The Monte Hall problem

[kirbivore]

 

[Forloveoforange]

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.

So you don't agree with kirbivore's video above?

 

[sutomcat]

So you don't agree with kirbivore's video above?

If you look at this from the perspective of the original game, when there were 3 doors to choose from, yes, the 2 of 3 is correct for the remaining rooms.

That isn’t how I am looking at it but that perspective is perfectly legit and I have no problem agreeing with it.

 

[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?

I'll try to answer this better. Well, obviously, showing a goat next to your game of 2 doors does not change the game. That seems to be separate. You had 50-50, as you know. You either win or you lose. No option for another door. If they just open a third door, you simply don't know if you won or you lost, until they open one of your doors. Then its over.

But the three door option has been explained, and in that video, too. Any luck understanding yet why Door 2 has a 2/3 chance?

Again, the difference is that Monty used the 2/3 chance of Doors 2 and 3 and told you where the 2/3 chance is. Still stays 2/3, and Door 3 was eliminated. Its the whole non-random deal. You take the luck factor away, and that's what you have left. Full value of both doors. If you have Monty guessing which door, then the last 2 doors are 50-50, (and no need to switch choices, but ok to switch, also--wouldn't change your odds). His knowledge made it stay at 66 and 2/3%, that not only Door 2 was the right door, but of course 66 and 2/3 chance that Door 1, your original choice, is the wrong door. Non-random vs guessing. His knowledge makes the difference between 50% and 66 and 2/3%.

Hope this helps.

 

[Forloveoforange]

Here's how 4-door would work. You guess Door 1, you have a 25% chance. There is a 75% chance in Doors 2, 3, and 4 combined. He shows Door 4 is a goat. Now the chances are 25% for your Door 1, 37 and 1/2% for Door 2 and 37 and 1/2% for Door 3. Now he shows Door 3 is a goat. Now you have 25% for Door 1 and 75% for Door 2. Now, its 3 times as likely that Door 2 is your door. In the 3-door game, your chances were 2 times as good for Door 2. I'd have to think about it a little more if you were offered to switch more than once.

Edit:
So... I thought about the extra switch. Door 1 stays 25%. You see, even though there are only 3 doors left, your odds stay as they Were. Doors 2 and 3 hold 75%. If you switch to one of those, say Door 2, then it gets a little complicated. Especially if he then shows Door 1. So he shows Door 1's goat, and you have Doors 2 and 3 left. So he eliminated only 25%. You end up with equal odds of 50-50 on Doors 2 and 3. Switching does not automatically have value.

But, if he had chosen a door out of the 75% "pile", and kept Door 1 at 25%, then your odds would have been 75%. So, actually your immediate switch after he opened Door 4 cost you a large chance difference--75% compared to 50%. Looks like it best to wait until down to 2 doors. You don't want to lose that low chance Door 1 as a factor until its your final move.

 

[STEVEHOLT]

Ok. Maybe a bit simplistic. But i think it’s as simple as switching is equivalent to giving you a second choice. Switching is choosing . You’ve now chosen 2 out of the 3 doors.

 

[SWC75]

I'll try to answer this better. Well, obviously, showing a goat next to your game of 2 doors does not change the game. That seems to be separate. You had 50-50, as you know. You either win or you lose. No option for another door. If they just open a third door, you simply don't know if you won or you lost, until they open one of your doors. Then its over.

But the three door option has been explained, and in that video, too. Any luck understanding yet why Door 2 has a 2/3 chance?

Again, the difference is that Monty used the 2/3 chance of Doors 2 and 3 and told you where the 2/3 chance is. Still stays 2/3, and Door 3 was eliminated. Its the whole non-random deal. You take the luck factor away, and that's what you have left. Full value of both doors. If you have Monty guessing which door, then the last 2 doors are 50-50, (and no need to switch choices, but ok to switch, also--wouldn't change your odds). His knowledge made it stay at 66 and 2/3%, that not only Door 2 was the right door, but of course 66 and 2/3 chance that Door 1, your original choice, is the wrong door. Non-random vs guessing. His knowledge makes the difference between 50% and 66 and 2/3%.

Hope this helps.

