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  #1  
Old 11-26-2011, 01:06 AM
friedo friedo is offline
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Monty Hall problem on MythBusters

The recent episode of MythBusters featured four fan-submitted myths, one of which was perennial Dope fave The Monty Hall problem.

I was impressed with the way they handled it. They divided it into two problems: first, that there is a strong psychological impulse to stick with your initial choice (assuming you aren't aware of the true probabilities) and second that switching is always a better bet (which it is.)

For the first, they had a horde of 20 volunteers play "pick a door" on a genuine three-door game show stage, with Jamie randomly resetting the prize door and Adam being Monty Hall and offering the switch. Every one of the 20 chose to stick!

For the second, they built a very clever door-picking-game simulator in which Adam and Jamie played 100 games, with Jamie always sticking and Adam always switching. Naturally, the results indicated that switching was a vastly superior strategy.
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  #2  
Old 11-26-2011, 01:21 AM
Sage Rat Sage Rat is offline
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My dad asked me the Monty Hall question when I was a kid and I got the answer right the first time. I used the same explanation then as they used. You only had a 1 in 3 chance of getting it the first time, so there's a 2 in 3 chance with the other two doors. If you remove one of the other two doors, so it's just one door, then it takes the full 2 in 3 odds on itself.
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Old 11-26-2011, 02:10 AM
Lord Feldon Lord Feldon is offline
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I didn't understand it (I accepted it, but I didn't understand it), until I realized that switching always reverses the outcome.
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Old 11-26-2011, 02:54 AM
GuanoLad GuanoLad is online now
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I've read of this many times before, and even the practical demonstration didn't make sense to me. So I still don't understand the maths behind it.
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Old 11-26-2011, 02:57 AM
friedo friedo is offline
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Originally Posted by GuanoLad View Post
I've read of this many times before, and even the practical demonstration didn't make sense to me. So I still don't understand the maths behind it.
Lord Feldon's insight is a really good one. Switching always reverses the outcome. If you picked the winner, then obviously switching will cause you to lose. If you picked a loser, the other loser will be eliminated when Monty opens the door, so switching guarantees that you will win. The probability of picking a loser is 2/3, so the "always switch" strategy gives you a 2/3 chance of winning.
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Old 11-26-2011, 04:03 AM
Grumman Grumman is offline
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Originally Posted by GuanoLad View Post
I've read of this many times before, and even the practical demonstration didn't make sense to me. So I still don't understand the maths behind it.
There are three decisions being made:

Which door does the contestant pick?
Which door does the host open?
Does the contestant switch doors?

The argument that switching doors is the best tactic relies entirely upon the host's decision not being random. In other words, you're not playing Three-card Monte, you're playing poker.
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Old 11-26-2011, 05:01 AM
Alka Seltzer Alka Seltzer is offline
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I didn't understand it (I accepted it, but I didn't understand it), until I realized that switching always reverses the outcome.
That's a nice way of explaining it.
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  #8  
Old 11-26-2011, 05:26 AM
Martian Bigfoot Martian Bigfoot is offline
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For me, at least, the the problem is easier to visualize if it's scaled up: Imagine a really rubbish game show where instead of three doors, there are 100 doors. You choose one, and then Monty says: "I'll make this easy for you, because this is the worst game show ever. I'll now open 98 of the remaining doors, where there is no prize. Only one of them remains closed. You can keep your door, or switch to my door instead."

Now, think. There are really only two sets of doors: The one door you picked, and the set of doors you didn't pick. With 100 doors instead of three, your chance of picking the right door on your first pick was just one percent. The chance of the prize being behind a door somewhere in the set of the doors you didn't pick, is 99 percent.

When Monty asks if you want to switch, he's really asking: "Do you want to keep the door that you initially picked, or switch to the full set of doors that you didn't? I've already opened 98 of those doors, so if the prize was anywhere in this set from the get go, and you switch, you'll win, guaranteed."

In other words, if you picked the right door to begin with, you win if you keep your door. If you picked the wrong door to begin with, you'll win if you switch. But your chances of picking the right one at the start was only one percent!

In this case, the answer should be a no-brainer. Doh, switch every time! You'll get a 99 percent chance of winning!

