5 ms·

It doesn't make a difference what causes Monty to reveal a goat. A goat behind a door the contestant did not choose is eliminated. That is all that matters. It

It doesn't make a difference what causes Monty to reveal a goat. A goat behind a door the contestant did not choose is eliminated. That is all that matters. It could have been done by a Heath-Robinson machine, or a passing lonely shrubber, or elves. Monty isn't a floating variable in the puzzle who makes choices. His choice is fixed. Which I imagine is why vos Savant adds the actually extraneous information that Monty is fully aware what is going on behind the scenes -- to underscore the concept that Monty isn't a variable. The fact that a goat behind an unchosen door was revealed is what is crucial to the setup of the entire puzzle. And that -- despite jncfhnb's protestations -- is information that makes the puzzle determinate.

rspeele 3y ago
> It doesn't make a difference what causes Monty to reveal a goat. Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door. The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.
- deleted 3y ago
  [deleted]
- bemusedthrow75 3y ago
  Nope. But I am going to leave this to someone else to explain. I'm tired out now.
  - jncfhnb 3y ago
    Suppose Car is C. The only possible options are (equally likely) You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B 4 options. You lose 50% of the time. The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat. You choose A and get shown C or you choose B and get shown C are forbidden. 50/50 probability on the dot. If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3
  - rspeele 3y ago
    They are two different problems. Map out all the scenarios exhaustively and you'll find the difference. In both cases we were originally 1/3 chance of being right. That is not in dispute. In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car. In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that: 1. We picked the car and saw Alice 2. We picked the car and saw Bob 3. We picked Alice and saw Bob 4. We picked Bob and saw Alice Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others. This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box. [0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
  - rspeele 3y ago
    Since this bugged me all day, and I suspect you are the kind of person where it bugged you all day, too, here is a better description of the "fall"/"hall" distinction. I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible configurations of the remaining prizes. My Door | Door 2 | Door 3 Car | G1 | G2 Car | G2 | G1 G1 | Car | G2 G1 | G2 | Car G2 | Car | G1 G2 | G1 | Car With the "Monty Hall" problem, Monty uses his knowledge to always open a goat door. Thus we see the following revealed options and resulting 2/3s probability of switching succeeding. This is the classic version of the problem. My Door | Door 2 | Door 3 | Monty Reveals | Switch Result Car | G1 | G2 | Either | Lose Car | G2 | G1 | Either | Lose G1 | Car | G2 | G2 | Win G1 | G2 | Car | G2 | Win G2 | Car | G1 | G1 | Win G2 | G1 | Car | G1 | Win With "Monty Fall", the first thing that happens is a randomly chosen door, that isn't our own, reveals a goat. This is interesting. In the classic problem we were always going to see a goat next, because those are the rules Monty plays by. But in this case, the fact that we randomly found one wasn't guaranteed. Essentially, you are blindfolded and throw a dart at the 2x6 grid of cells under the headers "door 2" and "door 3", and I tell you that the cell you've hit is a goat. What do you know about the row you hit being a switch-or-stay row? Well, half the possible goats you might've hit are in the first 2 scenarios where you should stay, and half the possible goats are in the last 4 scenarios where you should switch. So you're at 50/50. You don't have any new information to switch on. My Door | Door 2 | Door 3 Car | G1(a) | G2(b) Car | G2(c) | G1(d) G1 | Car | G2(e) G1 | G2(f) | Car G2 | Car | G1(g) G2 | G1(h) | Car You are just as likely to be looking at (a), (b), (c), or (d) (so you should stay) as you are to be looking at (e), (f), (g), or (h) (so you should switch). It is 50/50 in this version of the problem.[footnote] This may make it confusing going back to the original. I seem to have shown that both ways make sense but still, how is it different? Imagine it like Monty is doing a random dice roll for which door to open, and he simply juices the outcome by correcting it to the goat door when a car door is selected, since he can't reveal a car and spoil the game. Now we have these equally possible scenarios (a) through (l) for his fair dice roll... My Door | Door 2 | Door 3 Car | G1(a) | G2(b) Car | G2(c) | G1(d) G1 | Car(e) | G2(f) G1 | G2(g) | Car(h) G2 | Car(i) | G1(j) G2 | G1(k) | Car(l) Which he corrects, avoiding cars, to: My Door | Door 2 | Door 3 Car | G1(a) | G2(b) Car | G2(c) | G1(d) G1 | Car | G2(f,e) G1 | G2(g,h)| Car G2 | Car | G1(i,j) G2 | G1(k,l)| Car Now we are back to the original game scenario where we see a goat no matter what. And we can see that 8 of the possible ways we might have arrived at seeing this goat come from "switch" rows while 4 come from "stay" rows.
jncfhnb 3y ago
Noooo It is not the fact that a goat is behind the door. It is the fact that a goat door would always have been opened! These two facts are not the same!
- deleted 3y ago
  [deleted]
- bemusedthrow75 3y ago
  I'm sorry what now? Stop second guessing the puzzle which is clearly stated. I'm so, so done with this now that I am actually going to render inoperable my only HN account so I cannot possibly come back to this, or any other thread.
  - jncfhnb 3y ago
    It is NOT STATED in THIS VERSION OF THE PROMPT that monte will always open a door and that it will always be a goat door. It is simply stated that on one play of the game, Monte CHOSE to open a door and it was a goat. If this is truly all you know, you have learned nothing. It is only useful information if monte explicitly opens a goat door for you. If he opens a door at random and gets goat your information gain goes away even if it’s the same door! I’ve literally mapped it out for you else where. There’s only 4 outcomes. You can check yourself. 2:2

rspeele 3y ago

> It doesn't make a difference what causes Monty to reveal a goat. Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door. The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.

