Current Events > My brother cannot comprehend the Monty Hall problem.

(+) Yeah! (+)

tiornys posted...

If you're in scenario 2, the host is indirectly telling you where the car is (because he has no choice about which door he opens).

But he's not since he's not telling me anything. All he did was eliminate one. I could still very much be right. How is at all indirectly telling me where it is?

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i dont get it either

dave_is_slick posted...

And that knowledge is?

The host knows where car is, @dave_is_slick.

Let's put it this way. There are a THOUSAND (1000) doors you can choose from, 999 doors each hiding a goat, and 1 door with keys to your dream [______]. You open one, and then the host opens NINE HUNDRED NINETY EIGHT (998) doors, all revealing goats, leaving a single door closed.

Are you still gonna stick with your initial pick?
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VideoboysaysCube posted...

At this point, your odds are going to be 50/50, whether you switch or not. But if the host picked a door because he knew it was a goat, then it makes sense to switch, because he's actually eliminate a bad choice instead of making a random pick.

And this is what I'm stuck on. Everybody who ever explains this just simply says it's better to switch because a wrong choice was eliminated. WHY? Expand on that part in simple terms.

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dave_is_slick posted...

But he's not since he's not telling me anything. All he did was eliminate one. I could still very much be right. How is at all indirectly telling me where it is?

Well, it's not guaranteed that he's telling you something, which is where a lot of the confusion comes from. He's only telling you where the car is if you didn't pick the car to begin with.

Sad_Face posted...

The host knows where car is, @dave_is_slick.

Let's put it this way. There are a THOUSAND (1000) doors you can choose from. You open one, and then the host opens NINE HUNDRED NINETY EIGHT (998) doors, all revealing goats, leaving a single door closed.

Are you still gonna stick with your initial pick?

Yeah. All the other wrong choices have been eliminated. I am either wrong or right at this point.

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tiornys posted...

He's only telling you where the car is if you didn't pick the car to begin with.

Behind one of two doors, with my door possibly being it. I'm not switching.

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Going from 100+ doors to 2 doors isn't really the same thing. You were most likely going to pick the wrong door before, and going down to 2 basically reveals the right door, since your first pick was very unlikely to get the right door. There's no reason to not switch after such a huge change in probability.

Going from 3 to 2 doors, you had a pretty high 33.3% chance you were right the first time, and switching is technically a better chance, but you had a decently high chance to begin with, so you are much more likely to be wrong if you switch, compared to the 100 to 2 example.

dave_is_slick posted...

Behind one of two doors, with my door possibly being it. I'm not switching.

No. If you missed the car originally, he's guaranteeing you that it's behind the door you could switch to.

tiornys posted...

No. If you missed the car originally, he's guaranteeing you that it's behind the door you could switch to.

Or it's behind what I already picked. You are either right or wrong once everything else has been eliminated.

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dave_is_slick posted...

Yeah. All the other wrong choices have been eliminated. I am either wrong or right at this point.

Here's the thing, you have a 1/1000 chance of picking it the first time around. When the host comes in, he kills 998 doors, all of them incorrect with a 100% guarantee. And he leaves one door alone. Remember, you had a 1/1000 chance of picking the right one, what would the percentage chance of that other door be? Remember, the total probability of all choices has to add up to unity; 100% or 1000/1000 if you will.
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dave_is_slick posted...

Or it's behind what I already picked. You are either right or wrong once everything else has been eliminated.

Yes, you're either right or wrong, but the odds are no longer 50/50.

If your initial door is right, you were right at the initial choice, which you get right 1/3 of the time. If your initial door is wrong, you were wrong at the initial choice, which happens 2/3 of the time.

Raikuro posted...

Going from 100+ doors to 2 doors isn't really the same thing. You were most likely going to pick the wrong door before, and going down to 2 basically reveals the right door, since your first pick was very unlikely to get the right door. There's no reason to not switch after such a huge change in probability.

The purpose of using arbitrarily large numbers is to help visualize the thought experiment. While it's not intuitive, it's always in your best interest to switch doors after the host opens the other door(s) with goats. The more doors opened, the larger the percentage chance gain when you switch doors.

Using arbitrarily large or small numbers in general is how mathematicians and other theorists visualize and conclude their derivations. This is the heart of what taking the limit is (take the limit of Y = f(x) as x approaches _____ ).
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Sad_Face posted...

