The Monte Hall problem

[SWC75]

This started on the Eddie Lampkin thread and had nothing to do with it so I thought I'd put it in the portal to here. It's about a game show and math so it's kind of an "other sport".

It started here:

MS Orange: Exactly. I'm well aware that Lampkin has some limitations in his game, but I wasn't at all excited about most of the options being mentioned behind door #2.

My jokey response:

Monty Hall problem - Wikipedia

en.wikipedia.org

Ottoingrotto: This is a good post.

Me: Two theories:

- You take away one option, you are left with two options. It must be 50-50, right?

- You have a 1/3 chance it's behind the first door and that doesn't change just because it's not behind the third door.

I will go to my grave thinking that the first theory makes the most sense. What if you started out with two doors, had the contestant chose one of them, then wheeled out a third door, opened it and there was a goat there. Would that have made it a good idea for the contestant to switch? Isn't that the same situation?

I decided to test it with a deck of cards. I separated the black and red cards. The black cards are cars, the red goats. I combined one black card with two red cards. I shuffled the three cards as best I could. I took the first three and put them face down. I 'chose' #1 and flipped the third card over. if it was black, I put those three cards aside. If it was red, I then flipped the second card and recorded when it was black and when it was red. I had, as I recall 10 of 13 sets where the third card was red and the second card was black 7 times. I reshuffled them and did it a gain a couple more times and the 2/3 percentage held. I then realized that I was wrong but still don't understand why.

Skeeter223: And somehow, "Let's Make a Deal" became one of the most intriguing parts of the discussion.

Forloveoforang(e?): It took me 20 hours to figure out that it is absolutely twice as likely that its door number 2!

Door number one is a 1/3 chance. Doors 2 and 3 combined are a 2/3 chance. Monte does the contestant a huge favor by saying that Door number 3 is not it. So that means that number 2 has the full 2/3 chance.

Showing a non-car door does not change the 1/3 chance for door number 1. Again the bottom line is that doors 2 and 3 combined are a 2/3 chance, and Monte let the contestant know that the 2/3 chance is in door number 2 by opening door number 3.

That's quite a tough one to get your head around!

ROCuse: The first time you pick out of three doors the chance of success is 33%. The next time you pick out of two doors the chance of success is 50%. The scenario is are independent of each other, so given the chance to switch you should always switch to the “roll” with the best odds for success.

Ragman: This is incorrect. Look up the Money Hall problem explained. The correct answer is 67% chance of meeting correct if you switch, not 50.

Fat Tire 13: Hurts my brain too - and if you want to go down the Wiki rabbit hole….

Forloveoforang: Yes, depending on how the host is playing the game...There is a 2/3 chance that Monte did not pick door number 2 because the car was behind it, right? That's why the 2 doors are not equal 50-50 chances.

Me: Yeah, that's the second option. But why does it rule over the first option?

AZOrange: Take the issue to an absurd level. Let’s say you had 100 suitcases in them 1 with $1mm in it. You pick 1 and the host reveals 98 losers on the other side. Does it make any sense that you have a 50/50 chance now when you started with 1%?

Me: Yes. That doesn't make it true, (as I explained), but you are left with two suitcases, one of which has $1mm in it. I understand both arguments. I just don't understand why the second one rules the situation.

Forloveoforang: Like I said, I understand it is hard to understand. But I think the key is knowing that the combined 2 and 3 doors hold a 2 out of 3 chance. When its known that 3 is not it, it swings all of that 2 out of 3 chance to the 2nd door. Its like ah ha, the host had to pick over the 2nd door ( the scenario being that he knows which door has the car). The 2 remaining doors are not created equally, anymore. The second door absolutely holds the 66 and 2/3 % chance that is missing. (There is still a 33 and 1/3 % chance that it is Door number 1). Its hard to get your head around that fact.

JustPheru: I’ll find Jimmy Hoffa before I can understand this.

Capt Tuttle: Took me a while to get it, too.
The scenarios aren’t independent. They exist on a continuum. The 50/50 would only exist if Monty had the curtain closed so they could move the prizes around (or not) between the 2 choices. Because that doesn’t happen, door #1 will always have a1/3 probability.
Another way to think of it is after choosing door #1, there is a 2/3 probability the car is behind one of the other two. That doesn’t change with the elimination of door 3.
The determinating condition is that there were 3 choices for the original placement of the car. That doesn’t change.

