[nettneu]

	Originally Posted by D4P To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Am I mistaken somewhere?

Yes. For example, in the scenarios when you chose the wrong door the first time, the probability of getting the right door when you switch to the one that the presenter didn't open isn't 1/2, it's 1.

	Originally Posted by D4P To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Does this make me look stupid?

No. The vast majority of people get this wrong if they haven't encountered similar problems before.

[The Dan Keizer]

Wouldn't you just stick always?

[Malle]

	Originally Posted by INTJRyan To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Of course the original odds are still 1/3. But once one is revealed, I'm not understanding how it's not a 50/50 guess if you do not consider the revealed door at that point. There are only two choices remaining at that point. Is not the game reduced to a coin flip? Perhaps I don't get the game.

I must apologize, I was in error. See, even though I've dealt with this problem before, it's easy to get confused in it
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However, it's not because of the fact that there are only two doors left that it's a coin flip, so to speak, it's because of how he choses which door to reveal.


Okay, so, here's a schematic overview of all the possible outcomes of a game where there is 1 prize behind the doors (as shown in Stage 1: Setup, the checkmark being the prize, the crosses being duds), you choose one door, and since you don't know where the prize is, you're equally likely to pick any of the three possible doors (as shown in Stage 2: Pick a door) after which Monty Hall will reveal one of the remaining doors picking uniformly random between both remaining doors regardless of what's behind them (as shown in Stage 3: Reveal a door)

( Show Spoiler )

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The p = number indicates the probability that if you are in the previous state, you will go to this state next. E.g. the p = 1/3 in stage 2 indicates that the probability that you'll pick a specific door is 1/3 and the p = 1/2 in stage 3 indicates that Monty Hall is equally likely to open either door. To know the probability of a specific outcome, multiply all probabilities leading up to that outcome. Since the probabilities are all the same here, all outcomes in stage 3 have p_tot = 1/3 * 1/2 = 1/6

Now, what happens when Monty opens a door (still at uniform random between theremaining doors) and behind it is a dud? Well, we know that any outcome that has Monty revealing a prize simply cannot happen. Thus, we have this:

( Show Spoiler )

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Now there are 4 equally likely outcomes left. In two of them you win by sticking, two of them you win by switching.

And, for full clarity, if Monty always picks a door behind which there is no prize, we get the following schematic:

( Show Spoiler )

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Now, consider the case where if the door you picked is a dud, Monty Hall reveals the prize with a probability P and the dud with a probability 1-P. Then we get the following:

( Show Spoiler )

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If we play this and Monty Hall reveals a dud, is it better to stick or switch? The probabilities for the outcomes where we stick to win (the two left-most outcomes) are still 1/6 each for a total of 1/3. The possible outcomes where we switch to win are now each 1/3*(1-P) for a total of 2/3*(1-P). These are equal when 1/3 = 2/3*(1-P) or P = 1/2, which is what was shown before (if Monty Hall chooses at uniform random, you're equally likely to win if you stick or switch if he has revealed a dud). However, if P > 1/2, then you're better of sticking and if P < 1/2 you're better of switching. Note that the classical Monty Hall game is the case of P = 0 < 1/2 and we're better of switching.

[roninpro]

One way to understand the Monty Hall problem that might help to clarify the situation is to consider an extreme: suppose there are one million doors, one with a valuable prize, and the other 999999 with junk. You pick a door, keeping in mind that you had one in a million chance that it was the prize. Then, Monty Hall opens up 999998 doors to reveal junk. Now you have either your original door or one other door. Chances are, your original door contains junk, so the rational choice would be to switch doors. (And in particular, you have a 999999/1000000 chance of winning your prize by making the switch.)

This example helped to clarify the situation for me. I hope it helps others.

[PlatoHagel]

	Originally Posted by Tocsin To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I thought the "Monty Hall Problem" was whether or not to grope the show girls and risk a civil lawsuit. To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Only if Monty goes the whole way
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Seriously though "
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" issue came up more then the show.......I don't know why. But as I read the basis of the argument is located in the
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. So the older versions are the same question in Lets Make a Deal?