Yeah, i know the 1/3 vs. 2/3 thing, which I explained in my opening post. And I know it works that way because I tested it with the cards. What I don't know is why the odds don't change when you eliminate one possibility. You're going from a 3 door game to a 2 door game but he odds remain from the 3 door game.

 

[Forloveoforange]

Yeah, i know the 1/3 vs. 2/3 thing, which I explained in my opening post. And I know it works that way because I tested it with the cards. What I don't know is why the odds don't change when you eliminate one possibility. You're going from a 3 door game to a 2 door game but he odds remain from the 3 door game.

Its the same game, so don't look at it as a new game.

You just have to understand a couple things.

You have to understand that your original choice is a group in itself, and it has the obvious odds that will stay with it, because it is the same game (1 out of 3). Do you understand that?

Next, you have to understand that the remaining doors are 2 out of 3 odds, as a group. Do you understand that?

The host shows that one door of the 2 is not it. But that door group keeps the same 2 out of 3 chance. So Door 2 keeps the same 2 out of 3 chance. I think this is the one that you are not understanding. It is the same group. It loses Door 3, but it is the same group. It is not a 100% chance, but it is 66 and 2/3 %, and it is because the host had to pick a blank door. The odds change because it was not a luck factor pick. This is why it stays a 3 door game, basically, because it is not a luck factor game. The doors have unequal value because they are picked differently due to formula. It is the original game. It doesn't change. The host in the know helps the informed contestant, as long as the contestant understands the original group values.

So, I think you understand the game when you understand that the 2-door group stays at 2 out of 3 chance, and you know which specific door has that individual chance when the other door is taken away.

 

[Forloveoforange]

Ok. Maybe a bit simplistic. But i think it’s as simple as switching is equivalent to giving you a second choice. Switching is choosing . You’ve now chosen 2 out of the 3 doors.

Yes, you have chosen the other part of the equation, the 2-door side. Don't be disappointed by the loss of Door 3. One of those 2 doors was going to be shown 100% of the time. It still has the value of the 2 doors. Just pretend that you have those 2 doors and you haven't even been shown the door 3.

 

[Forloveoforange]

What makes the odds stay the same, and what makes it the same game, is that the host knows. If it were random guesses on the hosts part, then the doors would keep the same face value (in relation to the other unopened doors). The host doesn't have the option of picking random doors, so that is what keeps it the same game, and what keeps Door 1 from never having more than a 1 in 3 chance. Otherwise it would actively get better, because his luck-driven guesses would narrow the choices down equally. Face value. The doors look the same, but the non-randomness changes everything. If there were a bunch of doors, his luck in finding the car would equal the luck of your first guess, and you could keep getting better and better odds on your guess. This is why your guess stays the same, and why the other doors don't change in relation to your door. Your side 1 in 3 and the other side 2 in 3, or if 4 door game, 1 in 4 and 3 in 4, etc.

Switching is unimportant in a random game. You would keep comparing equally to what is left. Every door would have the same face value, so no reason to switch, or not to switch. You take away doors in a random way, that makes the odds go up for what is left, including your original choice. So that's it. I guess that is the key that is mind-boggling.

I think some people think that switching automatically freshens a game, but if its a RANDOM game, then I don't see it. The luck factor should keep your guess increasing in value. If you have 100 doors and you're down to 2 equal chances, you have gotten yourself to 50-50. But even then, you really had only a 2% chance at getting to this "50-50" spot in the game. But remember that is if the game is RANDOM. RANDOM can be a bit mind-boggling too.

 

[SWC75]

Its the same game, so don't look at it as a new game.

You just have to understand a couple things.

You have to understand that your original choice is a group in itself, and it has the obvious odds that will stay with it, because it is the same game (1 out of 3). Do you understand that?

Next, you have to understand that the remaining doors are 2 out of 3 odds, as a group. Do you understand that?

The host shows that one door of the 2 is not it. But that door group keeps the same 2 out of 3 chance. So Door 2 keeps the same 2 out of 3 chance. I think this is the one that you are not understanding. It is the same group. It loses Door 3, but it is the same group. It is not a 100% chance, but it is 66 and 2/3 %, and it is because the host had to pick a blank door. The odds change because it was not a luck factor pick. This is why it stays a 3 door game, basically, because it is not a luck factor game. The doors have unequal value because they are picked differently due to formula. It is the original game. It doesn't change. The host in the know helps the informed contestant, as long as the contestant understands the original group values.