Here's the thing, you have a 1/1000 chance of picking it the first time around. When the host comes in, he kills 998 doors, all of them incorrect with a 100% guarantee. And he leaves one door alone. Remember, you had a 1/1000 chance of picking the right one, what would the percentage chance of that other door be? Remember, the total probability of all choices has to add up to unity; 100% or 1000/1000 if you will.

Maybe it's because percentages were never my strong suit, but I still fail to see why I should switch when it eventually comes down to one or the other. Honestly, if it was more than one door left alone I could see it but I just don't see why when the final choice is ultimately 50/50. I don't get why it matters what the probability was before, it is no longer that once all others have been eliminated.

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When I taught math for non-majors, this usually got the same thing across without running into some of the hang ups people have with Monty Hall for whatever reason:

Suppose you are on vacation and you travel to a city with three hotels. You go to the first hotel but it seems kind of junky, so you drive to the second hotel. This one is somehow worse than the first!

It's too late to go back to the first hotel (someone behind you was going to take the last room), so you have two choices: stay at the second hotel or try your luck with the third one.

Which option leads to a better hotel stay (on average)?

(Note: for the purpose of this puzzle we're assuming that there is a strict ordering of the quality of the hotels without ties)

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Let's take the host out of the problem.

Here are three doors. One has a car, two have goats. You can pick one.

Now, I'll let you keep your one door, or switch to both of the other two doors. Do you want to switch? (assume I will always offer this choice--if I only sometimes offer the choice, that changes the problem again).

dave_is_slick posted...

Yeah. All the other wrong choices have been eliminated. I am either wrong or right at this point.

Just because there are two options (right or wrong) does not mean those options have the same probability of happening. If I put my chosen door off to the side and put the other two doors together in a set, that set of doors has a combined chance of 2/3 of having a car. If a door from that set is opened to reveal a goat, that set still has 2/3 chance of having a car, while my single door only has a 1/3 chance. That 1/3 chance for my door does not change simply because a door was eliminated.

Using your "right or wrong" logic, my chances of winning the lottery are 50/50, either I do or I don't.

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dave_is_slick posted...

Maybe it's because percentages were never my strong suit, but I still fail to see why I should switch when it eventually comes down to one or the other. Honestly, if it was more than one door left alone I could see it but I just don't see why when the final choice is ultimately 50/50. I don't get why it matters what the probability was before, it is no longer that once all others have been eliminated.

That's the thing, it's not an even decision. You have a 1/1000 chance of picking. The host, knowing where all the goats are, consequently knows where the keys to the dream _____ is. So he is in control of the 999/1000 doors. When he opens the doors to the goats, the percentage chance of your door being right doesn't change. It's still 1/1000. His door, because it's lumped with the eliminated goat doors, is a 999/1000 chance of being right.
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dave_is_slick posted...

Or it's behind what I already picked. You are either right or wrong once everything else has been eliminated.

Using the million door version, in order for you to be right, you had to guess one in a million. Obviously it's incredibly unlikely that you hit that one in a million. Now when the host removes all doors except door number 359,228 and tells you either that door is correct, or your one in a million door is correct. You always switch, everytime. You can't boil down the lottery to "you either win or lose, so it's 50-50".

Another way to think of it is you vs the host. The host is a cheater and will always, 100% of the time pick the winning door unless you originally picked it. So you pick your one in a million door, completely random, and then the host can pick the winning door if it's still available, which has a 999,999/1,000,000 chance to still be available. Can you say this game is fair and that it's a 50-50 on if the host wins or I do? No, the host knows the correct door, so by switching your basically taking their door from them, that they cheated to get.

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Anteaterking posted...

When I taught math for non-majors, this usually got the same thing across without running into some of the hang ups people have with Monty Hall for whatever reason:

Suppose you are on vacation and you travel to a city with three hotels. You go to the first hotel but it seems kind of junky, so you drive to the second hotel. This one is somehow worse than the first!

It's too late to go back to the first hotel (someone behind you was going to take the last room), so you have two choices: stay at the second hotel or try your luck with the third one.

Which option leads to a better hotel stay (on average)?

(Note: for the purpose of this puzzle we're assuming that there is a strict ordering of the quality of the hotels without ties)

Sticking with the second one. The third one could either be better or worse but you don't know. The first has been eliminated. I'm sticking with the junky one I know instead of risking it for something worse.