deleted 3y ago
[deleted]
bemusedthrow75 3y ago
Nope. But I am going to leave this to someone else to explain. I'm tired out now.
- jncfhnb 3y ago
  Suppose Car is C. The only possible options are (equally likely) You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B 4 options. You lose 50% of the time. The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat. You choose A and get shown C or you choose B and get shown C are forbidden. 50/50 probability on the dot. If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3
- rspeele 3y ago
  They are two different problems. Map out all the scenarios exhaustively and you'll find the difference. In both cases we were originally 1/3 chance of being right. That is not in dispute. In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car. In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that: 1. We picked the car and saw Alice 2. We picked the car and saw Bob 3. We picked Alice and saw Bob 4. We picked Bob and saw Alice Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others. This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box. [0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
- rspeele 3y ago
  Since this bugged me all day, and I suspect you are the kind of person where it bugged you all day, too, here is a better description of the "fall"/"hall" distinction. I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible configurations of the remaining prizes. My Door | Door 2 | Door 3 Car | G1 | G2 Car | G2 | G1 G1 | Car | G2 G1 | G2 | Car G2 | Car | G1 G2 | G1 | Car With the "Monty Hall" problem, Monty uses his knowledge to always open a goat door. Thus we see the following revealed options and resulting 2/3s probability of switching succeeding. This is the classic version of the problem. My Door | Door 2 | Door 3 | Monty Reveals | Switch Result Car | G1 | G2 | Either | Lose Car | G2 | G1 | Either | Lose G1 | Car | G2 | G2 | Win G1 | G2 | Car | G2 | Win G2 | Car | G1 | G1 | Win G2 | G1 | Car | G1 | Win With "Monty Fall", the first thing that happens is a randomly chosen door, that isn't our own, reveals a goat. This is interesting. In the classic problem we were always going to see a goat next, because those are the rules Monty plays by. But in this case, the fact that we randomly found one wasn't guaranteed. Essentially, you are blindfolded and throw a dart at the 2x6 grid of cells under the headers "door 2" and "door 3", and I tell you that the cell you've hit is a goat. What do you know about the row you hit being a switch-or-stay row? Well, half the possible goats you might've hit are in the first 2 scenarios where you should stay, and half the possible goats are in the last 4 scenarios where you should switch. So you're at 50/50. You don't have any new information to switch on. My Door | Door 2 | Door 3 Car | G1(a) | G2(b) Car | G2(c) | G1(d) G1 | Car | G2(e) G1 | G2(f) | Car G2 | Car | G1(g) G2 | G1(h) | Car You are just as likely to be looking at (a), (b), (c), or (d) (so you should stay) as you are to be looking at (e), (f), (g), or (h) (so you should switch). It is 50/50 in this version of the problem.[footnote] This may make it confusing going back to the original. I seem to have shown that both ways make sense but still, how is it different? Imagine it like Monty is doing a random dice roll for which door to open, and he simply juices the outcome by correcting it to the goat door when a car door is selected, since he can't reveal a car and spoil the game. Now we have these equally possible scenarios (a) through (l) for his fair dice roll... My Door | Door 2 | Door 3 Car | G1(a) | G2(b) Car | G2(c) | G1(d) G1 | Car(e) | G2(f) G1 | G2(g) | Car(h) G2 | Car(i) | G1(j) G2 | G1(k) | Car(l) Which he corrects, avoiding cars, to: My Door | Door 2 | Door 3 Car | G1(a) | G2(b) Car | G2(c) | G1(d) G1 | Car | G2(f,e) G1 | G2(g,h)| Car G2 | Car | G1(i,j) G2 | G1(k,l)| Car Now we are back to the original game scenario where we see a goat no matter what. And we can see that 8 of the possible ways we might have arrived at seeing this goat come from "switch" rows while 4 come from "stay" rows.

jncfhnb 3y ago

Noooo It is not the fact that a goat is behind the door. It is the fact that a goat door would always have been opened! These two facts are not the same!

deleted 3y ago
[deleted]
bemusedthrow75 3y ago
I'm sorry what now? Stop second guessing the puzzle which is clearly stated. I'm so, so done with this now that I am actually going to render inoperable my only HN account so I cannot possibly come back to this, or any other thread.
- jncfhnb 3y ago
  It is NOT STATED in THIS VERSION OF THE PROMPT that monte will always open a door and that it will always be a goat door. It is simply stated that on one play of the game, Monte CHOSE to open a door and it was a goat. If this is truly all you know, you have learned nothing. It is only useful information if monte explicitly opens a goat door for you. If he opens a door at random and gets goat your information gain goes away even if it’s the same door! I’ve literally mapped it out for you else where. There’s only 4 outcomes. You can check yourself. 2:2