So, I think you understand the game when you understand that the 2-door group stays at 2 out of 3 chance, and you know which specific door has that individual chance when the other door is taken away.

Obviously, I understand that, since I already said it when I introduced this subject. And I already said that I proved to myself that that's how it works out. I don't understand why it's the same game.

 

[SWC75]

What makes the odds stay the same, and what makes it the same game, is that the host knows. If it were random guesses on the hosts part, then the doors would keep the same face value (in relation to the other unopened doors). The host doesn't have the option of picking random doors, so that is what keeps it the same game, and what keeps Door 1 from never having more than a 1 in 3 chance. Otherwise it would actively get better, because his luck-driven guesses would narrow the choices down equally. Face value. The doors look the same, but the non-randomness changes everything. If there were a bunch of doors, his luck in finding the car would equal the luck of your first guess, and you could keep getting better and better odds on your guess. This is why your guess stays the same, and why the other doors don't change in relation to your door. Your side 1 in 3 and the other side 2 in 3, or if 4 door game, 1 in 4 and 3 in 4, etc.

Switching is unimportant in a random game. You would keep comparing equally to what is left. Every door would have the same face value, so no reason to switch, or not to switch. You take away doors in a random way, that makes the odds go up for what is left, including your original choice. So that's it. I guess that is the key that is mind-boggling.

I think some people think that switching automatically freshens a game, but if its a RANDOM game, then I don't see it. The luck factor should keep your guess increasing in value. If you have 100 doors and you're down to 2 equal chances, you have gotten yourself to 50-50. But even then, you really had only a 2% chance at getting to this "50-50" spot in the game. But remember that is if the game is RANDOM. RANDOM can be a bit mind-boggling too.

Wasn't my card experiment a random game? And wasn't I the host? Maybe my removing the instances from the experiment where the black card was the third option eliminated the randomness and caused it to be a 1/3 to 2/3 game.

 

[Forloveoforange]

Wasn't my card experiment a random game? And wasn't I the host? Maybe my removing the instances from the experiment where the black card was the third option eliminated the randomness and caused it to be a 1/3 to 2/3 game.

Yes, your card experiment was random, and that is a problem. I guess you should have taken the first card away, face down, then looked at the last 2 cards and chose a red one to put face up. Then you should have moved your choice to the other card of those last 2. I don't think that you did that, did you?

I mean think about what you would be doing here. Its still the same old question, and you might get it about now. It becomes quite clear you are choosing the 2 cards as opposed to your first choice. And since you are choosing 2 cards over 1 card it is so simple that the 2 cards have double the value, and double the chance of getting a black card.

It really is the same as if the host asked at the beginning, "Would you like to pick one card or 2 cards?" As long as you switch you are getting the 2 cards.

 

[SWC75]

Yes, your card experiment was random, and that is a problem. I guess you should have taken the first card away, face down, then looked at the last 2 cards and chose a red one to put face up. Then you should have moved your choice to the other card of those last 2. I don't think that you did that, did you?

I mean think about what you would be doing here. Its still the same old question, and you might get it about now. It becomes quite clear you are choosing the 2 cards as opposed to your first choice. And since you are choosing 2 cards over 1 card it is so simple that the 2 cards have double the value, and double the chance of getting a black card.

It really is the same as if the host asked at the beginning, "Would you like to pick one card or 2 cards?" As long as you switch you are getting the 2 cards.

Actually, what I did produced the 2/3 ratio and convinced me that that was the reality, even if I didn't understand why. I'm suggesting that if I'd counted all 13 opportunities, it might have come to 50-50 between #1 and #2 but my throwing out the ones were the black card was #3 made me the equivalent of the host who knows that the third door has a goat behind it and set up the 2/3 situation.

 

[SWC75]

I have a box with two decks of cards in it. i took out both decks, separated the black and red cards and made 26 piles of two red cards and 1 black. I shuffled them as best I could, (enough that I could not have known with was the black and put the three cards from each, face down, side by side. The clack card was the 1st one 8 times, the section one 7 times and the third one 11 times. If I'd tossed out the 11 times it was the third one, the score is 8-7 for the first card. Not the result I got when I first did this years ago, when it was 3-7 with one deck. I'll try it again later.

