize behind it. Now you are asked it you want to change your choice of door, with only the two remaining doors unopened. Part of the Mythbusters test was to see if people would stick with their original choice, or change. The claim is that most people will stick to their original choice. In fact, in the test, 100% decided to stick with their first choice, confirming that part of the test.

The part that confuses me is the second part. Mythbusters states that it is statistically wiser to switch doors. How can that be? What have I missed? It seems like with two doors remaining, you have a 50/50 chance of winning no matter which door you pick. Why is it that they state that statistically you should always switch doors for a better chance?

Are they saying that there is a 1/3 chance in the initial selection and that once one door is opened, it makes leaves the remaining door with a 2/3 chance and the original door still has a 1/3 chance? This is very difficult to see.

Besides simulation, is there a good way to prove it? It would be useful in creating gambling propositions, if it holds true.

tends to throw me as well. just saying i think zg nailed it. one thing the problem demonstrates is the power of information when it comes to decision making. if the door was opened but a new decision was not allowed then that information would be worthless.
consider if there were a multitude of doors, what?, say a hundred, whatever. and pose the same offer and information (ie. one door opened revealing a goat and the offer to let you change your first choice). the less doors involved in the problem the more valuable the offer and information, no? but for the multitude of doors problem, start opening more and more doors and allowing decisions, makes for more value relative to that information,edit: (where the limiting factor would be getting down to two doors and no more information allowed, sorta thing), no?
edit: aside from actually having an advantage, information is everything, no?

[Canceler]

Are they saying that there is a 1/3 chance in the initial selection and that once one door is opened, it makes leaves the remaining door with a 2/3 chance and the original door still has a 1/3 chance? This is very difficult to see.

I typed out my reply before reading the paragraph I quoted, which nails the answer. Here's my explanation anyway.

The Prize is behind door 1, 2, or 3. Each door has an equal 1/3 chance that The Prize is behind it.

Say you pick door 1. Now, doors 2 & 3 together have a 2/3 chance of having The Prize behind them. WHAT HAPPENS NEXT WILL NOT CHANGE THAT!

I open door 3, and show you The Prize is not behind it. Now door 2 by itself has the 2/3 chance that The Prize is behind it. You should change your guess from door 1 to door 2.

[aslan]

I typed out my reply before reading the paragraph I quoted, which nails the answer. Here's my explanation anyway.

The Prize is behind door 1, 2, or 3. Each door has an equal 1/3 chance that The Prize is behind it.

Say you pick door 1. Now, doors 2 & 3 together have a 2/3 chance of having The Prize behind them. WHAT HAPPENS NEXT WILL NOT CHANGE THAT!

I open door 3, and show you The Prize is not behind it. Now door 2 by itself has the 2/3 chance that The Prize is behind it. You should change your guess from door 1 to door 2.

Thank you. That is the only thing I could think of, but the way you phrased it made it even more clear why that is the case. Opening a door doesn't change a thing, only dumps the odds on the one remaining door.


I decided to prove it and then it became crystal clear. I set up three columns representing three doors in Excel and using a random number function between 1 and 10,000 to generate numbers for 24 rows. I awarded the highest number in each row the winner of the row so that I had 24 winners in the 72 numbers generated. I made column A (Door #1) the choice of the contestant in each case. I then eliminated one of the numbers in either Column B or Column C where it wasn't the highest number of the three. Then when I saw that the contestant had only won 8 times it hit me like a ton of bricks. The other 16 winners had to be contained in the remaining 24 doors that I had not eliminated. Eight wins was all that one would expect a person to have in picking 1 of 3 doors 24 times. It's exactly 1/3 of the time. This means that when the contestant is asked whether he wants to switch doors, he should always switch because his chances of being correct by switching are now 2 to 1, not 1 to 1. Sorry ZG, your wording almost nailed it, but Canceler was right. To be truthful, I could not see clearly why, until I tried my experiment. It's a heck of an illusion and could be a very profitable gambling proposition, or what I would call a trick bet.

