Riddle Me This, Batman!

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AuthorTopic: Riddle Me This, Batman!
The Establishment
Member # 6
Profile #0
I asked this puzzle a while back, but it came up again. Let's see how people around here fare this time around. Try to answer before reading the posts below (or search for the answer) as the answer might come up.

You are on a game show and you are given the option to select 1 of 3 doors. Behind one of the doors is a big prize and behind the other two is nothing.

You select door number 1. The host, as per the rules of the game, then opens a door that does not contain a prize nor the one you selected. He opens door number 3, revealing an empty door. Do you:

1) Stay with door 1
2) Switch to door 2
3) It does not matter either way

[ Tuesday, August 15, 2006 04:47: Message edited by: *i ]

Poll Information
This poll contains 1 question(s). 55 user(s) have voted.
You may not view the results of this poll without voting.

function launch_voter () { launch_window("http://www.ironycentral.com/cgi-bin/ubb/ultimatebb.cgi?ubb=poll;d=vote;pollid=WKLWhoaQtcCs"); return true; } // end launch_voter function launch_viewer () { launch_window("http://www.ironycentral.com/cgi-bin/ubb/ultimatebb.cgi?ubb=poll;d=view;pollid=WKLWhoaQtcCs"); return true; } // end launch_viewer function launch_window (url) { preview = window.open( url, "preview", "width=550,height=300,toolbar=no,location=no,directories=no,status,menubar=no,scrollbars,resizable,copyhistory=no" ); window.preview.focus(); return preview; } // end launch_window IMAGE(votenow.gif)     IMAGE(voteresults.gif)

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Your flower power is no match for my glower power!
Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Agent
Member # 2759
Profile Homepage #1
It's worth pointing out that the great mathematician Paul Erdős famously got this problem wrong. And he never truly accepted the solution, even when it was demonstrated using logic tables.

It is certainly quite difficult to explain convincingly to the layman (I've tried). Bayesian statistics will always be somewhat controversial.

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"I can't read this thread with that image. But then, that's not a complaint." -Scorpius

Geneforge 4 stuff. Also, everything I know about Avernum | Avernum 2 | Avernum 3 | Avernum 4
Posts: 1104 | Registered: Monday, March 10 2003 08:00
Agent
Member # 3349
Profile Homepage #2
Am I allowed to explain my answer, or should I wait until others have voted?

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And everybody say....Yatta!
Posts: 1287 | Registered: Thursday, August 14 2003 07:00
Infiltrator
Member # 4826
Profile #3
Post what you want. It's other peoples' problem if they want to look at the posts before voting.

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Bring back TM or [i]DIE.[/i]

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Posts: 458 | Registered: Friday, August 6 2004 07:00
Law Bringer
Member # 2984
Profile Homepage #4
This was my absolute nightmare in my stochastic course. I'm still not convinced my answer is wrong. My interpretation has always been that it doesn't matter: Regardless of which door you picked, the show master could have pulled this trick. And the prize door doesn't change either.

--

Another question: Is your chance of winning this game 33%, or 50%?

Edit: I'm almost convinced now. So apparently your chance is 66% if you switch, and 33% if you stay. It doesn't make any sense! I have to try this out...

[ Saturday, August 12, 2006 12:18: Message edited by: Drow ]

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Posts: 8752 | Registered: Wednesday, May 14 2003 07:00
Agent
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It's 67%, if you answered *i's riddle correctly.

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"I can't read this thread with that image. But then, that's not a complaint." -Scorpius

Geneforge 4 stuff. Also, everything I know about Avernum | Avernum 2 | Avernum 3 | Avernum 4
Posts: 1104 | Registered: Monday, March 10 2003 08:00
Post Navel Trauma ^_^
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Profile Homepage #6
As it stands, there isn't enough information. We need to know whether the host would always open a door with no prize, or only in some circumstances.

Assuming the former, in my experience the easiest way to explain the result to people who don't get it is to increase the number of doors to 1000, and the host reveals 998 incorrect doors after you've picked yours.

[ Saturday, August 12, 2006 12:14: Message edited by: Khoth ]

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Barcoorah: I even did it to a big dorset ram.

desperance.net - Don't follow this link
Posts: 1798 | Registered: Thursday, October 4 2001 07:00
Lifecrafter
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Profile #7
(Warning: this post has spoilers.)

But that doesn't make any difference. You're treating the first game (choose 1 door out of 3 or 1000) and the second game (choose 1 door out of 2) as the same.

When the host opens the third door, a new game begins, and there's a 1 in 2 probability there's a prize behind either door.

I honestly can't believe that the accepted consensus among statisticians here is what it is, because that consensus is manifestly wrong.

In the classical form of the problem (Monty Hall), there are two games. In the first, you select one door out of three, with each door having a 1 in 3 chance of containing a car and a 2 in 3 chance of containing a goat.

In the second game, there's a 1 in 2 chance either door has the car behind it.

The revealing of a goat door has no influence on either game - it's the segue between game #1 and game #2.

