1282: "Monty Hall"

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Re: 1282: "Monty Hall"

Postby gmalivuk » Fri Oct 25, 2013 10:40 pm UTC

Since those sounds are not unique to English, I can't imagine how it would only work for native English speakers.
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Re: 1282: "Monty Hall"

Postby J L » Fri Oct 25, 2013 10:43 pm UTC


Thanks for posting the original link to vos Savant's page, I missed this when first skipping through the posts.

some Anthony Tamalonis wrote:You have taken over our Mathematics and Science Departments! We received a grant to establish a Multimedia Demonstration Project using state-of-the-art technology, and we set up a hypermedia laboratory network of computers, scanners, a CD-ROM player, laser disk players, monitors, and VCR’s. Your problem was presented to 240 students ...

Oh, the humanity!

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Re: 1282: "Monty Hall"

Postby Eebster the Great » Sat Oct 26, 2013 12:25 am UTC

I never got the McGurk Effect. To me, the single syllable sounds very clearly like "ba," while the lips are totally indistinct (could be ba, pa, ma, ta, da, ga, ka, whatever). Maybe I'm just especially bad at reading lips.

Klear wrote:
Eebster the Great wrote:Of course, if you don't know whether the host knew in advance or just got lucky, then you might as well switch anyway, because staying with the first choice will never be beneficial.


Not in the case when the host only ever offers you the switch if you've chosen the car in your initial pick, as a schmuck bait.

That's a good point. By adjusting Monty's behavior, you can make the odds whatever you want.

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Re: 1282: "Monty Hall"

Postby XTCamus » Sat Oct 26, 2013 4:24 am UTC

Where are all the empiricists hiding? How do we know whether Monty Hall always offered a switch or only when some shmuck picked the car? By reviewing the actual data from every episode of the show! (No, I can't be arsed to check myself either, I'm busy making a point. Though anecdotally I never recall him NOT offering a chance to switch...)

Similarly, and more importantly, how do we know whether it would be better to switch if you were ever in this situation? You really think a model would be a reliable guide for such an empirical question? Again, somebody please check the tape, how many times on the show did someone switch and how did they fare relative to those who stayed put?

What I do like about the mathematical solution is how it belies the intuition that you are always better to stick with your first guess. Which leads me to a related point, if you were one of those highly intuitive types, aka a psychic, you wouldn't need particularly good psychic skills to stick with your first guess and still beat out the switchers.

Putting the above two together, if switchers on the actual show did NOT clearly outperform the stick-with-your-first-guessers, then this would be evidence for psychic powers. Either that or it would be evidence that not only did Monty Hall know, but so did the contestants, meaning the entire game show was rigged. Say it ain't so, Joe!

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Re: 1282: "Monty Hall"

Postby Eebster the Great » Sat Oct 26, 2013 6:03 am UTC

XTCamus wrote:Where are all the empiricists hiding? How do we know whether Monty Hall always offered a switch or only when some shmuck picked the car? By reviewing the actual data from every episode of the show! (No, I can't be arsed to check myself either, I'm busy making a point. Though anecdotally I never recall him NOT offering a chance to switch...)

Your anecdote is ineffective because the game described was never actually played on Let's Make a Deal. It's fictional.

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Re: 1282: "Monty Hall"

Postby Editer » Sat Oct 26, 2013 6:20 am UTC

If Monty Hall opens a door and offers a switch 100% of the time the contestant chose the car and 50% of the time the contestant chose a goat, then the choice whether to switch is exactly 50-50. I suspect that in fact, if this game was indeed played as described, he winged it.

Monika wrote:They show the McGurk effect with a fairly boring example, ba/fa. The ba/da/ga example is so much better! http://www.youtube.com/watch?v=jtsfidRq2tw


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Re: 1282: "Monty Hall"

Postby Kit. » Sat Oct 26, 2013 7:02 am UTC

Klear wrote:
Monika wrote:
bilkie wrote:I learned about the Monty Hall problem about three years ago, the same week I learned about the McGurk Effect.
Now there is a week to remember.

They show the McGurk effect with a fairly boring example, ba/fa. The ba/da/ga example is so much better! http://www.youtube.com/watch?v=jtsfidRq2tw


Didn't work for me. I heard "pa pa pa", but if I would have to choose between ba and da (as I knew I should from your post), I'd choose ba.

Might be some oddity in my brain, or possibly it only works for native English speakers...

Are you nearsighted and not wearing glasses when communicating with people?

"a a a" for me in the second video. "pa pa pa" in the first, but the guy puts so much effort into trying to show that he says "fa" that it's almost believable.