Cuse181: I think that is sort of the simple way to go about it. You are basically being given a second door. All the doors had 33% chance, you picked 2, he gave you 3 the idea (I think) is you’re going into your next decision with a 66% of being right because you KNOW 3 is a goat and 2 was your choice. You implicitly received information that 2 is not 1 of the 2 non-car doors, thus giving you the improved chance of being right.

AZOrange: You odds don’t change just because an answer is revealed. When this ”problem” came out many disputed it until they did a large number of testing of the hypothesis.

In summary, I understand the 1/3 vs. 2/3 argument and the alternative "now it's 50-50" scenario and have tested it and found that 1/3 vs. 2/3 is the one that works. I just don't understand why it does.

 

[mscanlo]

I think of it this way

When you select the one door out of three, you have a 1/3 chance you are right and there is a 2/3 chance the host wins

While the host still has 2/3 chance of winning, they open one of their doors (that they know doesn’t have the prize behind it),

After the host opens the door, you have the option to “switch spots with the host” and trade your door for them, and thus trading your 1/3 odds in favor of their 2/3 odds

 

[sutomcat]

I think of it this way

When you select the one door out of three, you have a 1/3 chance you are right and there is a 2/3 chance the host wins

While the host still has 2/3 chance of winning, they open one of their doors (that they know doesn’t have the prize behind it),

After the host opens the door, you have the option to “switch spots with the host” and trade your door for them, and thus trading your 1/3 odds in favor of their 2/3 odds

I don't think this is quite right.

Can we all agree that when you make your initial door selection your odds of getting a car are 1 of 3?

That shouldn't be controversial. Your odds of getting the car are 1 of 3.

When Monty shows a goat behind one of the doors you did not choose, it doesn't change your odds of getting a car with the door you initially chose. You could take the other two rooms and incinerate them. The odds that you chose the room with the car remain 1 of 3.

The reason you should always take the other door in this scenario is because if you do, you increase your probability of getting the car from 1 in 3 to 1 in 2. If you switch, and only if you switch to the other room, you change the game you are playing and are now there are only 2 rooms in play. So your odds of getting the room with the car go from 1 in 3 to 1 in 2.

It is counter intuitive but it makes sense if you think about it.

 

[MikeSU02]

AZOrange: Take the issue to an absurd level. Let’s say you had 100 suitcases in them 1 with $1mm in it. You pick 1 and the host reveals 98 losers on the other side. Does it make any sense that you have a 50/50 chance now when you started with 1%?

SWC75 this is how I got it ^^

The way someone hammered it to me is like this. Let's say there are 100 suitcases sitting on a beach and only one has a million bucks in it and there is a host that is saying you get to pick one to win.

You choose one.

There are two little tiny islands about 50 feet out from the beach. They take the one you chose out to the island on the left and the 99 other suitcases go out to the tiny island on the right.

After this, if you were given the choice to choose either to stay with your choice of one suitcase on the left island vs the right with 99 suitcases, you would obviously choose the right. We all would. Give me 99% chance every day.

BUT, that wouldn't make for good tv.

So the host says, "I'm going to open 98 empty suitcases on the island with 99 total suitcases... and once I do, you will then have the chance to switch your choice. Do you want to stay with your choice on the left island or switch it to right island?"

That is the same island that everyone would have jumped at to choose before the host opened up the 98 losers. It's holding that probability.

I know this just may be redundant to the convo you already had, so apologies. But that is what gave me clarity for it.

If you've ever seen the movie 21, they do this with the three doors/game show host in the beginning of the movie.

[EDIT: AZOrange please correct me if I summarized what you were trying to relay incorrectly.]

 

[mscanlo]

I think some guy did a thesis on this, and faced a lot of pushback for saying switching the door gets a 2/3 chance

 

[mscanlo]

It

I don't think this is quite right.

Can we all agree that when you make your initial door selection your odds of getting a car are 1 of 3?

That shouldn't be controversial. Your odds of getting the car are 1 of 3.

When Monty shows a goat behind one of the doors you did not choose, it doesn't change your odds of getting a car with the door you initially chose. You could take the other two rooms and incinerate them. The odds that you chose the room with the car remain 1 of 3.

The reason you should always take the other door in this scenario is because if you do, you increase your probability of getting the car from 1 in 3 to 1 in 2. If you switch, and only if you switch to the other room, you change the game you are playing and are now there are only 2 rooms in play. So your odds of getting the room with the car go from 1 in 3 to 1 in 2.