[aslan]

Here's an illustration:

	DOOR #1	DOOR #2	DOOR #3
1	WIN	LOSE	ELIMINATE
2	LOSE	WIN	ELIMINATE
3	LOSE	WIN	ELIMINATE
4	LOSE	WIN	ELIMINATE
5	WIN	LOSE	ELIMINATE
6	LOSE	ELIMINATE	WIN
7	LOSE	WIN	ELIMINATE
8	WIN	LOSE	ELIMINATE
9	LOSE	WIN	ELIMINATE
10	LOSE	ELIMINATE	WIN
11	LOSE	ELIMINATE	WIN
12	LOSE	ELIMINATE	WIN
13	WIN	ELIMINATE	LOSE
14	WIN	ELIMINATE	LOSE
15	LOSE	ELIMINATE	WIN
16	LOSE	WIN	ELIMINATE
17	WIN	ELIMINATE	LOSE
18	LOSE	ELIMINATE	WIN
19	LOSE	WIN	ELIMINATE
20	LOSE	ELIMINATE	WIN
21	WIN	ELIMINATE	LOSE
22	WIN	LOSE	ELIMINATE
23	LOSE	ELIMINATE	WIN
24	LOSE	ELIMINATE	WIN

[sagefr0g]

.......... Sorry ZG, your wording almost nailed it, but Canceler was right. To be truthful, I could not see clearly why, until I tried my experiment. It's a heck of an illusion and could be a very profitable gambling proposition, or what I would call a trick bet.

yes i think you are right, ZG almost nailed it and Canceler did. me i still can't see clearly in a physically rational way how the process ends up as it does, but really nothing physically mysteriously changes (edit: albeit the process of the information change and even the process of our observation is a physical process ) only the process of the mathematical expectation changes as our information changes. but yes a heck of an illusion, it reminds me of the conundrum of the split slit quantum dual particle/wave experiment and of Schrodinger's alive/dead cat thang.
so lemme see here, if there were a thousand doors, with 999 goats and one prize., and i was at first allowed to choose one door., then monty came and with pre-knowledge opened 998 doors with goats behind them., then i'm also offered the option to switch doors., then if i stay with my original door., i'm 1/1000 chances for the prize but if i switch doors., i'm 999/1000 chances for the prize?, wowser

[zengrifter]

Oops

[aslan]

Oops

You still made the right decision to switch. PSI?

[SoFlaLaw]

I typed out my reply before reading the paragraph I quoted, which nails the answer. Here's my explanation anyway.

The Prize is behind door 1, 2, or 3. Each door has an equal 1/3 chance that The Prize is behind it.

Say you pick door 1. Now, doors 2 & 3 together have a 2/3 chance of having The Prize behind them. WHAT HAPPENS NEXT WILL NOT CHANGE THAT!

I open door 3, and show you The Prize is not behind it. Now door 2 by itself has the 2/3 chance that The Prize is behind it. You should change your guess from door 1 to door 2.

I have reviewed this Monty Hall Bayes Theorem problem thousands of times. Although I understand the alleged probabilities solution through the application of Bayes Theorem to is underlying facts, I always come back to wondering how this does not become 50:50 after 1 option is displayed without the prize being behind that initial option.

Let me explain my confusion:

1. We have 3 doors, behind 1 there is a prize, and behind 2 there are booby (no jokes please) prizes or no prizes. So, there exists 3 possible winning venues.

2. We select 1 of the doors (for simplicity purposes, assume I select Door 1), hoping to win the prize, at which time we have a 1 out of 3 chance of success. Remember, there are 3 possible winning choices (Door 1 - 33 1/3% chance; Door 2 - 33 1/3% chance; and Door 3 - 33 1/3% chance).