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Re: 1282: "Monty Hall"

Postby Klear » Sat Oct 26, 2013 8:22 am UTC

XTCamus wrote:Where are all the empiricists hiding? How do we know whether Monty Hall always offered a switch or only when some shmuck picked the car? By reviewing the actual data from every episode of the show! (No, I can't be arsed to check myself either, I'm busy making a point. Though anecdotally I never recall him NOT offering a chance to switch...)

Similarly, and more importantly, how do we know whether it would be better to switch if you were ever in this situation? You really think a model would be a reliable guide for such an empirical question? Again, somebody please check the tape, how many times on the show did someone switch and how did they fare relative to those who stayed put?

What I do like about the mathematical solution is how it belies the intuition that you are always better to stick with your first guess. Which leads me to a related point, if you were one of those highly intuitive types, aka a psychic, you wouldn't need particularly good psychic skills to stick with your first guess and still beat out the switchers.

Putting the above two together, if switchers on the actual show did NOT clearly outperform the stick-with-your-first-guessers, then this would be evidence for psychic powers. Either that or it would be evidence that not only did Monty Hall know, but so did the contestants, meaning the entire game show was rigged. Say it ain't so, Joe!


Read this.

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Re: 1282: "Monty Hall"

Postby Monika » Sat Oct 26, 2013 9:08 am UTC

Eebster the Great wrote:I never got the McGurk Effect. To me, the single syllable sounds very clearly like "ba," while the lips are totally indistinct (could be ba, pa, ma, ta, da, ga, ka, whatever). Maybe I'm just especially bad at reading lips.

(In the second video I assume?)
Well, as the sound actually is da da da, and you hear ba ba ba, the McGurk Effect worked for you, too.

Klear wrote:
Monika wrote:
bilkie wrote:I learned about the Monty Hall problem about three years ago, the same week I learned about the McGurk Effect.
Now there is a week to remember.

They show the McGurk effect with a fairly boring example, ba/fa. The ba/da/ga example is so much better! http://www.youtube.com/watch?v=jtsfidRq2tw


Didn't work for me. I heard "pa pa pa", but if I would have to choose between ba and da (as I knew I should from your post), I'd choose ba.

Might be some oddity in my brain, or possibly it only works for native English speakers...

Close enough, it shows the effect anyway, as the sound is "da da da", so if you hear something else it worked.

It does not only work for native English speakers. My native language is German. I originally saw the ba/da/ga demonstration at the Expo in Germany. It would probably not work or not work the same way for speakers of a language that lacks one of the sounds b, d, g and also doesn't have a sound that sounds and looks close enough like p, t, k. Does your native language lack b but contain p? (I thought of Arabic first, but that contains b and lacks p.)

Kit. wrote:"a a a" for me in the second video. "pa pa pa" in the first, but the guy puts so much effort into trying to show that he says "fa" that it's almost believable.

When I first watched the second video I also heard a a a but then when I watched it again I heard ba ba ba.

BTW when I first started watching Star Trek TNG (without having watched TOS before) I had a hard time figuring out when they are saying warp and when they are saying Worf (German synced version).
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Re: 1282: "Monty Hall"

Postby Klear » Sat Oct 26, 2013 9:26 am UTC

Monika wrote:
Klear wrote:
Monika wrote:They show the McGurk effect with a fairly boring example, ba/fa. The ba/da/ga example is so much better! http://www.youtube.com/watch?v=jtsfidRq2tw


Didn't work for me. I heard "pa pa pa", but if I would have to choose between ba and da (as I knew I should from your post), I'd choose ba.

Might be some oddity in my brain, or possibly it only works for native English speakers...


Close enough, it shows the effect anyway, as the sound is "da da da", so if you hear something else it worked.


The sound in your video is "ba ba ba". They say it several times, and they say that if you hear "da da da", you are in a majority. Check it out again.

Af for the native language thing, it's not that we lack these sounds, but the accent and everything is different. It's possible that the mouth movements are a bit different too.

On the other hand, I watched "Hush", the silent episode of Buffy recently, and was pleased to see that I could lip-read English pretty good.

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Re: 1282: "Monty Hall"

Postby Eebster the Great » Sat Oct 26, 2013 10:35 am UTC

Klear wrote:Read this.

This article had everything I didn't want to see in the first five paragraphs. I stopped reading after the claim of a "228 point IQ" (impossible by definition). The Monty Hall problem is not some great question in mathematics, it's just a trivial but counterintuitive puzzle. And now they have "realistically simulated" it thirty times, and decided that was enough.