To say, as the consensus does, that there's a 2 in 3 chance of there being a goat behind the door you go into game #2 having selected by default is to pretend that somehow you're still playing game #1.

...

On a related note, I trust Erdos's judgement on this one. The consensus answer to this problem seems like an elaborate logical shell game without an actual grounding in reality; further, I trust explicitly one of the most brilliant mathematicians of the last century before someone whose claim to fame is the belief she is the smartest person alive.

[ Saturday, August 12, 2006 12:26: Message edited by: The Worst Man Ever ]
Posts: 794 | Registered: Tuesday, October 11 2005 07:00
Agent
Member # 2759
Profile Homepage #8
The two games aren't independent.

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"I can't read this thread with that image. But then, that's not a complaint." -Scorpius

Geneforge 4 stuff. Also, everything I know about Avernum | Avernum 2 | Avernum 3 | Avernum 4
Posts: 1104 | Registered: Monday, March 10 2003 08:00
Electric Sheep One
Member # 3431
Profile #9
Huh. Interesting point, Khoth. If an ignorant audience member opened the door, and there happened to be no prize behind it, we really wouldn't be any better off switching; it would be 50/50. Only if we know that the opened door could not have had the prize does it pay off to switch.

The argument for switching is that your initial guess was right only 1/3 of the time, so 2/3 of the time the prize was behind one of the other doors -- and after the door is opened, you know which one it would have to be if your initial 1/3 chance didn't come up, so you get 2/3 by switching.

From a Bayesian point of view, if you like, the fact that a randomly opened other door has no prize boosts your confidence that your initial guess was lucky, so you have to revise it upwards from 1/3 -- to 50%, in fact. A door that was opened non-randomly, and was guaranteed to show no prize, has no implications for the correctness of your initial choice. So you get no Bayesian boost, and your chances from sticking are still 1/3.

This little remark by Khoth should clarify the fallacy in Alec's analysis. In the classic version with the canny host, the two games are not independent, because the host chose which door to open based on his prior knowledge and your choice. But if 998 doors were opened randomly, and still showed no prize, you'd be certain that an incredible fluke had occurred; but the two possible incredible flukes would be equally likely.

Fun, fun, fun.

[ Saturday, August 12, 2006 12:53: Message edited by: Student of Trinity ]

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We're not doing cool. We're doing pretty.
Posts: 3335 | Registered: Thursday, September 4 2003 07:00
Lifecrafter
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quote:
Originally written by Micawber:

The two games aren't independent.
What prevents them from being independent, exactly? Even though at first blush it would seem they're related, they're completely independent.
Posts: 794 | Registered: Tuesday, October 11 2005 07:00
Electric Sheep One
Member # 3431
Profile #11
Let's put it this way, Alec. We'll play the 1000-door version, with me as the host. It'll cost you $10 to play each time, and you have to stick to your initial choice every time. But the prize is $100.

Easy money, right? For $10 you get a 50% chance at $100!

Step right up.

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We're not doing cool. We're doing pretty.
Posts: 3335 | Registered: Thursday, September 4 2003 07:00
Lifecrafter
Member # 6388
Profile #12
But if I have to stick to my first choice every time, there isn't a second game, now is there?

Also, I'd dispute the idea that the 1000-door version and the 3-door version are equivalent - in fact, I'm certain it's wrong, but have no idea how - but that's neither here nor there.

[ Saturday, August 12, 2006 13:21: Message edited by: The Worst Man Ever ]
Posts: 794 | Registered: Tuesday, October 11 2005 07:00
Shaper
Member # 32
Profile #13
It depends upon whether or not the host has information that the player doesn't. Did he intentionally reveal a door without the prize?

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Lt. Sullust
Cogito Ergo Sum
Polaris
Posts: 2462 | Registered: Wednesday, October 3 2001 07:00
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Well this is what I came up with. I'm probably overlooking something. Spoilers.

1/3 of the time you'll pick the correct door on your first try. The host will eliminate one of the two other doors at random since both are incorrect, which one he chooses is irrelevant. If you stick with your first choice you win. Change and you lose.

2/3 of the time you'll pick incorrectly on your first try. The host will have 2 doors left, one with a prize and one with nothing. He'll have to open the remaining prizeless door in order to keep the game going, and make sure you have a shot in round 2. This leaves you with your original choice(nothing) and what the host left for you (Jackpot). If you stick with your first choice you lose. Change and you win.

The second scenario happens 2/3 of the time, so you're better off changing. This assumes that the host behaves in the way I wrote, instead of choosing at random. It also assumes that I didn't overlook something.

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Guaranteed to blow your mind.

Frostbite: Get It While It's...... Hot?
Posts: 900 | Registered: Monday, August 8 2005 07:00
The Establishment
Member # 6
Profile #15
quote:
It depends upon whether or not the host has information that the player doesn't. Did he intentionally reveal a door without the prize?
One can assume the host knows the answer ahead of time. If he revealed the right door, it would sort of make the game uninteresting when you make your second choice. :P

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Your flower power is no match for my glower power!
Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Infiltrator
Member # 2836
Profile #16
quote:
Originally written by Lazarus.:

This assumes that the host behaves in the way I wrote, instead of choosing at random. It also assumes that I didn't overlook something.
But if the host truly behaves randomly, isn't there a chance that he will accidentally show you the door containing the car (provided you haven't picked that one)?

