I don't think so since nowhere is it claimed to be random. It spells it out in the narrative.

The reason that the host's selection not being random is so important is that it makes the second choice dependent on the first. This is what I couldn't get the first time I saw this. I kept insisting that the second choice was independent of the first when it actually isn't due to the host knowing he is picking a door with nothing (or a goat) and the fact that he cannot pick the door you chose the first time.

It's not a trick, it's an integral part of the problem. The fact that an incorrect door is eliminated tilts the odds in your favor. It's often described as the "monty hall" problem, even though Monty never did anything exactly like this.

If you'd like a full explanation, check out this link.
http://en.wikipedia.org/wiki/Monty_Hall_problem

Off the subject, but is this how 3 Card Monty is done to steal people's money in Nawlins?

After much stewing i have come to this conclusion:

I was under the idea that this was supposed to be some random thing that defies probabilities and statistics and is very counterintuitive.

The truth is that it isnt random at all, and it is setup to turn out the way it does.

It just deals with the odds that you picked the wrong door are 2/3 so if you choose A, then 2/3 of the its B or C. If B is taken away its more likely that it is C.

Funny thread.

Ton of stubborn ppl.

This probably isn't adding much to the conversation, but just to type my $.02:

You have 3 doors. 1 has a prize. You pick a door. You have a 1/3 chance you picked the right door. The hosts opens a door without the prize. The original door you picked still has a 1/3 chance of being correct. That is why, there is a 2/3 chance if you switch, you win.

The whole 1000 door thing is a great way of showing it.

So it looks like Howdy was right it isn't random, that's why the proof works.

It's all about how the question is asked. If it were asked another way, then the choices would be mutually exclusive.

somebody write an app to prove this.

The key here is that the game show host always picks a door that you have not picked. This provides you with more information about the remaining door than it does your original choice, and ultimately skews the game to your favor. If the game show host was equally likely to open your first choice as he is the other two, then the second choice would be a 50/50 toss up.

Think of it this way: you pick 1 door. The game show host tells you you have the decision to stick with your choice or chose BOTH of the other doors. Which one are you going to pick?

[This message has been edited by BusterAg (edited 10/19/2005 11:54p).]

It works because the Host ALWAYS picks a wrong door. If the Host picked a door at random then the two choices would be mutually exclusive.

I don't really see what the big deal is about this paradox simply because it's obvious that it isn't random. Howdy was right from the start, it only works because it isn't random.

You got it Vander.

I studied this stupid thing for like 3 hours the other day before i figured out it was basically a trick, set up to work out the way it does. It isnt random thats why it works.

so in "Deal or No Deal" should you always switch your case at the end?

I remember talking about this question with some friends a few years ago. I don't want to go into that **** again.

quote:

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so in "Deal or No Deal" should you always switch your case at the end?

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Yes.

The odds of that case you first picked is 1mil is 1/24

The probability of it not being your first case is 23/24.

(It is 24 cases right?)

2 years later and I'm still posting on this thread.

quote:

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so in "Deal or No Deal" should you always switch your case at the end?

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Nope: You have a 50/50 chance. It's a completely different game.

How is it different?

head hurts! no likey!

will the case take off?

the door is moot!

Think about it in reverse and you may understand better.

If you first get to pick 2 doors, then your chances are 2/3. You would not want to switch because you had a 2/3 shot when you picked the first 2 doors.

The Deal or No Deal problem is exactly the same. You always switch.

this is a simple Bayes Rule problem. You losers should all take ECON 449 with Truocy and go find something better to do with your lives. Yawn.

quote:

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this is a simple Bayes Rule problem. You losers should all take ECON 449 with Truocy and go find something better to do with your lives. Yawn.

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Got an A in it... just like the rest of my econ classes.

Are the three doors on a conveyor?

It's an issue with semantics, nothing else.

Yes if you have 3 doors and you choose 1, you have a 2/3 chance of being wrong (and a 1/3rd chance of being correct).

But to then say you should choose one of the other two doors because they have a 2/3rd chance of winning is incorrect because you are only gettin to choose ONE out of the remaining 2 doors (.6666 * .5 = .333 or 1/3rd --- the same chance you originally had with your first pick).

It's only the wording that makes it hard for people to understand, as it makes them think that they are better off choosing from one of the other two doors, when thats not the case at all (they would still only have a 33.3% chance of picking the right door).

Your chances only improve to 2/3rds if you are allowed to choose BOTH doors that you didn't choose.

But you ARE choosing the other two doors, since one of those two doors is opened to show that it is not the correct one.

[This message has been edited by RockOn (edited 2/1/2008 1:06a).]

I realize that, but the way the article is initially written (and why it is confusing to most) is that it makes it sound like that without knowing what is behind any of the 3 doors, you are better off swapping the one you initially "picked" for one of the two that you didn't pick.

TexAgs is dense.

You missed the whole question part, which is in the thread.