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Re: 1282: "Monty Hall"

Postby Klear » Sat Oct 26, 2013 10:54 am UTC

Eebster the Great wrote:
Klear wrote:Read this.

This article had everything I didn't want to see in the first five paragraphs. I stopped reading after the claim of a "228 point IQ" (impossible by definition). The Monty Hall problem is not some great question in mathematics, it's just a trivial but counterintuitive puzzle. And now they have "realistically simulated" it thirty times, and decided that was enough.


Read on, there's an interview with Monty Hall, describing how exactly it worked on the actual show - something you wanted to know.

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Re: 1282: "Monty Hall"

Postby Uzh » Sat Oct 26, 2013 2:32 pm UTC

I always had my problems with the goat in the Monty Hall Problem, since in the german show "Geh auf's Ganze!" there was an actual ZONK http://www.paw-mc.de/ebay/2011Juni/11.06./Zonk1.JPG, which was a red-greyish plush mouse. You actually won this mouse.

Once a big Zonk (about 1.5 m high) sat on the parking lot next to my office which caused quite a lot of fuzz and many people on the windows. I don't know, whether someone stole a car and left this ZONK behind...

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Re: 1282: "Monty Hall"

Postby Extragorey » Sat Oct 26, 2013 5:10 pm UTC

Of course, if you got the car you could buy many more goats. And hire people to clean up after them. :wink:
But realistically, the car has got to be behind the leftmost or rightmost door, because there wouldn't be room for a whole car between the other two rooms that are filled with goats.
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Re: 1282: "Monty Hall"

Postby Monika » Sat Oct 26, 2013 8:24 pm UTC

Klear wrote:The sound in your video is "ba ba ba". They say it several times, and they say that if you hear "da da da", you are in a majority. Check it out again.

Ah, I mixed that up.
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Re: 1282: "Monty Hall"

Postby Showsni » Sun Oct 27, 2013 12:58 am UTC

I think everyone's got the Monty Hall problem sorted by now. How about we solve some other gameshows?

In the end game of Strike it Rich, there are ten sets of three screens. Each set contains a hot spot, an arrow and a question arranged at random - get the question wrong, and it becomes another hot spot, get it right and it becomes an arrow. You have to pick one screen from each set going from left to right - but before you start, you have to decide which jackpot to try and play for. Choose one:
£10,000 (you can't hit more than two hot spots)
£7,000 (you can't hit more than 3 hot spots)
£5,000 (you can't hit more than 4 hot spots)

However! For each different jackpot, if you fail to make it all the way across, they give you a set amount of money for each screen you passed without a Hot Spot on. In the same order as above, they are -
£500 per screen cleared
£350 per screen cleared
£250 per screen cleared

For example, if you chose to go for the £10,000 and picked Hot Spot, Arrow, Arrow, Question -> Arrow, Hot Spot, Hot Spot, you'd stop there and win £1,500.

The questions are pretty easy, so assume you always get them right (or maybe like 90%). Then, what are your expected winnings with each jackpot choice? Which jackpot should you attempt to go for?

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Re: 1282: "Monty Hall"

Postby Karilyn » Sun Oct 27, 2013 2:47 am UTC

When I get a particularly stubborn person who insists it's the same whether or not you switch doors, I usually draw out the following for them.

They usually start steaming in rage once they see it in front of their face. I've yet to see someone who's able to present an argument against it a visual aid like this.

Spoiler:
Image


A common misconception is that the host needed to know which door the goat was behind. That isn't true. All that it requires to be advantageous to switch is for the host to have selected the goat in the past tense. As long as the host has already selected the goat, it's better to switch. That's it. All you need is the following two events in order:

1. You select a door.
2. The host selects a door and it wasn't a car.

The host's knowledge is irrelevant. As long as he didn't select a car, it's always better to switch. (Unless the host is cheating and always selects the car if you didn't pick it, and is thus forced to select the goat only if you picked the car, but that's another matter entirely)

To simulate it up with the host selecting a door at random:

You select a door: Your chance of having selected the car: 1/3rd
The host selects a door: The chance of him getting the car is 1/3rd. Instant loss:
1/3rd of all games end in the host opening a door and finding a car. What about the other 2/3rds?
The host selects a door: The chance of him getting a goat is 2/3rds. The game continues
You are now offered the chance to switch: The chance of having selected the car is 1/3rd. The chance of having NOT selected it is 2/3rds. So you switch.

End result:
No switch strategy:
1/3rd the time you get the car, 1/3rd the time the host opened the car. 2/3rd the time he opened a goat.

Switch strategy:
1/3rd the time the host opened the car door. 4/9ths of the time switching wins the game. 2/9ths of the time switching loses the game.