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GATTJT posted...

Using your "right or wrong" logic, my chances of winning the lottery are 50/50, either I do or I don't.

I'm not. What I am doing is seeing that it ultimately comes to two choices at the end. That is not the case with the lottery.

Sad_Face posted...

That's the thing, it's not an even decision. You have a 1/1000 chance of picking. The host, knowing where all the goats are, consequently knows where the keys to the dream _____ is. So he is in control of the 999/1000 doors. When he opens the doors to the goats, the percentage chance of your door being right doesn't change. It's still 1/1000. His door, because it's lumped with the eliminated goat doors, is a 999/1000 chance of being right.

How is it not a decision? You switch or you don't. That is a decision. All others have been eliminated. Then you have to make a decision and since it comes down to only two at the end, I'm either right or wrong. Now, if it wasn't only two doors at the end I might see it but as it is I don't.

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dave_is_slick posted...

How is it not a decision? You switch or you don't. That is a decision. All others have been eliminated. Then you have to make a decision and since it comes down to only two at the end, I'm either right or wrong. Now, if it wasn't only two doors at the end I might see it but as it is I don't.

The function of the host opening a door is to let you pick two doors by switching.

NeoShadowhen posted...

I understand the argument. Has it been tested? Seems like it would be wildly easy to program.

Yes and it turns out you do in fact win 66% of the time approximately.


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dave_is_slick posted...

How is it not a decision? You switch or you don't. That is a decision. All others have been eliminated. Then you have to make a decision and since it comes down to only two at the end, I'm either right or wrong. Now, if it wasn't only two doors at the end I might see it but as it is I don't.

I suppose I had poor phrasing there. It's a decision but not an even decision. Not an even decision in a sense that it's not 50:50. It's 1:999. But do you accept my point in that your door's percentage chance is 1/1000 when you first choose? And your door is still 1/1000 after the host eliminates the doors?
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tiornys posted...

dave_is_slick posted...

How is it not a decision? You switch or you don't. That is a decision. All others have been eliminated. Then you have to make a decision and since it comes down to only two at the end, I'm either right or wrong. Now, if it wasn't only two doors at the end I might see it but as it is I don't.

The function of the host opening a door is to let you pick two doors by switching.

Yep, that's the simplest and quickest way to explain it. You are effectively getting to choose two doors instead of one. There's a reason that when you test it, you win 2/3rds of the time with when you switch, and 1/3rd of the time when you don't.
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dave_is_slick posted...

Sticking with the second one. The third one could either be better or worse but you don't know. The first has been eliminated. I'm sticking with the junky one I know instead of risking it for something worse or better.

So here are the possible rankings from first visit to second visit

1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1

How many of these have the first one better than the second one? Only these three:

1 2 3
1 3 2
2 3 1

How many of these three have the third hotel being better?

1 3 2
2 3 1

So you should go to the third hotel.

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dave_is_slick posted...

I am either wrong or right at this point.

this is not how probability works

there being only two outcomes does not mean it is 50/50

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I legit don't know how people don't understand this.

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Another way to look at it.

When you make your first choice, 2 of the 3 doors have goats. So statistically, you will pick a goat 2 out of 3 times.

The host must expose a goat after your first pick. Since there is a higher probability that your first pick was a goat, there is a higher probability that the host has exposed the second goat. Following that, there is a higher possibility that the final door has the prize. That's why you switch.
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Bad_Mojo posted...

I legit don't know how people don't understand this.

It goes completely against immediate common sense. I consider myself to be pretty smart and perceptive, but it took diagramming it out on a whiteboard before it clicked for me.
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What if you would prefer to have a goat? Much less expensive to maintain.

Hello_Hello_Hey posted...

What if you would prefer to have a goat? Much less expensive to maintain.

Then you should switch to the open door.

dave_is_slick posted...

How is it not a decision? You switch or you don't. That is a decision.

Sad_Face didn't say it's not a decision, he said it's not even.

dave_is_slick posted...

Then you have to make a decision and since it comes down to only two at the end, I'm either right or wrong.

You're trying to choose between two options that don't matter in this problem. It's switch or stay, not right or wrong. Switching always has better probability of getting you the car, because you chose a door (which the game host knows what it has behind it) when there were more to choose from. If at the start there were only 2 doors, then that is 50/50, but with 3 doors it's 33/33/33.