 

[Forloveoforange]

I have a box with two decks of cards in it. i took out both decks, separated the black and red cards and made 26 piles of two red cards and 1 black. I shuffled them as best I could, (enough that I could not have known with was the black and put the three cards from each, face down, side by side. The clack card was the 1st one 8 times, the section one 7 times and the third one 11 times. If I'd tossed out the 11 times it was the third one, the score is 8-7 for the first card. Not the result I got when I first did this years ago, when it was 3-7 with one deck. I'll try it again later.

I was hoping you would take note of what I said about the game being the same as if You pick a card. Then the host, if you switch, gives you the other 2 cards. That is what is really happening. Consider it winning 2 random cards instead of one, if you like .

It is silly how simple it is, really. If it doesn't click here how simple it is, then you still are not getting it. Once you get it, then you get it. You don't have to do experimentation to know that 2 cards will, in the end, double your success, over one card. There is nothing magic about it.

 

[SWC75]

I was hoping you would take note of what I said about the game being the same as if You pick a card. Then the host, if you switch, gives you the other 2 cards. That is what is really happening. Consider it winning 2 random cards instead of one, if you like .

It is silly how simple it is, really. If it doesn't click here how simple it is, then you still are not getting it. Once you get it, then you get it. You don't have to do experimentation to know that 2 cards will, in the end, double your success, over one card. There is nothing magic about it.

it's like Scheffler vs. the field in a major, huh?

The second time I did it, The third card was black 5 times. Of the other 21 times, the first card was black 13 times. That's 21 times in 36 "Monte Hall" situations.

 

[Forloveoforange]

it's like Scheffler vs. the field in a major, huh?

The second time I did it, The third card was black 5 times. Of the other 21 times, the first card was black 13 times. That's 21 times in 36 "Monte Hall" situations.

But do you get it?

I'm not really following what you're doing in your Monty Hall situations, but like I said, there is no need to do experimentation, really, when you understand that it is a straight 1 card random vs 2 cards random situation.

Its strange, I know, to call it random, when I've been saying that it is not random, but in a way it is random, based on how you pick the first card randomly and automatically make the other 2 cards the other choice.

You have to understand that Monty uncovering one of the doors does not change that you are really getting the value of both other doors. If you don't understand this, then you are at the whim of your findings on your chance card game. Its an absolute 33 and 1/3%/ 66 and 2/3 % chance, but your small pool of card games will not necessarily show you that.

 

[SWC75]

But do you get it?

I'm not really following what you're doing in your Monty Hall situations, but like I said, there is no need to do experimentation, really, when you understand that it is a straight 1 card random vs 2 cards random situation.

Its strange, I know, to call it random, when I've been saying that it is not random, but in a way it is random, based on how you pick the first card randomly and automatically make the other 2 cards the other choice.

You have to understand that Monty uncovering one of the doors does not change that you are really getting the value of both other doors. If you don't understand this, then you are at the whim of your findings on your chance card game. Its an absolute 33 and 1/3%/ 66 and 2/3 % chance, but your small pool of card games will not necessarily show you that.

From my initial post:

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.

I've understood the 1/3 vs. 2/3 theory all along. When I first checked it with the card game some years back, I saw that that's how it worked out and accepted it but I still didn't see why the game doesn't change to a two door game with a 50-50 chance when you eliminate one door. My current efforts, which are now a much bigger sample than I looked at then, suggests it does change. Each time the third card comes up red, (a goat), it comes down to how often the black card, (the car) is the first card or the second. That's the Monte Hall situation. I just played the card game again and, of 26 hands, 9 of them had the car behind the third door and thus were not the Monte Hall situation. Of the other 17 hands, the second card was the black one 9 times. The first card was black 8 times. Virtually 50-50. For the three games, I'm 29-24 in Monte Hall situations, (the #1 card has bene black 29 times when the third card was red).

 

[Forloveoforange]

From my initial post:

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.