It is counter intuitive but it makes sense if you think about it.

they call it a “stastical illusion” or paradox

If you like that one, you should look up Bertrand’s Box paradox

 

[sutomcat]

It

they call it a “stastical illusion” or paradox

If you like that one, you should look up Bertrand’s Box paradox

I am familiar with it. Computer science major. This stuff is all right up my alley...

 

[SWC75]

What about my original complaint. You start out with two doors. You pick door #1. It's 50-50. They wheel out another door, open it and there's a goat there. Did the odds ever change from 50-50? How is that different from having three doors to begin with?

 

[Forloveoforange]

What about my original complaint. You start out with two doors. You pick door #1. It's 50-50. They wheel out another door, open it and there's a goat there. Did the odds ever change from 50-50? How is that different from having three doors to begin with?

Here is the thing with the three doors. You pick door 1. The host then shows door 3. You like to think that your odds have turned to 50/50, but in Reality, the host was always going to show you a door with a goat, so that activity can not change anything. You had a 1 in 3 chance and that remains. The goat is either behind door 1 or it isn't. Its still a 1 in 3 chance, since there were 3 doors to begin with, and that was your guess, which door out of 3 doors.

So the second thing is, where is the other 2 in 3 chance, right? Well it was in the combination doors 2 and 3, but host tells you it isn't 3, so now you have 2 out of 3 chance on door 2, since door 1 odds are not going higher than 1 in 3, and door 2 has the full odds of the doors 2 and 3 combination.

Door 2 and 3 combined at 66 and 2/3 % chance becomes door 2 at 66 and 2/3 % chance.

 

[Forloveoforange]

I don't think this is quite right.

Can we all agree that when you make your initial door selection your odds of getting a car are 1 of 3?

That shouldn't be controversial. Your odds of getting the car are 1 of 3.

When Monty shows a goat behind one of the doors you did not choose, it doesn't change your odds of getting a car with the door you initially chose. You could take the other two rooms and incinerate them. The odds that you chose the room with the car remain 1 of 3.

The reason you should always take the other door in this scenario is because if you do, you increase your probability of getting the car from 1 in 3 to 1 in 2. If you switch, and only if you switch to the other room, you change the game you are playing and are now there are only 2 rooms in play. So your odds of getting the room with the car go from 1 in 3 to 1 in 2.

It is counter intuitive but it makes sense if you think about it.

No, switching doors goes from 1 in 3 to 2 in 3 odds. You can't have one scenario at 1 in 3 and the other at 1 in 2. They have to add up to 3 in 3.

 

[SWC75]

Here is the thing with the three doors. You pick door 1. The host then shows door 3. You like to think that your odds have turned to 50/50, but in Reality, the host was always going to show you a door with a goat, so that activity can not change anything. You had a 1 in 3 chance and that remains. The goat is either behind door 1 or it isn't. Its still a 1 in 3 chance, since there were 3 doors to begin with, and that was your guess, which door out of 3 doors.

So the second thing is, where is the other 2 in 3 chance, right? Well it was in the combination doors 2 and 3, but host tells you it isn't 3, so now you have 2 out of 3 chance on door 2, since door 1 odds are not going higher than 1 in 3, and door 2 has the full odds of the doors 2 and 3 combination.

Door 2 and 3 combined at 66 and 2/3 % chance becomes door 2 at 66 and 2/3 % chance.

I know. I'm wondering why the odds don't change. There were three doors to begin with but there aren't any more.

 

[Melancer46]

The part I struggle to wrap my head around is this all only works if the host knows where the car is. Everything I’ve read about this problem says if the host just randomly opens a door, it’s no longer advantageous to switch. This example probably doesn’t work with just 3 doors since the host would randomly reveal the car a decent amount of the time but with the 100 suitcase example, how would randomly revealing a few goats not tilt the odds but purposefully revealing a few goats would?

That’s the part I personally struggle with even though I understand the randomness is a major factor. I even kinda understand the math that explains the difference but it is admittedly still hard for me to wrap my head around.

 

[Knicks411]

I like one of the explanations in the Wiki, which is if you initially pick the right door and then switch, you lose. And if you initially pick the wrong door and switch, you win. And you have a 1/3 chance of initially picking the right door, so...