Now, in the actual Monty Hall problem, the situation is exactly the same, just with different numbers. Your chance of picking the right door to begin with was 1/3. The chance of picking the wrong one was 2/3. Switching gives you a 2/3 chance of winning.

Last edited by Martian Bigfoot; 11-26-2011 at 05:31 AM..
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Old 11-26-2011, 05:32 AM
GuanoLad GuanoLad is online now
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But surely what happens at the start is irrelevant. The final decision you make is 1:2 because those are your only options.
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Old 11-26-2011, 05:33 AM
Martian Bigfoot Martian Bigfoot is offline
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But surely what happens at the start is irrelevant. The final decision you make is 1:2 because those are your only options.
And they ask me why I drink.
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  #11  
Old 11-26-2011, 06:02 AM
Grumman Grumman is offline
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The final decision you make is 1:2 because those are your only options.
No. If you had no information with which to make your decision, it would be a fifty-fifty chance. But you do have information: you know that the host knows which door is the right door, and you know that the host will act in a known, non-random manner.
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Old 11-26-2011, 06:10 AM
Sitnam Sitnam is online now
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Imagine 100 doors, you pick one, then the host tells you it's not these 98 doors. Do you change? Of course! Because now you're really choosing 99 doors all at once versus the 1 in 100 chance you had before. But if you walked into the studio and had no idea was just happened, picking either closed doors is just as likely.

Mythbusters is really stretching, tell me they at least blew up some doors at the end. Modern Marvels is losing it too, I expect an episode on the paperclip soon.

Last edited by Sitnam; 11-26-2011 at 06:13 AM..
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  #13  
Old 11-26-2011, 06:17 AM
Martian Bigfoot Martian Bigfoot is offline
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Mythbusters is really stretching, tell me they at least blew up some doors at the end.
[hijack] Actually, please tell me they didn't. There have been way too many shows lately where they've just gone: "Oh, we couldn't think of any good myths this week, but that's OK - we'll just blow something up or shoot a really big gun, and surely no one will notice." [/hijack]
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Old 11-26-2011, 06:21 AM
Little Nemo Little Nemo is offline
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One thing I noticed was that Jamie said there was no point in telling us how many of the first group won or lost because they all picked to stay so their results wouldn't compare the two strategies. But he was wrong. It's a game with only two outcomes - so everyone who lost by staying would have won by switching (and vice versa if that had happened). So if we knew, for example, that seven people stayed and won and thirteen people stayed and lost, then we'd know that switching would have won in those other thirteen cases.
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Old 11-26-2011, 06:22 AM
friedo friedo is offline
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Originally Posted by Sitnam View Post
Mythbusters is really stretching, tell me they at least blew up some doors at the end. Modern Marvels is losing it too, I expect an episode on the paperclip soon.
Don't worry - Monty was only 1/4 of the episode. The other parts involved grenades, gangsta-style shootouts, and a hatchback loaded with two tons of building supplies.
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  #16  
Old 11-26-2011, 06:24 AM
Grumman Grumman is offline
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Imagine 100 doors, you pick one, then the host tells you it's not these 98 doors. Do you change? Of course! Because now you're really choosing 99 doors all at once versus the 1 in 100 chance you had before.
That is not inherently true. It could also mean the host is trying to trick you into moving away from the correct answer. Like I said, it's not about probabilities, it's about knowing how the host responds to information he can see that you can't, to work out what he knows.

Last edited by Grumman; 11-26-2011 at 06:29 AM..
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Old 11-26-2011, 07:05 AM
Heller Highwater Heller Highwater is offline
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That's the thing, though. The way the problem is usually set up (at least, every time I've ever seen it) is that you know there are three doors (one having the prize behind it), and you know that the host knows where the prize is, and you know that the host will always open a door with no prize behind it after you make your initial selection. Of course, this may be different than what happened on the actual TV show, but the "Monty Hall problem" is not meant to exactly copy the circumstances of the TV show. Rather, it's to set up a specific mathematical thought experiment.