3. Game show host opens 1 of the 2 doors we did not select (either of Door 2 or Door 3, and will assume for example purposes, it was Door 2), and shows that behind it, there was no prize or a booby prize (be nice). We are offered the opportunity to switch from our choice, Door 1, to the remaining door. Should we switch doors and hope for a doubled chance from success from our 33 1/3% chance with Door 1?

Now, applying Bayes Theorem to this dilemna, as Canceler explained, there was a 66 2/3% chance that the prize was behind 1 of Door 2 or Door 3 (which I will refer to as "Door 2/3") that we did not select, and that does not change, so if given a choice to switch, we should switch from our 1/3 (33 1/3% choice of Door 1) to the 2/3 (66 2/3% choice of Door 3 - keeping Door 2's percentage).

Being a lawyer and (amateur economist - earning my B.S. Degree in Economics) I cannot accept the validity of Bayes Theorem being the solution, for the following reasons:

1. There existed three realities under which I would find the prize when making my initial door selection (i.e. - Door 1, Door 2, and Door 3).

2. Each had attached an equal probability of success (33 1/3% each).

3. When the game show host reveals 1 of either Door 2/3, and it does not have the desired prize behind it, then, the initial reality of the prize being behind it is revealed to have been a false reality.

4. The fact that it was a non-option was exposed, and its impact upon the initial decision is now revealed to include a null set (0%). It is as if it never existed.

Consequently, at that point in time, the new reality (actually the ongoing reality) is that the prize was behind either Door 1 (which I selected) or Door 2/3 (whichever being the one not exposed by the game show host, in this example, Door 2 exposed with no prize, Door 3 remaining). Consequently, Door 1 and Door 3 (the remaining door) now each have a 1:1 (50%) likelihood of having the prize behind them, as it has now been proven that the exposed door (Door 2) never existed as a legitimate option for success.

All available data has to be accounted for, and the fact that one of the doors is removed from the equation, changes the equation such that the removed door never existed for purposes of determining which door the prize is actually behind (both now, and when I made my initial selection). To NOT take this information (the nullification of one of the doors having the prize behind it), would be like not taking a third of a shoe into account when keeping a running and/or true count (forgetting unbalanced card counting systems for the moment).

Please explain why my thought process and analysis is incorrect. I would appreciate any educational value that can be imparted such that I see why I should not consider the fact of the removal of 1 option when determining my odds/chances for success, based upon my now existing options.

[sagefr0g]

I have reviewed this Monty Hall Bayes Theorem problem thousands of times. Although I understand the alleged probabilities solution through the application of Bayes Theorem to is underlying facts, I always come back to wondering how this does not become 50:50 after 1 option is displayed without the prize being behind that initial option.

Let me explain my confusion:

1. We have 3 doors, behind 1 there is a prize, and behind 2 there are booby (no jokes please) prizes or no prizes. So, there exists 3 possible winning venues.

2. We select 1 of the doors (for simplicity purposes, assume I select Door 1), hoping to win the prize, at which time we have a 1 out of 3 chance of success. Remember, there are 3 possible winning choices (Door 1 - 33 1/3% chance; Door 2 - 33 1/3% chance; and Door 3 - 33 1/3% chance).

3. Game show host opens 1 of the 2 doors we did not select (either of Door 2 or Door 3, and will assume for example purposes, it was Door 2), and shows that behind it, there was no prize or a booby prize (be nice). We are offered the opportunity to switch from our choice, Door 1, to the remaining door. Should we switch doors and hope for a doubled chance from success from our 33 1/3% chance with Door 1?

Now, applying Bayes Theorem to this dilemna, as Canceler explained, there was a 66 2/3% chance that the prize was behind 1 of Door 2 or Door 3 (which I will refer to as "Door 2/3") that we did not select, and that does not change, so if given a choice to switch, we should switch from our 1/3 (33 1/3% choice of Door 1) to the 2/3 (66 2/3% choice of Door 3 - keeping Door 2's percentage).