I've understood the 1/3 vs. 2/3 theory all along. When I first checked it with the card game some years back, I saw that that's how it worked out and accepted it but I still didn't see why the game doesn't change to a two door game with a 50-50 chance when you eliminate one door. My current efforts, which are now a much bigger sample than I looked at then, suggests it does change. Each time the third card comes up red, (a goat), it comes down to how often the black card, (the car) is the first card or the second. That's the Monte Hall situation. I just played the card game again and, of 26 hands, 9 of them had the car behind the third door and thus were not the Monte Hall situation. Of the other 17 hands, the second card was the black one 9 times. The first card was black 8 times. Virtually 50-50. For the three games, I'm 29-24 in Monte Hall situations, (the #1 card has bene black 29 times when the third card was red).

If you understood the game, you would know why it doesn't change to a 2 door game.

Aren't you just basically taking sets of 3 cards and looking at how they show up in order?

So, what you do is, you take ANY one of the 3 cards and place it to the side.

Next, you look at the other 2 cards. They will be either red and black or red and red. Now be like Monty Hall and place a red card face up to the side. This card is out of the game.

So, NOW SWITCH to the other card you peeked at. That's it. You have effectively taken advantage of 2 cards. 2 cards, no matter what 2 cards, 2 cards at face value will ALWAYS have a 2/3 chance of having a black card in them. That is the Monty Hall scenario and you win a car unless your first choice was the car, which was a 1/3 chance.

What if Monty didn't even show a goat? What if he just asked you to pick a 2nd door to add to your first? He would be giving you the same odds as what he did by showing a goat. The game could be random and you would still get the same high odds, since he let you choose 2 random doors.

 

[SWC75]

If you understood the game, you would know why it doesn't change to a 2 door game.

Aren't you just basically taking sets of 3 cards and looking at how they show up in order?

So, what you do is, you take ANY one of the 3 cards and place it to the side.

Next, you look at the other 2 cards. They will be either red and black or red and red. Now be like Monty Hall and place a red card face up to the side. This card is out of the game.

So, NOW SWITCH to the other card you peeked at. That's it. You have effectively taken advantage of 2 cards. 2 cards, no matter what 2 cards, 2 cards at face value will ALWAYS have a 2/3 chance of having a black card in them. That is the Monty Hall scenario and you win a car unless your first choice was the car, which was a 1/3 chance.

What if Monty didn't even show a goat? What if he just asked you to pick a 2nd door to add to your first? He would be giving you the same odds as what he did by showing a goat. The game could be random and you would still get the same high os, since he let you choose 2 random doors.

I shuffle the three cards multiple times- when I first put them in groups of three. When I take a group of three for the next round and when I put them down, which is not in the order they were stacked in at that point. I look at the third card first. if it's black, it's not the Monte Hall scenario. I put those cards aside. If it's red, then the goat is behind the third door. That's the Monte Hall scenario. I then overturn the second card, then the first and record how many times the second card is black and how many times the first card is black. Through 53 times so far, the first card has been black 29 times. In the 1/3 vs. 2/3 scenario I should count the times the third card was black as if the third card was the second card. The third card has been black 25 times. That makes it 29-49, the 2/3 chance that you are looking for. But in 25 of those 49 times, the black card was the third one.

When I first dealt with this, (it was 2008), I came up with three arguments for why the odds would change:

The supposed answer to this is that when Monte opens the third door the odds that the car is behind door #2 go up but the odds that the car is behind door #1 don’t. What if Monte then opened door #2 and there was a goat there? Would the odds that the car was behind door #1 remain the same? Of course not- they would go to 100%. Therefore the opening of a door to reveal a goat changes the odds of the car being behind door #2. The opening of door #3 must have also changed the odds that the car was behind door #1.

What if there were two doors? One car and one goat. The odds are obviously 50-50 that the car is behind door #1. Suppose then that Monte rolls up a third door, opens it and there’s a goat. What difference would that make? What’s different between this and there being three doors initially and Monte opens door #3 and there’s a goat there?