I'll also mention that there was a short scene on the Mythbusters episode with Adam establishing precisely those conditions I described above. So in other words, they were testing the standard set-up for the problem. If those conditions are not expressly given when posing the problem, then it will change the answer.
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  #18  
Old 11-26-2011, 07:37 AM
Left Hand of Dorkness Left Hand of Dorkness is online now
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That is not inherently true. It could also mean the host is trying to trick you into moving away from the correct answer. Like I said, it's not about probabilities, it's about knowing how the host responds to information he can see that you can't, to work out what he knows.
Rules of the following game:
1) I think of a number between 1 and 100.
2) I ask you to guess a number between 1 and 100.
3) I tell you that I'm going to rule out 98 numbers, one of which is the number you chose. If you chose correctly in the first place, then I'll choose another number randomly not to rule out. If you chose incorrectly in the first place, then I won't rule out the correct number.
4) I give you the opportunity to switch to the other number, or stick with yours.
5) I reveal whether your final choice is the number I thought of in step 1.

Let's try it out; I'll fill in your guess more-or-less randomly in step 2.

1) I've got my number, but I'm not telling you what it is.
2) You guess that it's 47.
3) I tell you that it's not 1-46, 48-72, or 74-100. It's either 47 or 73.
4) You can now switch your guess to 73, or you can stick with 47.
5) If you stuck with 47, you'd be wrong. If you switched to 73, you'd be right.

Step 3 is important. I'm never going to rule out the correct number, any more than Monty is ever going to open the door with the prize and say, "Oops, too bad!" I'm ALWAYS going to rule out an incorrect choice. That move on my part is what skews the odds so heavily in favor of switching for you.

If you're still not convinced, play the game I described above with a friend. If you follow the rules correctly, if you switch every time, you'll win 99 times out of 100. If you stick with your original number every time, you'll win 1 time out of 100.
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Old 11-26-2011, 07:38 AM
John DiFool John DiFool is offline
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For the first, they had a horde of 20 volunteers play "pick a door" on a genuine three-door game show stage, with Jamie randomly resetting the prize door and Adam being Monty Hall and offering the switch. Every one of the 20 chose to stick!
The fascinating part of this ep. isn't the math involved, but the psychology. I expected beforehand that, yeah, the majority would stick with their choice, but maybe 1/3rd would switch, if for no other reason than to be contrary. I was absolutely blown away when not a single solitary soul changed his/her mind; they invariably gave excuses like "I sticking with my gut". Even if I didn't know the math involved going in, I know from playing boardgames and such that your strategy should be random so as to not leave yourself vulnerable to your opponent correctly gauging your tendencies (which in the case of the original Monty Haul Hall game show he would most certainly do if given the chance). In other words I would expect the host to expect me to stick, like most everyone else would, so I would thus confound his strategy (assuming there was one) by switching.

Last edited by John DiFool; 11-26-2011 at 07:40 AM..
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  #20  
Old 11-26-2011, 08:58 AM
Súil Dubh Súil Dubh is offline
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Originally Posted by Martian Bigfoot View Post
For me, at least, the the problem is easier to visualize if it's scaled up: Imagine a really rubbish game show where instead of three doors, there are 100 doors.
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Originally Posted by Left Hand of Dorkness View Post
...If you're still not convinced, play the game I described above with a friend. If you follow the rules correctly, if you switch every time, you'll win 99 times out of 100. If you stick with your original number every time, you'll win 1 time out of 100.
Martian Bigfoot and the Left Hand of Dorkness explain the maths behind the "Monty Hall" problem perfectly. If you're still having problems trying to understand it, you've got to try and "think big".

Instead of 100 doors, try visualizing 1000. Or a thousand million. What's one thousand million times infinity? They'll try and tell you that there's no number larger than infinity, but they're wrong.

So imagine three infinite sets of infinite doors. And behind each door is one hundred thousand years, one hundred thousand lies, one hundred thousand eyes. Each set of doors takes an infinite amount of time to open. When the eyes open, you must avert your gaze unless you burst into flame.

And the flames of justice burn bright! They will tell you that they are trying to "bust" or dispell the "myth". Was Enoch a myth when he saw the "burning wheel"? The wheel that always revolves at an "infinite" amount of time? They will try and "tell" you "no".

Seven times seven angels dance. Can you read the signs? One of them stops your car as you drive. She gives you a choice: to continue down the road you are on, or follow one of the new paths that he illuminates. But only one! While you think, the "others" try to interfere. They are stopped and turned away.