4/9th > 1/3rd

You still lose 1/3rd the time either way, cause the host has a 1/3rd chance of randomly selecting the car. But you still have a higher chance from switching than not switching. It's not as big of a chance of winning as the "Host always selects the goat" but it's still a bigger chance thans taying with your original 1/3rd chance of winning.
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Re: 1282: "Monty Hall"

Postby Eebster the Great » Sun Oct 27, 2013 7:11 am UTC

Karilyn wrote:A common misconception is that the host needed to know which door the goat was behind. That isn't true. All that it requires to be advantageous to switch is for the host to have selected the goat in the past tense. As long as the host has already selected the goat, it's better to switch. That's it. All you need is the following two events in order:

1. You select a door.
2. The host selects a door and it wasn't a car.

The host's knowledge is irrelevant. As long as he didn't select a car, it's always better to switch. (Unless the host is cheating and always selects the car if you didn't pick it, and is thus forced to select the goat only if you picked the car, but that's another matter entirely)

But that's wrong. If the host just happened to pick a goat, that affects the prior probability. The host is more likely to have picked the goat by chance if there were two goats to choose from (100%) than if there was a goat and a car (50%). Thus the former has twice the prior probability as the latter. This completely negates the advantage.

As was pointed out earlier, manipulating the host's behavior can produce any desired probability.

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Re: 1282: "Monty Hall"

Postby Wnderer » Sun Oct 27, 2013 1:21 pm UTC

Wait. On Let's Make a Deal, Isn't Beret Guy supposed to have a costume?

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Re: 1282: "Monty Hall"

Postby drewder » Sun Oct 27, 2013 2:32 pm UTC

J L wrote:In any case, it seems that problem made a lot of people very angry, and very confused:

"Vos Savant wrote in her first column on the Monty Hall problem that the player should switch since the first door has a 1/3 chance of winning, so the second door has a 2/3 chance as the host always opens a losing door on purpose. (vos Savant 1990a) She went on to explain her answer by asking the reader to visualize the player selecting #1 amongst a million doors. Following the constraints, the host opens all remaining doors except door #777,777. Her conclusion: "You’d switch to that door pretty fast, wouldn’t you?"

She received thousands of letters from her readers; 92% of the general public, 65% of universities, and many with PhDs, were against her answer."

http://en.wikipedia.org/wiki/Monty_Hall_problem

(Warning: Following those other paradox/problem links is a serious time sink.)

Problem with judging it based on the response people wrote in is very few people will write in to say that a columnist is right. You'll have some people who initially belived that they were wrong but then were persuaded or did experiment to prove her right but I think the number who write in to say she did a good job was probably fairly small even if the number who thought she did a good job was very large.

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Re: 1282: "Monty Hall"

Postby stickler » Sun Oct 27, 2013 4:43 pm UTC

I have a nasty feeling that I am wrong but:

There are two choices A or B.
You can choose between them. One of them represents a win.
The chance of getting a win is 50/50.


After the goat has been picked that is the situation.

I have a feeling that the reason why the proofs for swap are wrong might be related to gamblers fallacy.

Edit: I finally get it! A minute into [url="http://www.youtube.com/watch?v=7u6kFlWZOWg"]this video[/url] it all makes sense.
Last edited by stickler on Mon Jun 02, 2014 1:49 pm UTC, edited 1 time in total.

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Re: 1282: "Monty Hall"

Postby gmalivuk » Sun Oct 27, 2013 4:58 pm UTC

If the goat door is opened before you do anything, you're right. But since the goat door is always picked from among the two you didn't choose, it's the 1/3-2/3 situation.
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Re: 1282: "Monty Hall"

Postby stickler » Sun Oct 27, 2013 5:22 pm UTC

Having picked a door earlier doesn't mean anything, as in all cases the situation will resolve into what I put above.

I don't see how having A and B labelled keep and swap makes any difference.

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Re: 1282: "Monty Hall"

Postby Klear » Sun Oct 27, 2013 6:33 pm UTC

stickler wrote:Having picked a door earlier doesn't mean anything, as in all cases the situation will resolve into what I put above.

I don't see how having A and B labelled keep and swap makes any difference.


Read through the thread, there have been sever very nice illustrations of why is the 1:3 thing true.

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Re: 1282: "Monty Hall"

Postby gmalivuk » Sun Oct 27, 2013 6:40 pm UTC

stickler wrote:Having picked a door earlier doesn't mean anything, as in all cases the situation will resolve into what I put above.

I don't see how having A and B labelled keep and swap makes any difference.