How one arrives at getting to a clearer point would have meant a revision of the original question with some insight to the original problem? Explained, in Let make a deal? Hmmmm......

( Show Spoiler )

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Ideas On Quantum Interrogation?

[fokalina]

I remember learning about this when i was little. I thought it was cool.

Also, people, please read.

---------- Post added 06-17-2012 at 11:54 PM ----------

	Originally Posted by CrudeHypothesis To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	You can see through this easier if there was 1,000,000 doors. You pick one and the host opens 999,998 of them, leaving just your door and another door. Your chances of picking it right first were 1 in 1,000,000 for some reason out of 999,999 doors, that other door wasn't opened. Still feel like sticking?

Originally Posted by roninpro To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

One way to understand the Monty Hall problem that might help to clarify the situation is to consider an extreme: suppose there are one million doors, one with a valuable prize, and the other 999999 with junk. You pick a door, keeping in mind that you had one in a million chance that it was the prize. Then, Monty Hall opens up 999998 doors to reveal junk. Now you have either your original door or one other door. Chances are, your original door contains junk, so the rational choice would be to switch doors. (And in particular, you have a 999999/1000000 chance of winning your prize by making the switch.) This example helped to clarify the situation for me. I hope it helps others.

It's a good illustration, sure. But it's even the exact same number.

Sorry for being bothered.
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[roninpro]

	Originally Posted by fokalina To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	It's a good illustration, sure. But it's even the exact same number. Sorry for being bothered. To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Ah, I had skimmed the previous posts to see if it was there, but I guess I missed it. CrudeHypothesis beat me to it.

[endless]

I was explaining this to my partner recently, it's curious, I had a hunch that maybe people who play a game want to defy the authoritarian figure that the gameshow host represents by not changing their mind but when presented with a thousand doors in the same manner, one car, 999 goats they would but I'm no longer sure thats the case.

[spronston]

As I tend to prefer visual explanations over purely textural ones, I find the following diagram clearly explains the solution by drawing each of possible outcomes:

( Show Spoiler )

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When you switch (shown in yellow) you will have a 2 in 3 chance of the good prize -- but if you stick (shown in green) you will only have a 1 in 3 chance of the good prize

(I first saw this solution in the wonderful book The Curious Incident of the Dog in the Night-time by Mark Haddon.)

[AlfredSchnittke]

What bearing does this problem have on the Sleeping Beauty Problem?

[scorpiomover]

I love this problem. I've been trying to work on it. But it's incredibly deep.

The first time I came across a problem like it, it was in university. I was in the 1st year. A 3rd year told me about decision theory, and explained the same basic solution using matches. Apparently, it's really, really useful for making decisions of any kind.

Really don't know why INTJs don't love decision theory. After all, it's all about making decisions that are as optimal as possible.

[nettneu]

	Originally Posted by AlfredSchnittke To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	What bearing does this problem have on the Sleeping Beauty Problem?

Not a lot. The Sleeping Beauty Problem is boring because it doesn't have a prize. And once you give it one, it tends to become easy. For example, suppose she gets asked whether the coin was heads or tails, and gets a prize for every right guess, then she does best to say "tails" every time (that only gives her a 50% chance of winning anything, but when she does win she gets two prizes). On the other hand, if the rules are that one right guess is enough to let her marry the handsome prince, then she should make sure that her answers are random - maybe toss a coin herself to choose an answer, rather than risking that any subconscious bias might make her choose the same answer twice. That way she has a 62.5% chance of winning. Or if the only way she can marry the prince is to reach the end of the experiment without a wrong answer, then for herself it doesn't matter what she does, but for the sake of the prince it's probably best if she resolves to always say "tails", because that avoids the possibility of putting him through the extra agony of winning on Monday and then having his hopes dashed on Tuesday.

[EdmontonAspie]

Originally Posted by JTG To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

This problem is named after the host and developer of the old game show "Let's Make a Deal" The rules of the game are as follows: You are presented with three doors, one concealing a good prize You choose one door as yours The host then reveals a consolation prize behind one of the unchosen doors You then choose whether you want to keep your door or swap to the remaining other Obviously without the glimpse of a consolation prize behind one of the other doors, your chance of winning the grand prize would always be 33%. However, does the revelation change your chances? After the swap, is there strategy to be employed, or are you still shooting in the dark except now with a 50/50 chance?