You stop time and return. The maths and lies and myths are left behind, like childish children's childish crimes. The angels breathe, and kiss you a final gift. "We will close all the doors", they sing and sigh and sign, "all but the won you dearly wish to take".

To choose and win? Your final fate. The mountain looms ahead. The mountain hall and hall of kings, the King that Enoch tried to spell, the myth that larger man must surely dread.

Pretty simple, really.
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  #21  
Old 11-26-2011, 10:15 AM
Bryan Ekers Bryan Ekers is offline
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Huh... this has morphed into Monty Hall of the Mountain King.
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  #22  
Old 11-26-2011, 11:06 AM
standingwave standingwave is offline
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Originally Posted by GuanoLad View Post
But surely what happens at the start is irrelevant. The final decision you make is 1:2 because those are your only options.
It seems irrelevant but it isn't because the door you pick initially has some bearing on which door Monty will reveal (because he always reveals a loser).
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  #23  
Old 11-26-2011, 11:20 AM
Sandra Battye Sandra Battye is offline
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In other words, you're not playing Three-card Monte, you're playing poker.
I saw what you did there.
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  #24  
Old 11-26-2011, 11:22 AM
Nametag Nametag is offline
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I think people stick because they assume a far simpler mechanism is at work -- Monty opens a door to try to get you to switch. He wouldn't do that if you had picked the wrong door in the first place (except for a few to throw off suspicion). The only way to assess the accuracy of this hypothesis is to watch hundreds of games and determine whether picking the right door the first time is correlated with the reveal-and-offer.
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  #25  
Old 11-26-2011, 11:24 AM
The Other Waldo Pepper The Other Waldo Pepper is online now
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Guy A picks Door #1.
Guy B picks Door #2.
Monty throws open Door #3 and shows it's empty.
Guy A and Guy B get asked if they want to switch.
Should they both say yes?
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Old 11-26-2011, 11:28 AM
Little Nemo Little Nemo is offline
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Okay, another way to visualize it for those having problems seeing the logic.

Standard set-up of one prize and three doors. You pick a door at random. But Monty's lazy in this game and doesn't feel like opening doors. He just offers you a choice - you can keep the door you have or you can switch and take both of the other doors.

Now do you stay with the one door you have or do you switch and take two doors?

Mathematically this game is identical to the standard game.
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  #27  
Old 11-26-2011, 11:32 AM
Little Nemo Little Nemo is offline
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Originally Posted by The Other Waldo Pepper View Post
Guy A picks Door #1.
Guy B picks Door #2.
Monty throws open Door #3 and shows it's empty.
Guy A and Guy B get asked if they want to switch.
Should they both say yes?
No, different situation. This scenario is only possible if one of the players picked the prize door - if both had picked losing doors then the unpicked door would be the prize door and Monty couldn't have opened it without revealing the prize. So here the odds are equal and each player has a 50/50 chance of winning.
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  #28  
Old 11-26-2011, 11:32 AM
ZenBeam ZenBeam is offline
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Originally Posted by The Other Waldo Pepper View Post
Guy A picks Door #1.
Guy B picks Door #2.
Monty throws open Door #3 and shows it's empty.
Guy A and Guy B get asked if they want to switch.
Should they both say yes?
The important fact to note here is that Monty can't do this if neither picked the winning door. That happens 1/3 of the time, and that accounts for the chance of a particular contestant winning being only 50/50, instead of 2/3.
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  #29  
Old 11-26-2011, 11:39 AM
The Other Waldo Pepper The Other Waldo Pepper is online now
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The important fact to note here is that Monty can't do this if neither picked the winning door. That happens 1/3 of the time, and that accounts for the chance of a particular contestant winning being only 50/50, instead of 2/3.
Well, yeah: in the two-guy scenario, Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.

But make it a one-guy scenario by removing Guy B: Guy A picks Door #1 -- and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.

Or make it a one-guy scenario by removing Guy A: Guy B picks Door #2 -- and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.
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  #30  
Old 11-26-2011, 11:49 AM
garygnu garygnu is offline
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The only time you lose by switching is if you chose the winning door to begin with.
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  #31  
Old 11-26-2011, 12:02 PM
Martian Bigfoot Martian Bigfoot is offline
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Originally Posted by The Other Waldo Pepper View Post
Well, yeah: in the two-guy scenario, Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.