 

[Forloveoforange]

I know. I'm wondering why the odds don't change. There were three doors to begin with but there aren't any more.

The odds don't change, because like Melancer said after you, the key part of the equation is that the host is not randomly opening doors like a clueless bystander. If that were the case, then yes, the fewer the doors, the better the odds. Up to 50-50. If you had many doors, picked at random, your odds would steadily improve to 50-50, or until the magic door accidentally got opened.

At least I think the odds should improve that way. I mean the other doors would then be using the same luck factor, so the doors left over would require the same luck. 50-50. Someone let me know if that's wrong.

But since, part of the equation is that he will always open a non-car door, how could your odds change from 1 out of 3? You picked a door, he showed a horse door, blah, blah, blah. That's the equation. Still the same odds when the equation tells you he's going to open a non-car door. 1 out of 3.

 

[Forloveoforange]

I like one of the explanations in the Wiki, which is if you initially pick the right door and then switch, you lose. And if you initially pick the wrong door and switch, you win. And you have a 1/3 chance of initially picking the right door, so...

Yes, one of the keys is knowing that there must be a 2/3 in the other door, which is the accumulation of doors 2 and 3 combined. So at the beginning of the game, would you rather pick door #1 or would you rather pick DOORS #2 AND #3? That is the TRUE effect of the game.

 

[sutomcat]

No, switching doors goes from 1 in 3 to 2 in 3 odds. You can't have one scenario at 1 in 3 and the other at 1 in 2. They have to add up to 3 in 3.

I respectfully disagree.

When Monty opens door 3 and shows the goat, and offers the alternative to keeping the door originally suggested, he is offering a chance to switch to a different game of chance.

If you switch, there are now only 2 doors to choose from. The third option has clearly been eliminated. You can't include it as an option.

If you stick with your first choice, your odds remain 1 of 3.

If you switch to the other remaining door, the odds go from 1 of 3 to 1 of 2.

 

[Knicks411]

I respectfully disagree.

When Monty opens door 3 and shows the goat, and offers the alternative to keeping the door originally suggested, he is offering a chance to switch to a different game of chance.

If you switch, there are now only 2 doors to choose from. The third option has clearly been eliminated. You can't include it as an option.

If you stick with your first choice, your odds remain 1 of 3.

If you switch to the other remaining door, the odds go from 1 of 3 to 1 of 2.

Isn't the problem the math just doesnt work out? Or am I being an idiot here?
You stick, you're right 33% of the time. You switch, you're right 50%. So where is the other 17%? I think you have to consider them the same game of chance because you know after you make your first selection they're going to eliminate one of the wrong choices.

Though I do see the point you are making and it underscores why switching is the obvious strategy

 

[Forloveoforange]

The part I struggle to wrap my head around is this all only works if the host knows where the car is. Everything I’ve read about this problem says if the host just randomly opens a door, it’s no longer advantageous to switch. This example probably doesn’t work with just 3 doors since the host would randomly reveal the car a decent amount of the time but with the 100 suitcase example, how would randomly revealing a few goats not tilt the odds but purposefully revealing a few goats would?

That’s the part I personally struggle with even though I understand the randomness is a major factor. I even kinda understand the math that explains the difference but it is admittedly still hard for me to wrap my head around.

I think you might call it the luck factor. If random luck eliminates choices, then that is the same luck that could help you to pick the right one to begin with. Luck is equal to luck. 98 suitcases opened in a lucky way will give the same luck factor as you picking potentially the right one to begin with. But if the host says he's going to open 98 empty suitcases, there is no luck involved.

I respectfully disagree.

When Monty opens door 3 and shows the goat, and offers the alternative to keeping the door originally suggested, he is offering a chance to switch to a different game of chance.

If you switch, there are now only 2 doors to choose from. The third option has clearly been eliminated. You can't include it as an option.

If you stick with your first choice, your odds remain 1 of 3.

If you switch to the other remaining door, the odds go from 1 of 3 to 1 of 2.

That's fine, thank you, for respectfully disagreeing, but same here. In my opinion, it has to add up to 100%. 33 and 1/3% plus 50% equals 83 and 1/2%. What happens the other 17 and 1/2 %?

Isn't the problem the math just doesnt work out? Or am I being an idiot here?
You stick, you're right 33% of the time. You switch, you're right 50%. So where is the other 17%? I think you have to consider them the same game of chance because you know after you make your first selection they're going to eliminate one of the wrong choices.