Picking your door first matters because it means the host won't pick that door to open, so there's no longer an equal chance of picking either goat. (If you picked a goat, there's a 0% chance the host will pick that one, and a 100% chance he'll pick the other one, and thus a 100% chance that you'll win by switching.)
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Re: 1282: "Monty Hall"

Postby drewder » Sun Oct 27, 2013 7:14 pm UTC

stickler wrote:I have a nasty feeling that I am wrong but:

There are two choices A or B.
You can choose between them. One of them represents a win.
The chance of getting a win is 50/50.


After the goat has been picked that is the situation.

I have a feeling that the reason why the proofs for swap are wrong might be related to gamblers fallacy.


What is happening is that there are two subsets of doors
Subset 1 the door you picked winning probability 1:3
Subset 2 the doors you didn't pick winning probability 2:3
Both of which together comprise the full set 3:3

When you select your first door you have a one in three chance of getting a win or a two in three chance of getting a lose. When Monty opens a door he doesn't pick from the full set but rather from subset two. Each subset's probability of winning does not change but the probability for a particular door in the second subset does.

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Re: 1282: "Monty Hall"

Postby orthogon » Sun Oct 27, 2013 8:44 pm UTC

"Four ninths? Where does four ninths come from?" I hear you cry. Well, four ninths is two thirds squared, don't you see? You ask a silly question, you get a silly answer.

With apologies to Tom Lehrer. (I like to think he reads this forum).
For what it's worth, you picking a goat and Monty picking a goat are not independent events, so I don't think you can multiply the two thirdses.
xtifr wrote:... and orthogon merely sounds undecided.

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Re: 1282: "Monty Hall"

Postby Eebster the Great » Sun Oct 27, 2013 8:53 pm UTC

orthogon wrote:
"Four ninths? Where does four ninths come from?" I hear you cry. Well, four ninths is two thirds squared, don't you see? You ask a silly question, you get a silly answer.

With apologies to Tom Lehrer. (I like to think he reads this forum).
For what it's worth, you picking a goat and Monty picking a goat are not independent events, so I don't think you can multiply the two thirdses.

It always bugs me endlessly when people put unreferenced quotes up allegedly from people I know that cannot be found on Google.

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Re: 1282: "Monty Hall"

Postby Klear » Sun Oct 27, 2013 8:57 pm UTC

Eebster the Great wrote:
orthogon wrote:
"Four ninths? Where does four ninths come from?" I hear you cry. Well, four ninths is two thirds squared, don't you see? You ask a silly question, you get a silly answer.

With apologies to Tom Lehrer. (I like to think he reads this forum).
For what it's worth, you picking a goat and Monty picking a goat are not independent events, so I don't think you can multiply the two thirdses.

It always bugs me endlessly when people put unreferenced quotes up allegedly from people I know that cannot be found on Google.


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Re: 1282: "Monty Hall"

Postby orthogon » Sun Oct 27, 2013 9:11 pm UTC

Eebster the Great wrote:
orthogon wrote:
"Four ninths? Where does four ninths come from?" I hear you cry. Well, four ninths is two thirds squared, don't you see? You ask a silly question, you get a silly answer.

With apologies to Tom Lehrer. (I like to think he reads this forum).
For what it's worth, you picking a goat and Monty picking a goat are not independent events, so I don't think you can multiply the two thirdses.

It always bugs me endlessly when people put unreferenced quotes up allegedly from people I know that cannot be found on Google.

I'm sorry. The original quote is from "New Math", when he's doing the subtraction in base 8 and gets to the sixty-fours: "sixty-four is eight squared, don't you see?". I had the same reaction when I read Karilyn's 4/9.

Edit: thanks, Klear. You ninja'd me but for some reason it didn't tell me. It's very lively here for a Sunday night!
Edit2: I spelled Karilyn wrong originally. Sorry!
Last edited by orthogon on Wed Oct 30, 2013 9:59 pm UTC, edited 2 times in total.
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Re: 1282: "Monty Hall"

Postby Klear » Sun Oct 27, 2013 9:28 pm UTC

orthogon wrote:Edit: thanks, Klear. You ninja'd me but for some reason it didn't tell me. It's very lively here for a Sunday night!


Happens sometimes. BTW, I wasn't aware of Tom Lehrer. I'll have to check him out, so thanks for bringing him to my attention. Reminds me of Roy Zimmerman.