But make it a one-guy scenario by removing Guy B: Guy A picks Door #1 -- and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.

Or make it a one-guy scenario by removing Guy A: Guy B picks Door #2 -- and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.
Let's see if I got this: Now you're looking at two instances of the original problem being played out, and in both the prize is, say, behind door 2. It's behind door 2 when guy A plays, he switches, and he wins. It's also behind door 2 when guy B plays, he also switches, and he loses.

So what? Keep playing the game enough times, and switching will still have you win two out of three times. So yes, both of the guys should switch, even though one of them end up losing and one of them end up winning, because all they can do, short of using telepathy or time travel, is to use what information they have to pick the best strategy in order max out their chances.
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  #32  
Old 11-26-2011, 12:16 PM
ZenBeam ZenBeam is offline
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Originally Posted by The Other Waldo Pepper View Post
Well, yeah: in the two-guy scenario, Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.

But make it a one-guy scenario by removing Guy B: Guy A picks Door #1 -- and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.

Or make it a one-guy scenario by removing Guy A: Guy B picks Door #2 -- and Door #1 or Door #2 is the winning door, which Monty reveals by opening Door #3.
So in those last two sentences, you're looking at a case where the prize is behind Door #1 or Door #2, but never behind Door #3. And then Monty reveals there is no prize behind Door #3 (which you already knew). So yes, in that case Door #1 and Door #2 have equal chances of winning, and there's no benefit to switching.

ETA: If this isn't what you mean, you need to specify what happens if the prize is behind Door #3.

Last edited by ZenBeam; 11-26-2011 at 12:17 PM..
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  #33  
Old 11-26-2011, 12:18 PM
Martian Bigfoot Martian Bigfoot is offline
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Originally Posted by ZenBeam View Post
So in those last two sentences, you're looking at a case where the prize is behind Door #1 or Door #2, but never behind Door #3. And then Monty reveals there is no prize behind Door #3 (which you already knew). So yes, in that case Door #1 and Door #2 have equal chances of winning, and there's no benefit to switching.
Hang on. How could anyone possibly know that there's no prize behind door 3 before the game starts? Because if you do know, then it's a different game, with only two doors, and not the original problem.

Last edited by Martian Bigfoot; 11-26-2011 at 12:21 PM..
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  #34  
Old 11-26-2011, 12:23 PM
The Other Waldo Pepper The Other Waldo Pepper is online now
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So in those last two sentences, you're looking at a case where the prize is behind Door #1 or Door #2, but never behind Door #3.
Well, no; I'm just saying it happened not to be behind #3 when folks picked #1 and #2. As to this:

Quote:
ETA: If this isn't what you mean, you need to specify what happens if the prize is behind Door #3.
I hadn't gotten that far; I'm feeling this out as I go. So let me twist it further out of shape:

* Figure there are four doors: three empty, one prize.
* Guy A picks one door.
* Guy B picks another door.
* Monty opens one of the two remaining doors.
* A and B are then given the chance to switch with each other. Should they?
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  #35  
Old 11-26-2011, 12:29 PM
Martian Bigfoot Martian Bigfoot is offline
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Look, just to make one thing clear: What you shouldn't do is overthink the problem, or imagine that there's some deep paradox at work or anything profound going on.

There isn't. It's dead simple. Your cat could understand it.

Sure, you can also make up new scenarios with various numbers of doors or contestants. Those scenarios will have different optimal strategies depending on various factors.

That doesn't have anything to do with the original problem, which has very clearly stated rules, and a definite answer.
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  #36  
Old 11-26-2011, 12:29 PM
ZenBeam ZenBeam is offline
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ETA: Sorry, didn't notice you specified four doors. Deleted my post.

Last edited by ZenBeam; 11-26-2011 at 12:31 PM..
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  #37  
Old 11-26-2011, 12:30 PM
ftg ftg is offline
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It was a bit disappointing. Jamie actually said at one point that they didn't care about the Math or proofs or some such. (But they did give a brief explanation later.)