Let's do it like this:

Goat/Goat/Car
Goat/Car/Goat
Car/Goat/Goat

In each scenario, we pick door 1. First scenario, we get a goat. Door 2 is opened. We stay, we get the goat. Second scenario, we get the goat. Door 3 is opened, we stay, we get the goat. Third scenario, we get the car. Here it doesn't matter which door is opened, we stay, we get the car. As you can see, in Scenario 1 and 2, if we switch after a door is opened, we get a car. Keep in mind that the game host is guaranteed to open a goat door.

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Sad_Face posted...

But do you accept my point in that your door's percentage chance is 1/1000 when you first choose?

Yes.

Sad_Face posted...

And your door is still 1/1000 after the host eliminates the doors?

No. I don't get how it's the same when all the others have been eliminated and don't matter anymore.

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ChocoboMogALT posted...

Doesn't matter.

What? Of course it does. What if the host opens the door with the car? The whole point is the host eliminates a goat door and leaves the car door.
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dave_is_slick posted...

No. I don't get how it's the same when all the others have been eliminated and don't matter anymore.

So I think at this point I'm confident in saying you're just pulling our legs.

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dave_is_slick posted...

No. I don't get how it's the same when all the others have been eliminated and don't matter anymore.

You're focusing too much on current conditions and dismissing your prior information. If I flip a card off of a deck, what are the odds that it is red? If I flip a second card off the deck, what are the odds that it's red if the first card was black? What if the first card was red?

I honestly didn't get it until one of these posts drilled in the fact that the first choice is much less likely to be the correct one. I very much thought like dave, but when you say there is a 99% chance that the door hasn't been picked by you (given one of the extreme examples) and the host then has to select it, it made it clear to me.

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dave_is_slick posted...

No. I don't get how it's the same when all the others have been eliminated and don't matter anymore.

Then don't think of it as the doors being eliminated, think of it as the host picking all those doors as a single group, and the host wins if the prize is in any of those doors. You get 1 door, the host gets 999 doors. The host then offers to trade his 999 doors for your single door. Obviously his group has a better chance of winning, so you switch.

Same with 3 doors. You get 1 door, the host gets 2. He offers to trade your single door for his 2 doors, and if either of those two doors have the prize, you win.

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Corrupt_Power posted...

It goes completely against immediate common sense. I consider myself to be pretty smart and perceptive, but it took diagramming it out on a whiteboard before it clicked for me.

it also kind of feels like a compulsory human instinct to stick with your initial pick to avoid possible second-guessing of yourself, because for most people being wrong after changing your answer would feel infinitely worse than just sticking to your guns and being wrong

it "feels" like more of a risk to switch

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Does the original problem state whether the host has knowledge? Because I think that's what everything depends on. If the host has knowledge and picks a goat on purpose, it's better to switch. If the host picked randomly, then it doesn't make a difference.

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VideoboysaysCube posted...

Does the original problem state whether the host has knowledge? Because I think that's what everything depends on

Yes, that IS the problem. Have you not seen the TV show? The host would always bring it down to a loser and a winner, they never gave a player 2 losers. And then they asked them to switch

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VideoboysaysCube posted...

Does the original problem state whether the host has knowledge? Because I think that's what everything depends on. If the host has knowledge and picks a goat on purpose, it's better to switch. If the host picked randomly, then it doesn't make a difference.

Correct.

I will say the best case solution is a bit hard to comprehend unless you have a firm grasp of statistics.

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Drawing a decision tree is also a good way to go about it. You'll find that eliminating one option only improves the chance of the door you didn't pick and wasn't revealed.

But the 1000 door example is probably a far better explanation. At that point they're just wiggling the answer in front of you

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If you chose the door with the goat, did you get to actually keep the goat?

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It's an easy concept to understand if you realize the 1/3 and 2/3 probabilities are locked in right away as soon as the player picks their door, NOT when the player decides to open their door.

The door the player picks initially has a 1/3 chance of containing a car, while the other 2 doors have a combined 2/3 chance of containing a car.

By removing one of the two doors, the last door STILL has the same 2/3 chance to contain a car because you know that no matter what, the door you initially picked will always have a 1/3 chance of containing a car.

Therefore, it's always better to switch.

FL81 posted...

CE doesn't understand it either

incoming 100 post topic

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