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Karilyn
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Re: 1282: "Monty Hall"

Postby Karilyn » Sun Oct 27, 2013 9:34 pm UTC

Eebster the Great wrote:
Karilyn wrote:A common misconception is that the host needed to know which door the goat was behind. That isn't true. All that it requires to be advantageous to switch is for the host to have selected the goat in the past tense. As long as the host has already selected the goat, it's better to switch. That's it. All you need is the following two events in order:

1. You select a door.
2. The host selects a door and it wasn't a car.

The host's knowledge is irrelevant. As long as he didn't select a car, it's always better to switch. (Unless the host is cheating and always selects the car if you didn't pick it, and is thus forced to select the goat only if you picked the car, but that's another matter entirely)

But that's wrong. If the host just happened to pick a goat, that affects the prior probability. The host is more likely to have picked the goat by chance if there were two goats to choose from (100%) than if there was a goat and a car (50%). Thus the former has twice the prior probability as the latter. This completely negates the advantage.

As was pointed out earlier, manipulating the host's behavior can produce any desired probability.


You're actually wrong. Sim it up.

Randomly select one out of 3 doors.
Randomly select one of the remaining two doors and open it.
If it's not a car, then swap.

You'll wind up with the following breakdown:

1/3rd the time the host will open the door with the car and you instantly lose without the option to switch.
4/9ths switching will win
2/9ths switching will lose

To continue my earlier drawing:

Spoiler:
Image


As you can see, the host selecting a door at pure random (Let's say a coin flip), doesn't actually change the fact that you have double the chance of winning if you switch. It just means you get an additional 1/3rd chance of losing when the host opens the door with the car in the first place.

Like I said, a very common misconception, and you're one of a majority of people in getting confused in thinking the host needs to intentionally open a goat door to make switching desirable. He doesn't, all that has to have happened is he opened a goat door at all, because in the times he opens a door with a car you have already lost. Whether or not the host knows, doesn't change the probability of a situation where the goat was opened, because the door with the goat being opened is the only way the game continues.

It's okay to be wrong, probability is a very unintuitive subject. But since this is such an easy problem, you can always draw it out and see the results are exactly what they are.

EDIT: For that matter, you don't even need a host. You can literally do this with yourself. Just select a door, any door. Then flip a coin, open one of the other two doors based on the coin flip, then switch doors if the coin flip didn't reveal the car. You'll get the 1/3rd, 4/9ths, 2/9ths, I can promise you that.
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Re: 1282: "Monty Hall"

Postby Klear » Sun Oct 27, 2013 9:53 pm UTC

Karilyn wrote:You're actually wrong.


I took the liberty of editing the middle part of your picture. Instead of the OR thing, I pictured both sets of possible choices:

Spoiler:
Image


The blue ovals are the host's picks. You are correct that he will pick the car 1/3 of the times and a goat 2/3. Since once he picks a car the game is over and you don't get to choose to switch or not, I've crossed these cases out, and you're left with four cases - in two of them you switch to a goat and in the other two you switch to the car.

Edit: I see now where you made your blunder - you forgot about the case where the host has a choice of two goats. In your picture, there's only the case where he picks one of the goats, but since if you picked the car he's 100% sure to pick a goat, both of the picks need to be taken into account.

In other words, if the host randomly chooses from two goats, even if they are both the same, it counts as two different cases.

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Re: 1282: "Monty Hall"

Postby rmsgrey » Sun Oct 27, 2013 10:08 pm UTC

Karilyn wrote:To simulate it up with the host selecting a door at random:

You select a door: Your chance of having selected the car: 1/3rd
The host selects a door: The chance of him getting the car is 1/3rd. Instant loss:
1/3rd of all games end in the host opening a door and finding a car. What about the other 2/3rds?
The host selects a door: The chance of him getting a goat is 2/3rds. The game continues
You are now offered the chance to switch: The chance of having selected the car is 1/3rd. The chance of having NOT selected it is 2/3rds. So you switch.

End result:
No switch strategy:
1/3rd the time you get the car, 1/3rd the time the host opened the car. 2/3rd the time he opened a goat.

Switch strategy:
1/3rd the time the host opened the car door. 4/9ths of the time switching wins the game. 2/9ths of the time switching loses the game.

4/9th > 1/3rd

You still lose 1/3rd the time either way, cause the host has a 1/3rd chance of randomly selecting the car. But you still have a higher chance from switching than not switching. It's not as big of a chance of winning as the "Host always selects the goat" but it's still a bigger chance thans taying with your original 1/3rd chance of winning.


There's something odd about your figures: in the scenario where you stick, you have a 1/3 chance of the car being behind your chosen door; in the scenario where you switch, you have a 2/9 chance of the car not being behind either the host's door or the door you switch to.

Either your decision whether to stick or switch moves the car, or there's something wrong with your figures.