They didn't quote the numbers of their simulated run, but the boards looked like Adam won far more than 2/3 of the time and Jamie far less than 1/3. Anybody know what their data was?

OTOH, they were mostly (but not completely) clear on the rules of the game. Failure to precisely state the conditions is extremely common. So people fill in with their own assumptions, and the 2/3, 1/3 result does not apply in all these cases. Cf. "Ask Marilyn".

The idea that just because there are two options, the odds have to be 1:2 is a damning indictment of the failure to teach people basic probability. I guess that's why lotteries, casinos, and bookies stay in business.
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Old 11-26-2011, 12:34 PM
ZenBeam ZenBeam is offline
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Originally Posted by The Other Waldo Pepper View Post
So let me twist it further out of shape:

* Figure there are four doors: three empty, one prize.
* Guy A picks one door.
* Guy B picks another door.
* Monty opens one of the two remaining doors.
* A and B are then given the chance to switch with each other. Should they?
I assume Monty knows where the prize is, and always opens an empty door. In this case, the prize has a 25% chance to be behind each of the doors one of the Guys picked, and a 50% chance to be behind the remaining door. Switching with each other won't help, but switching to the other remaining door will.

Last edited by ZenBeam; 11-26-2011 at 12:35 PM.. Reason: Clarifying
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  #39  
Old 11-26-2011, 12:36 PM
tim-n-va tim-n-va is offline
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If you make your choice and Monte offers you to keep your door or trade for the other two doors would you trade? That is essentially what you are being offered with the non-winning goats removed.
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  #40  
Old 11-26-2011, 12:40 PM
Quimby Quimby is offline
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I've read of this many times before, and even the practical demonstration didn't make sense to me. So I still don't understand the maths behind it.
The breakthrough for me that helped me understand the math is when I realized that by switching I am actually getting the other two doors. It's just that I know for sure one of the two is definitely a loser (which is a given anyway).
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  #41  
Old 11-26-2011, 12:42 PM
BigT BigT is online now
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Originally Posted by Nametag View Post
I think people stick because they assume a far simpler mechanism is at work -- Monty opens a door to try to get you to switch. He wouldn't do that if you had picked the wrong door in the first place (except for a few to throw off suspicion). The only way to assess the accuracy of this hypothesis is to watch hundreds of games and determine whether picking the right door the first time is correlated with the reveal-and-offer.
Thing is, knowing that's what people would usually do means that any smart person would do the opposite. It seems odd that most people would think the scenario part of the way through, and not all the way.

What's also odd is the version of the test not shown: where one person repeats the test over and over, with there always being an offer to switch. Even then, it takes humans longer to figure it out than it does birds. Even when we know it's random, we still stick to the hypothesis that they're trying to trick us.
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Old 11-26-2011, 12:46 PM
ultrafilter ultrafilter is offline
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Originally Posted by Martian Bigfoot View Post
There isn't. It's dead simple. Your cat could understand it.
Maybe not cats, but pigeons get it just fine. Y'all are smarter than pigeons, right?
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Old 11-26-2011, 01:01 PM
standingwave standingwave is offline
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Originally Posted by Quimby View Post
The breakthrough for me that helped me understand the math is when I realized that by switching I am actually getting the other two doors. It's just that I know for sure one of the two is definitely a loser (which is a given anyway).
Yeah, and they actually did a good job explaining that when they went through the mathematics towards the end of the segment.
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  #44  
Old 11-26-2011, 02:03 PM
OpalCat OpalCat is offline
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Quote:
Originally Posted by Martian Bigfoot View Post
For me, at least, the the problem is easier to visualize if it's scaled up: Imagine a really rubbish game show where instead of three doors, there are 100 doors. You choose one, and then Monty says: "I'll make this easy for you, because this is the worst game show ever. I'll now open 98 of the remaining doors, where there is no prize. Only one of them remains closed. You can keep your door, or switch to my door instead."

Now, think. There are really only two sets of doors: The one door you picked, and the set of doors you didn't pick. With 100 doors instead of three, your chance of picking the right door on your first pick was just one percent. The chance of the prize being behind a door somewhere in the set of the doors you didn't pick, is 99 percent.