Though I do see the point you are making and it underscores why switching is the obvious strategy

Exactly, you can't have a 33 and 1/3 % chance of rain and a 50% chance of no rain.

 

[upperdeck]

You always have a 50/50 chance. No matter which one you chose you never had a 1 in 3 because you can always change whether you were right or wrong.

its the added drama from choosing off the correct choice.

 

[sutomcat]

I think you might call it the luck factor. If random luck eliminates choices, then that is the same luck that could help you to pick the right one to begin with. Luck is equal to luck. 98 suitcases opened in a lucky way will give the same luck factor as you picking potentially the right one to begin with. But if the host says he's going to open 98 empty suitcases, there is no luck involved.

That's fine, thank you, for respectfully disagreeing, but same here. In my opinion, it has to add up to 100%. 33 and 1/3% plus 50% equals 83 and 1/2%. What happens the other 17 and 1/2 %?

Exactly, you can't have a 33 and 1/3 % chance of rain and a 50% chance of no rain.

Sigh.

These are two separate choices you can make as a game show participant.

You make the 1 of 3 choice, which I think we can all agree gives you a 33.3% (or thereabouts) chance of being right.

At this point, you have a 67.7% (or thereabouts) chance of being wrong.

If you then switch after one of the goat rooms is revealed, you are done with the 1 of 3 option. You have changed to a new game, where there are two rooms in play and you have a 1 of 2 chance to pick the right one. When you switch to playing the second game, you have a 50-50 chance of being right.

In both games, the possible options add to 100%.

The thing is, the only way you can change from the 1 of 3 game to the 1 of 2 game is if you switch rooms. That is why you should always switch.

 

[upperdeck]

we forget that when this same things happens in horse racing and you change off the original pick your odds go down 100% not up..

 

[Forloveoforange]

Sigh.

These are two separate choices you can make as a game show participant.

You make the 1 of 3 choice, which I think we can all agree gives you a 33.3% (or thereabouts) chance of being right.

At this point, you have a 67.7% (or thereabouts) chance of being wrong.

If you then switch after one of the goat rooms is revealed, you are done with the 1 of 3 option. You have changed to a new game, where there are two rooms in play and you have a 1 of 2 chance to pick the right one. When you switch to playing the second game, you have a 50-50 chance of being right.

In both games, the possible options add to 100%.

The thing is, the only way you can change from the 1 of 3 game to the 1 of 2 game is if you switch rooms. That is why you should always switch.

This is a new concept to me. We both agree that Door 1 has a 1 in 3 chance and Door 2 has a 2 in 3 chance. Why is there a reset to face value? You are throwing the history out the window. If you are correct, color me not understanding. There must be some reason you are saying this... , but I'm missing it.

 

[OburgOrange]

I was told there wouldn’t be math

 

[Forloveoforange]

Sigh.

These are two separate choices you can make as a game show participant.

You make the 1 of 3 choice, which I think we can all agree gives you a 33.3% (or thereabouts) chance of being right.

At this point, you have a 67.7% (or thereabouts) chance of being wrong.

If you then switch after one of the goat rooms is revealed, you are done with the 1 of 3 option. You have changed to a new game, where there are two rooms in play and you have a 1 of 2 chance to pick the right one. When you switch to playing the second game, you have a 50-50 chance of being right.

In both games, the possible options add to 100%.

The thing is, the only way you can change from the 1 of 3 game to the 1 of 2 game is if you switch rooms. That is why you should always switch.

Or... are you treating this "problem" as random. I'm treating it as the host not being random. If you are treating it as random, then that may explain why you're saying 50%.

 

[sutomcat]

Or... are you treating this "problem" as random. I'm treating it as the host not being random. If you are treating it as random, then that may explain why you're saying 50%.

I am not treating this as if the host picked a door with a goat randomly. Monty knew there was a goat behind the door he revealed.

I am saying that once Monty eliminates one of the three doors as a choice (he does this with the goat reveal), the probability that the car is behind one of the other doors changes from 1 of 3 to 1 of 2. From 33.3% to 50%. Because there are only two choices left. I don’t know why you think the probability the car is in one of the remaining rooms goes to 2 of 3. There is only one car and two rooms left. How can the probability a single car is in one of the two rooms be anything other than 1 of 2 (or 50%)?

It is an abstract concept that many clearly have trouble understanding.

And I must be bad at explaining. I apologize for that.