In fact, for the "stick" scenario, your figures either assume that the host never picks your door to open, or that you sometimes win by sticking when he opens your door. Meanwhile, for the "switch" scenario, your figures assume that the host picks your door to reveal the car 1/9 of the time - and presumably picks your door to reveal a goat 2/9 of the time since he doesn't know where the car is - in which case it's a little unfair to lump that in with the 2/9 of the time when he opens a different door from yours and you both picked goats...

If you allow the host to open the door you picked, then you have a 2/9 stick wins, 2/9 switch wins, 2/9 he opened your door to reveal a goat, and 1/3 he revealed the car. So that's equal chances for "stick" and "switch" when he doesn't open your door, and an unclear outcome when he reveals a goat behind your door...

The usual assumption is that the host picks a different door than you did, in which case it's 1/3 that he hits the car, 1/3 that you started with it, and 1/3 that it's the third door, so sticking and switching are equally likely.

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Re: 1282: "Monty Hall"

Postby Karilyn » Sun Oct 27, 2013 10:09 pm UTC

Edit: I see now where you made your blunder - you forgot about the case where the host has a choice of two goats. In your picture, there's only the case where he picks one of the goats, but since if you picked the car he's 100% sure to pick a goat, both of the picks need to be taken into account.


I actually didn't, that's the top row.

There's only 3 possible arrangements for doors.

Code: Select all

                DoorA DoorB DoorC
Arrangement A:  Car!  Goat  Goat
Arrangement B:  Goat  Car!  Goat
Arrangement C:  Goat  Goat  Car!


As you can see, for any potential arrangement of doors, the host has the same 1/3rd chance of picking a car at random as you do.

Let's say you pick DoorA. You don't know which arrangement it is. It could be Arrangement A, B, or C. If it's Arrangement A, the host has two goats to chose between. If it's arrangement B, he has a goat or a car. If it's arrangement C, he has a goat or a car. That's 4 goats, and 2 cars. A 1/3rd chance to pick a car.

This works for both the other doors too...

Spoiler:
You pick DoorB. You don't know which arrangement it is. It could be Arrangement A, B, or C. If it's Arrangement A, the host has a goat or a car. If it's arrangement B, two goats to chose between. If it's arrangement C, he has a goat or a car. That's 4 goats, and 2 cars. A 1/3rd chance to pick a car.

You pick DoorC. You don't know which arrangement it is. It could be Arrangement A, B, or C. If it's Arrangement A, the host has a goat or a car. If it's arrangement B, he has a goat or a car. If it's arrangement C, he has two goats to chose between. That's 4 goats, and 2 cars. A 1/3rd chance to pick a car.
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Re: 1282: "Monty Hall"

Postby J L » Sun Oct 27, 2013 10:20 pm UTC

drewder wrote:Problem with judging it based on the response people wrote in is very few people will write in to say that a columnist is right. You'll have some people who initially belived that they were wrong but then were persuaded or did experiment to prove her right but I think the number who write in to say she did a good job was probably fairly small even if the number who thought she did a good job was very large.

Quite true. Same with customer reviews, fan communities, and maybe 2/3 of the internet. I was mostly amused by the sheer amount of discussion (as has been said above, it's not like it's a great math problem), and some of the letters she published on her page border on the absurd in terms of conviction (or chauvinism).

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Re: 1282: "Monty Hall"

Postby Karilyn » Sun Oct 27, 2013 10:22 pm UTC

rmsgrey wrote:There's something odd about your figures: in the scenario where you stick, you have a 1/3 chance of the car being behind your chosen door; in the scenario where you switch, you have a 2/9 chance of the car not being behind either the host's door or the door you switch to.


Correct, it you select DoorA there's a 1/3rd chance the door you selected has a car. Then flip a coin and open a door you didn't pick and find a car. There is now a 100% chance that there is not a car behind DoorA.

You're not altering the probability of a car being behind your door, you're deducing the likelihood of specific arrangements of doors.

Code: Select all

                DoorA DoorB DoorC
Arrangement A:  Car!  Goat  Goat
Arrangement B:  Goat  Car!  Goat
Arrangement C:  Goat  Goat  Car!


All 3 arrangements are equally likely, there's a 1/3rd possibility of each. HOWEVER, by seeing what is behind a door you didn't pick, you can deduce which arrangement is most likely to be the arrangement that actually exists using probability.

In fact, for the "stick" scenario, your figures either assume that the host never picks your door to open

Not the point, nor relevant, we're operating on a single thing here.

1. You select 1 out of 3 doors at random.
2. The remaining two doors receive a coin flip, and the door is opened.
3. If the opened door isn't a car, you're given an opportunity to switch.