When Monty asks if you want to switch, he's really asking: "Do you want to keep the door that you initially picked, or switch to the full set of doors that you didn't? I've already opened 98 of those doors, so if the prize was anywhere in this set from the get go, and you switch, you'll win, guaranteed."

In other words, if you picked the right door to begin with, you win if you keep your door. If you picked the wrong door to begin with, you'll win if you switch. But your chances of picking the right one at the start was only one percent!

In this case, the answer should be a no-brainer. Doh, switch every time! You'll get a 99 percent chance of winning!

Now, in the actual Monty Hall problem, the situation is exactly the same, just with different numbers. Your chance of picking the right door to begin with was 1/3. The chance of picking the wrong one was 2/3. Switching gives you a 2/3 chance of winning.
Great explanation! It makes it all make so much more sense!


Quote:
Originally Posted by Little Nemo View Post
One thing I noticed was that Jamie said there was no point in telling us how many of the first group won or lost because they all picked to stay so their results wouldn't compare the two strategies. But he was wrong. It's a game with only two outcomes - so everyone who lost by staying would have won by switching (and vice versa if that had happened). So if we knew, for example, that seven people stayed and won and thirteen people stayed and lost, then we'd know that switching would have won in those other thirteen cases.

That bothered me, too. Why wouldn't those results be valid data? I asked my husband the same question at the time.
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Old 11-26-2011, 02:14 PM
Heller Highwater Heller Highwater is offline
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There's no reason why it wouldn't be. It's just not the point of that particular experiment. They were only testing the "do people stay or switch" part of the myth. I suspect they just wanted a more visual demonstration of "why it's better to switch", which they provided nicely with the comparison between Adam and Jamie.
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  #46  
Old 11-26-2011, 02:23 PM
enalzi enalzi is online now
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Quote:
Originally Posted by Heller Highwater View Post
There's no reason why it wouldn't be. It's just not the point of that particular experiment. They were only testing the "do people stay or switch" part of the myth. I suspect they just wanted a more visual demonstration of "why it's better to switch", which they provided nicely with the comparison between Adam and Jamie.
Yeah, the visual part of these experiments is part of what makes Mythbusters so successful. There have been plenty of urban legend shows in the past, none of which lasted very long because they just told you whether something was true or not. It's plenty easy to just say "It's better to switch," "The airplane will take off," or "The toilet will/will not blow up." But having the visual of all those red squares really sells the myth.
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  #47  
Old 11-26-2011, 03:41 PM
GuanoLad GuanoLad is online now
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Quote:
Originally Posted by Left Hand of Dorkness View Post
Let's try it out; I'll fill in your guess more-or-less randomly in step 2.

1) I've got my number, but I'm not telling you what it is.
2) You guess that it's 47.
3) I tell you that it's not 1-46, 48-72, or 74-100. It's either 47 or 73.
4) You can now switch your guess to 73, or you can stick with 47.
5) If you stuck with 47, you'd be wrong. If you switched to 73, you'd be right.
Okay, this explanation finally clarified things for me.

However, it still makes my head hurt.
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  #48  
Old 11-26-2011, 04:08 PM
Marley23 Marley23 is online now
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I thought they did a very good job with it, actually. I've read the Straight Dope column a couple of times and probably a few of our threads on the problem, but the answer never really sank in. The machine was a good demonstration - and of course when they finally got around to explaining the math it was extremely simple.
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  #49  
Old 11-26-2011, 05:22 PM
Little Nemo Little Nemo is offline
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Quote:
Originally Posted by OpalCat View Post
That bothered me, too. Why wouldn't those results be valid data? I asked my husband the same question at the time.
What he said was that they couldn't compare the success rate of the staying strategy against the switching strategy because nobody chose the switching strategy so they had no data. So they ran the second experiment where Jamie used the staying strategy while Adam used the switching strategy.

But their assumption was incorrect. Because the two strategies mirror each other, you can make the comparison even if you only have the results from one strategy. The success rate of one strategy will always be the complement of the other.
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  #50  
Old 11-26-2011, 05:24 PM
OpalCat OpalCat is offline
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Exactly. That's what I thought. Just look at how often they won by staying, and subtract that from 20 to see how often they would have won by switching. Simple.
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