Edit: *i beat me.

[ Saturday, August 12, 2006 14:15: Message edited by: The Stew Boy ]
Posts: 587 | Registered: Tuesday, April 1 2003 08:00
...b10010b...
Member # 869
Profile Homepage #17
If you still don't trust the arguments, collect some empirical evidence for yourself: try the simulation and see how many times you win by switching and how many times you win by not switching; I just tried out 100 runs for each and won 32 by staying and 62 by switching. Pretty convincing. (The source code is available if you don't trust that it's an actual simulation.)

quote:
But if I have to stick to my first choice every time, there isn't a second game, now is there?
If you choose to stick to your first choice every time, there's no second game either.

Believing that your chance of winning is 50% if you don't switch is magical thinking; it assumes that somehow information given to you by the host can benefit you even if you don't act on it.

Oh, and in answer to your previous question about why the two games aren't independent: it's because the host can't open a door that you've already picked in the first game. As a result, his opening a door gives you information about which of the doors you didn't pick is wrong, but no information about whether your door is right or wrong; therefore, your chance of winning if you stay with your original door remains at 1 in 3.

[ Saturday, August 12, 2006 14:51: Message edited by: Thuryl ]

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The Empire Always Loses: This Time For Sure!
Posts: 9973 | Registered: Saturday, March 30 2002 08:00
The Establishment
Member # 6
Profile #18
If you don't trust computer simulations, you can always do it with three cards. Suppose you have two 2s and an ace. Picking the ace is a win. Have a person who knows the order of the cards act as the host. Play the game again and again and you will eventually get the result.

Incidentally, how this first came up with me was last summer at my internship, one person (one of the brightest math/computer people out there I've been told*) proposed the question to some of the interns. The answer can be quite deceptive.

On the ride home, my roommate and I discussed this and it was difficult to convince him with theory. He's more of an experimental person and had to develop the above experiment to convince himself of the answer.

* As an aside, the person I speak of was able to conquer the Valley of the Kings. Basically, you go to Wendy's burger place in the US and eat a triple, double, single, double, and triple in under an hour.

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Your flower power is no match for my glower power!
Posts: 3726 | Registered: Tuesday, September 18 2001 07:00
Post Navel Trauma ^_^
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Alec: I'm thinking of a number between 1 and 1000. Please post your guess.

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Barcoorah: I even did it to a big dorset ram.

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Posts: 1798 | Registered: Thursday, October 4 2001 07:00
...b10010b...
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Okay, all y'all can stop poking Alec now; I've had a little talk with him over AIM and he's more or less convinced.

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The Empire Always Loses: This Time For Sure!
Posts: 9973 | Registered: Saturday, March 30 2002 08:00
Guardian
Member # 2238
Profile Homepage #21
Out of ten tries, I got it right ten times. Does this qualify me for some award/reward?

Damnation, the 11th failed me.

[ Saturday, August 12, 2006 15:51: Message edited by: d3m0n5L4y3r ]

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DEMON PLAY,
DEMON OUT!
Posts: 1582 | Registered: Wednesday, November 13 2002 08:00
Agent
Member # 3349
Profile Homepage #22
Right right.

The answer is yes. Your chances increase from 1/3 to 2/3 if you swap once there are only 2 doors left. It’s explained better in the Wikipedia article on the subject.

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And everybody say....Yatta!
Posts: 1287 | Registered: Thursday, August 14 2003 07:00
Law Bringer
Member # 6785
Profile #23
It depends upon whether the host knows which door has the prize.

If the host knows the answer then there is a 2/3 chance that your choice is wrong and that he only reveals the losing chocie and not the winning one that you didn't pick. If the host doesn't know the answer then there is a 2/3 chance that the door opened will be wrong and the chance that you have the right door is now 1/2. On the 1/3 chance that the opened door has the prize the game ends.

Let's Make a Deal used this premise in order to prolong the game since the host knew the prize position and tried to sway the person into a decision.

Deal or No Deal has the host knowing the answer, but the revealing of the prize in the briefcases is random (chosen by the contestant). There the contestant gains the new information and it changes the odds that the original decision is wrong.

[ Saturday, August 12, 2006 20:54: Message edited by: Randomizer ]
Posts: 4643 | Registered: Friday, February 10 2006 08:00
Infiltrator
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Nice thought experiment, *i. It took me a few minutes before I was convinced.

In the same spirit, here is a problem that confused me greatly several years back.

Joe and Bob go to two days of batting practice, Joe has the better batting average on both the first and second days. Bob has the better overall average after both days. How is this possible?

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"As our circle of knowledge expands, so does the circumference of darkness surrounding it." --Albert Einstein
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