That's it. We're talking randomnness here. The host doesn't get to offer you money to not take the door (unlike the actual show). There doesn't even need to be a host. We're talking pure randomness and probability here.

1/3rd chance you've already lost before you get the option to switch.
2/9ths chance if you switch you'll lose.
4/9ths chance if you switch you'll win.
Gelsamel wrote:If you punch him in the face repeatedly then it's science.

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Re: 1282: "Monty Hall"

Postby Karilyn » Sun Oct 27, 2013 10:39 pm UTC

orthogon wrote:
"Four ninths? Where does four ninths come from?" I hear you cry. Well, four ninths is two thirds squared, don't you see? You ask a silly question, you get a silly answer.

With apologies to Tom Lehrer. (I like to think he reads this forum).
For what it's worth, you picking a goat and Monty picking a goat are not independent events, so I don't think you can multiply the two thirdses.

The 4/9ths comes a 1/3rd chance of the host picking a car at true random, and all remaining probabilities taking place out of the 2/3rds chance he didn't pick a car. IF we were to keep consistent fractions across the entire thing, there's a 3/9ths chance of the host randomly picking a car, and a 6/9ths chance he wouldn't pick a car at random. Of the 6/9ths where he didn't pick a car, 2/9ths is you picked the car at random, 4/9ths is the car was the one the host didn't pick at random

Basically stop for a moment. What's behind your door is entirely irrelevant.

You have a 1/3rd chance of having gotten a car correct, the first time, right.

This means there's 3 possibilities:

The are two goats behind the randomly selected door.
There is one goat and one car behind the randomly selected door
There is one goat and one car behind the randomly selected door.

That's a 2/3rds chance for a goat, and a 1/3rd chance for a car (3/9ths).

Of the 2/3rds chance that the host selected the goat, 2/3rds of the time, the car will be behind the other door (4/9ths). The other 1/3rd the time, it'll be behind the door you already chose (2/9ths).

That's where the 4/9ths comes from.

1/3rd chance you got the car on your first try
2/3rd chance you didn't get the car on the first try

3/9ths chance the host got the car on his first try
4/9ths chance that if the host didn't get the car on his first try, that the door the host didn't select has a car behind it
2/9ths chance that if the host didn't get the car on his first try, the car was behind your door all along.
Last edited by Karilyn on Sun Oct 27, 2013 10:44 pm UTC, edited 1 time in total.
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Re: 1282: "Monty Hall"

Postby rmsgrey » Sun Oct 27, 2013 10:41 pm UTC

Karilyn wrote:Not the point, nor relevant, we're operating on a single thing here.

1. You select 1 out of 3 doors at random.
2. The remaining two doors receive a coin flip, and the door is opened.
3. If the opened door isn't a car, you're given an opportunity to switch.

That's it. We're talking randomnness here. The host doesn't get to offer you money to not take the door (unlike the actual show). There doesn't even need to be a host. We're talking pure randomness and probability here.

1/3rd chance you've already lost before you get the option to switch.
2/9ths chance if you switch you'll lose.
4/9ths chance if you switch you'll win.


Okay, so if I played 1800 times:

1) I select one of the three doors at random - 600 times I pick the car
2) I open one of the other two doors at random - 600 times I reveal the car
3) By your figures, of the 1200 games remaining, 400 of them (2/9 of 1800), the car is still behind the door I first picked, so 200 of the 600 games where I picked the car have been eliminated. Since a game is only eliminated by the car being behind the opened door, which is never the first door picked, the car must be behind two different doors at once in 1/9 of the games...

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Re: 1282: "Monty Hall"

Postby Karilyn » Sun Oct 27, 2013 10:45 pm UTC

... Oops.

Okay where the hell did I mess that up?

... Oh. Figured it out. Using the same arrangements of before.

Code: Select all

                DoorA DoorB DoorC
Arrangement A:  Car!  Goat  Goat
Arrangement B:  Goat  Car!  Goat
Arrangement C:  Goat  Goat  Car!


Of the 2/3rds time that the host selects a goat at random, there's a 2/4 chance that it's Arrangement A, and a 1/4th chance it's Arrangement B, and a 1/4th chance it's arrangement C. Making it a 50% chance switching will win and a 50% switching will lose.

Whoops.

How ironic. I made the exact opposite mistake as the people who insist the Monty Hall problem doesn't work. Damn probability, you a fickle mistress.

Okay now I'm tripping out over the idea that the host intentionally selecting the goat creates a different probability than if he randomly selects the goat. Fuck. My brain.
Last edited by Karilyn on Sun Oct 27, 2013 10:59 pm UTC, edited 5 times in total.
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