Being a lawyer and (amateur economist - earning my B.S. Degree in Economics) I cannot accept the validity of Bayes Theorem being the solution, for the following reasons:

1. There existed three realities under which I would find the prize when making my initial door selection (i.e. - Door 1, Door 2, and Door 3).

2. Each had attached an equal probability of success (33 1/3% each).

3. When the game show host reveals 1 of either Door 2/3, and it does not have the desired prize behind it, then, the initial reality of the prize being behind it is revealed to have been a false reality.

4. The fact that it was a non-option was exposed, and its impact upon the initial decision is now revealed to include a null set (0%). It is as if it never existed.

Consequently, at that point in time, the new reality (actually the ongoing reality) is that the prize was behind either Door 1 (which I selected) or Door 2/3 (whichever being the one not exposed by the game show host, in this example, Door 2 exposed with no prize, Door 3 remaining). Consequently, Door 1 and Door 3 (the remaining door) now each have a 1:1 (50%) likelihood of having the prize behind them, as it has now been proven that the exposed door (Door 2) never existed as a legitimate option for success.

All available data has to be accounted for, and the fact that one of the doors is removed from the equation, changes the equation such that the removed door never existed for purposes of determining which door the prize is actually behind (both now, and when I made my initial selection). To NOT take this information (the nullification of one of the doors having the prize behind it), would be like not taking a third of a shoe into account when keeping a running and/or true count (forgetting unbalanced card counting systems for the moment).

Please explain why my thought process and analysis is incorrect. I would appreciate any educational value that can be imparted such that I see why I should not consider the fact of the removal of 1 option when determining my odds/chances for success, based upon my now existing options.

it also needs to be accounted for that the original choice you made had only a 1/3 chance of being a prize winner or a 2/3 chance of being a prize loser. that fact remains a mathematical reality even after monty exposed the goat and offered another choice. if that fact remains then there is no fifty fifty chance only a 2/3 chance for the door that was not chosen by you and that was not opened by monty.
edit: so really #3 & #4 above isn't shown to be true by the revealing of the goat.
edit: now if there was a late comer to the show who was unaware of anything other than there are two doors remaining unopened, then his mathematical reality would be fifty fifty.
to me, all this seems paradoxical and confusing. i think the weird factor comes from the fact that we are forced to deal with mathematical expectations and perceived realities over time and physical changes. that coupled with the fact that predictions aren't always correct.

[Dodo]

3. When the game show host reveals 1 of either Door 2/3, and it does not have the desired prize behind it, then, the initial reality of the prize being behind it is revealed to have been a false reality.

4. The fact that it was a non-option was exposed, and its impact upon the initial decision is now revealed to include a null set (0%). It is as if it never existed.

The game show host is privy to "information" that prevents him from choosing and revealing the desired prize. If the game show host were to open a door at random (and that door were to reveal a junk choice), then your analysis would be correct. (If the game show host were to reveal the big prize, then the contestant would have a 0/3 chance of getting it whether or not the door choice is changed.) However, the revealed door never has the desired prize behind it and this information allows the advantage player to make a decision (switch door choice) that would increase the likelihood of choosing the best prize. This is why collusion between a table games dealer and player is "frowned upon" by the casino management and owners. The game show host (in this case) is effectively colluding with the contestant and providing a better-than-random-expectation game.

[Canceler]

I liked what Dodo had to say about information, except if this were to happen...

(If the game show host were to reveal the big prize, then the contestant would have a 0/3 chance of getting it whether or not the door choice is changed.)

... I'd change my guess to the door that's open that I can see has the prize behind it.

SoFlaLaw-
It might help to think about sagefr0g's 1000-door version of the situation. You pick one door. Of the remaining 999 doors, I show you the prize is not behind 998 of them. According to you, your chances are now 50/50 whether you change doors or not. You have to ask yourself, does this really make sense?