Suppose the producer of the show, in a production meeting before the start of the season said “We’re going to have 99 contestants on this year’s show. To be fair to them, we’re going to put the car behind door #1 33 times, behind door #2 33 times and behind door #3 33 times. So when Monte opens door #3 to reveal the goat, what does that tell us? It tells us that this isn’t one of the 33 times the car is behind door #3. Instead, it’s one of the 33 times it’s behind door #2…or one of the 33 times it was behind door #1. It can’t be one of the 66 times it’s behind door #2 because the producer didn’t decide to put it behind door #2 66 times.

I abandoned those arguments because I did the card game twice, with a single deck of cards and the results pointed to the 1/3 vs. 2/3 argument. I've now done it three times with two decks of cards and I'm at 29 for card #1, 24 for card #2 and 25 for card #3. Toss out card #3, (open that door), and it's 29-24.

I'll do it a third time: 26 hands, 10 times the black card was #1, 9 times it was #2 and 7 times it was #3. It was #1 10 times and 16 times it wasn't - but 7 of those times it was #3. So Monte would not have asked for door #3 to be opened those seven times. Would he have asked for door #2 to be opened the other 9 times? Maybe. But the opening of a door doesn't add 7 cars to door #2 or 9 cars to door #3. Anyway, through 72 Monte Hall situations, (a red card, meaning a goat) was #3), #1 has had the black card, (a car 39 times. #2 has had the black card 33 times.

Sherlock Holmes said “Eliminate all that is impossible, whatever remains is the explanation, however improbable.” I was willing to accept the 1/3 vs. 2/3 argument in 2008 because I did this with a deck of cards twice, (26 'hands'). I've now done it with two decks of card four times, (104 hands) and it doesn't hold up. With three doors, you are going to get a card behind each door about 1/3 of the time. Of the times, If it's not behind the third door, the odds it's behind either doors #1 or #2 both increase.

Try it yourself.

 

[Forloveoforange]

I shuffle the three cards multiple times- when I first put them in groups of three. When I take a group of three for the next round and when I put them down, which is not in the order they were stacked in at that point. I look at the third card first. if it's black, it's not the Monte Hall scenario. I put those cards aside. If it's red, then the goat is behind the third door. That's the Monte Hall scenario. I then overturn the second card, then the first and record how many times the second card is black and how many times the first card is black. Through 53 times so far, the first card has been black 29 times. In the 1/3 vs. 2/3 scenario I should count the times the third card was black as if the third card was the second card. The third card has been black 25 times. That makes it 29-49, the 2/3 chance that you are looking for. But in 25 of those 49 times, the black card was the third one.

When I first dealt with this, (it was 2008), I came up with three arguments for why the odds would change:

The supposed answer to this is that when Monte opens the third door the odds that the car is behind door #2 go up but the odds that the car is behind door #1 don’t. What if Monte then opened door #2 and there was a goat there? Would the odds that the car was behind door #1 remain the same? Of course not- they would go to 100%. Therefore the opening of a door to reveal a goat changes the odds of the car being behind door #2. The opening of door #3 must have also changed the odds that the car was behind door #1.

What if there were two doors? One car and one goat. The odds are obviously 50-50 that the car is behind door #1. Suppose then that Monte rolls up a third door, opens it and there’s a goat. What difference would that make? What’s different between this and there being three doors initially and Monte opens door #3 and there’s a goat there?

Suppose the producer of the show, in a production meeting before the start of the season said “We’re going to have 99 contestants on this year’s show. To be fair to them, we’re going to put the car behind door #1 33 times, behind door #2 33 times and behind door #3 33 times. So when Monte opens door #3 to reveal the goat, what does that tell us? It tells us that this isn’t one of the 33 times the car is behind door #3. Instead, it’s one of the 33 times it’s behind door #2…or one of the 33 times it was behind door #1. It can’t be one of the 66 times it’s behind door #2 because the producer didn’t decide to put it behind door #2 66 times.

I abandoned those arguments because I did the card game twice, with a single deck of cards and the results pointed to the 1/3 vs. 2/3 argument. I've now done it three times with two decks of cards and I'm at 29 for card #1, 24 for card #2 and 25 for card #3. Toss out card #3, (open that door), and it's 29-24.