No, it does not.

The door that Monty Hall reveals is one that does not have the grand prize
(Big Deal of the Day), and he already knows it. You had to make your decision
in the first place using a priori odds (1 out of 3), and these odds still apply.
So changing the door is the winning strategy (2:1 odds in your favour).

A more modern version of the Monty Hall problem would be the "Barker's Markers"
problem (after the pricing game on The Price Is Right, renamed "Make Your Mark"*). To win all three prizes
on offer in this pricing game, you (in effect) have to choose the one wrong price
(i.e. the one that doesn't match any prize) out of four prices presented to you
and leave it without one of Barker's markers. Bob Barker (now Drew Carey)
will reveal two of the three correct prices, and then offer you a chance to win an
additional $500 if you believe you are right the first time (and leave the marker where
it is). The alternative is to give the $500 back in order to change the third marker
to the other price. You had three chances out of four of being wrong the first time
(leaving a correct price without a marker) when you made the initial decision. So,
the odds are three to one in favour of giving back the $500 Barker offers and changing
the last marker to the other price.

*the story has it that this pricing game is no longer in circulation, so to speak.

[AlfredSchnittke]

Originally Posted by nettneu To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.

Not a lot. The Sleeping Beauty Problem is boring because it doesn't have a prize. And once you give it one, it tends to become easy. For example, suppose she gets asked whether the coin was heads or tails, and gets a prize for every right guess, then she does best to say "tails" every time (that only gives her a 50% chance of winning anything, but when she does win she gets two prizes). On the other hand, if the rules are that one right guess is enough to let her marry the handsome prince, then she should make sure that her answers are random - maybe toss a coin herself to choose an answer, rather than risking that any subconscious bias might make her choose the same answer twice. That way she has a 62.5% chance of winning. Or if the only way she can marry the prince is to reach the end of the experiment without a wrong answer, then for herself it doesn't matter what she does, but for the sake of the prince it's probably best if she resolves to always say "tails", because that avoids the possibility of putting him through the extra agony of winning on Monday and then having his hopes dashed on Tuesday.

But the sleeping beauty problem hasn't been "solved" officially.

The MHP pretty much has.

How can the SBP be more boring?

It's standard interpretation asks you whether to accept a bet based on the structure. Presumably you must determine some risk-reward.

There are also some issues with some of the standard answers, see:


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[PlatoHagel]

You people might want to check out game theory.

To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts. Game theory is the study of the ways in which strategic interactions among rational players produce outcomes with respect to the preferences (or utilities) of those players, none of which might have been intended by any of them. The meaning of this statement will not be clear to the non-expert until each of the italicized words and phrases has been explained and featured in some examples. Doing this will be the main business of this article. First, however, we provide some historical and philosophical context in order to motivate the reader for all of this technical work ahead......

If you ever have to go into negotiations this could come in handy?
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See also:
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This Nobel Prize award was of interest to me.


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"for having laid the foundations of mechanism design theory"

Leonid Hurwicz, Eric S.Maskin, Roger B. Myerson

[nettneu]

	Originally Posted by AlfredSchnittke To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	But the sleeping beauty problem hasn't been "solved" officially. The MHP pretty much has. How can the SBP be more boring?

I was looking at it from Sleeping Beauty's point of view. It makes no difference whether the thirders or the halfers or neither are right, she just gets drugged and interrogated and gets nothing out of it. I even wonder whether the reason it hasn't been "solved" is that a solution would have no practical significance anyway. Maybe "credence" is simply a flawed concept.

	Originally Posted by AlfredSchnittke To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	It's standard interpretation asks you whether to accept a bet based on the structure.

That's interesting, because the only descriptions I found that involved bets weren't actually the same problem; even people who disagreed on the "credence" value would agree on the relative merits of different bets, and then proceed to explain why the betting situation wasn't a true representation of the problem. But I admit I've not looked exhaustively. Do you have an example of it as a bet which still hasn't been solved (in the sense of identifying the extent to which different betting strategies are relatively profitable)?