I'll do it a third time: 26 hands, 10 times the black card was #1, 9 times it was #2 and 7 times it was #3. It was #1 10 times and 16 times it wasn't - but 7 of those times it was #3. So Monte would not have asked for door #3 to be opened those seven times. Would he have asked for door #2 to be opened the other 9 times? Maybe. But the opening of a door doesn't add 7 cars to door #2 or 9 cars to door #3. Anyway, through 72 Monte Hall situations, (a red card, meaning a goat) was #3), #1 has had the black card, (a car 39 times. #2 has had the black card 33 times.

Sherlock Holmes said “Eliminate all that is impossible, whatever remains is the explanation, however improbable.” I was willing to accept the 1/3 vs. 2/3 argument in 2008 because I did this with a deck of cards twice, (26 'hands'). I've now done it with two decks of card four times, (104 hands) and it doesn't hold up. With three doors, you are going to get a card behind each door about 1/3 of the time. Of the times, If it's not behind the third door, the odds it's behind either doors #1 or #2 both increase.

Try it yourself.

Why wouldn't it be 1/3 the way you are doing it? You are taking face value history of cards you are labeling Card 1, Card 2 and Card 3, right.?

The first part of the game is to move your choice card, hidden, to the side. But lets see what happens if you can actually see it. To keep it simple, you always pick 1 as your first pick.

If 1 is black, 2 is red, 3 is red

So monty shows red 3 , and you choose to switch to door 2, you lose.

Next, 1 is red, 2 is black, 3 is red

In this case Monty has to pick 3, you switch to 2, you win

Last game, 1 is red, 2 is red, 3 is black

In this final one Monty has to pick 2, you switch to 3 and win

You win 2/3 of the time. If you didn't switch at all, you would have won 1/3 of the time.

What I just did is EVERY scenario of the Monty Hall problem. You are absolutely not mimicking the game. You are not hand-picking and switching. You really only have 3 scenarios, if you always pick Card 1. If you keep the cards face up, 9 cards will tell you all you want to know.

 

[SWC75]

Why wouldn't it be 1/3 the way you are doing it? You are taking face value history of cards you are labeling Card 1, Card 2 and Card 3, right.?

The first part of the game is to move your choice card, hidden, to the side. But lets see what happens if you can actually see it. To keep it simple, you always pick 1 as your first pick.

If 1 is black, 2 is red, 3 is red

So monty shows red 3 , and you choose to switch to door 2, you lose.

Next, 1 is red, 2 is black, 3 is red

In this case Monty has to pick 3, you switch to 2, you win

Last game, 1 is red, 2 is red, 3 is black

In this final one Monty has to pick 2, you switch to 3 and win

You win 2/3 of the time. If you didn't switch at all, you would have won 1/3 of the time.

What I just did is EVERY scenario of the Monty Hall problem. You are absolutely not mimicking the game. You are not hand-picking and switching. You really only have 3 scenarios, if you always pick Card 1. If you keep the cards face up, 9 cards will tell you all you want to know.

I have proven that if the car is not behind door three, there's a 50-50 chance it will be behind #1 and #2. I would think that If Monte has door #2, rather than door #3 opened, it's the same situation.

 

[Forloveoforange]

I have proven that if the car is not behind door three, there's a 50-50 chance it will be behind #1 and #2. I would think that If Monte has door #2, rather than door #3 opened, it's the same situation.

I'll ask this in the form of a question. Did you pick your first pick, then hand-pick your "host" choice, then switch? I don't hear you doing any of these things, yet you think you are mimicking the game.

What you are doing absolutely DOES give 50-50 odds, because your picks are not getting the special treatment that Monty is giving the non-originally-picked doors.

Edit: You do realize that Monty is choosing around the odd doors, right? Sorry, we've been back and forth, so I'm not sure what you've acknowledged. You act like you don't realize what the host is doing, because the fact that you are not doing the picking around and switching, makes it seem that maybe you don't know that the host is 100% picking around and you are meant to react by switching. You just aren't duplicating that, and that keeps it as you say 50-50. That is the difference in what you are doing.

I spelled it out in the last post. I don't know why the 3 scenarios aren't your new proof, because they are what happen in the game, and they lead to 66% success when you make a habit of switching.

 

[SWC75]

I think our way out of this is to say that if someone is considering a theory that it don't matter if you switch and another that you ought to switch, you might as well go ahead an switch.