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#6: Post edited by user avatar DNB‭ · 2021-08-24T06:08:04Z (about 3 years ago)
  • The second quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? The first quotation below uses "transfer". But how can probabilities "shift" or "transfer"?
  • This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
  • 1. Judea Pearl, _The Book of Why _(2018). Page 181 of 402.
  • >![](https://i.imgur.com/zBGwqvk.jpg)
  • >&nbsp; &nbsp; &nbsp; As you can see from Figure 6.1, Door Opened is a collider. Once we
  • obtain information on this variable, all our probabilities become conditional
  • on this information. But when we condition on a collider, we create a spurious
  • dependence between its parents. The dependence is borne out in the
  • probabilities: if you chose Door 1, the car location is twice as likely to be
  • behind Door 2 as Door 1; if you chose Door 2, the car location is twice as
  • likely to be behind Door 1.
  • >&nbsp; &nbsp; &nbsp; It is a bizarre dependence for sure, one of a type that most of us are
  • unaccustomed to. It is a dependence that has no cause. It does not involve
  • physical communication between the producers and us. It does not involve
  • mental telepathy. It is purely an artifact of Bayesian conditioning: a magical
  • transfer of information without causality. Our minds rebel at this possibility
  • because from earliest infancy, we have learned to associate correlation with
  • causation. If a car behind us takes all the same turns that we do, we first think
  • it is following us (causation!). We next think that we are going to the same
  • place (i.e., there is a common cause behind each of our turns). But causeless
  • correlation violates our common sense. Thus, the Monty Hall paradox is just
  • like an optical illusion or a magic trick: it uses our own cognitive machinery
  • to deceive us.
  • >&nbsp; &nbsp; &nbsp; Why do I say that Monty Hall’s opening of Door 3 was a “transfer of
  • information”? It didn’t, after all, provide any evidence about whether your
  • initial choice of Door 1 was correct. You knew in advance that he was going
  • to open a door that hid a goat, and so he did. No one should ask you to change
  • your beliefs if you witness the inevitable. So how come your belief in Door 2
  • has gone up from one-third to two-thirds?
  • 2. [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • 3. [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
  • The second quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? The first quotation below uses "transfer". But how can probabilities "shift" or "transfer"?
  • This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
  • 1. Judea Pearl, _The Book of Why_ (2018). Page 181 of 402.
  • >![](https://i.imgur.com/zBGwqvk.jpg)
  • >&nbsp; &nbsp; &nbsp; As you can see from Figure 6.1, Door Opened is a collider. Once we
  • obtain information on this variable, all our probabilities become conditional
  • on this information. But when we condition on a collider, we create a spurious
  • dependence between its parents. The dependence is borne out in the
  • probabilities: if you chose Door 1, the car location is twice as likely to be
  • behind Door 2 as Door 1; if you chose Door 2, the car location is twice as
  • likely to be behind Door 1.
  • >&nbsp; &nbsp; &nbsp; It is a bizarre dependence for sure, one of a type that most of us are
  • unaccustomed to. It is a dependence that has no cause. It does not involve
  • physical communication between the producers and us. It does not involve
  • mental telepathy. It is purely an artifact of Bayesian conditioning: a magical
  • transfer of information without causality. Our minds rebel at this possibility
  • because from earliest infancy, we have learned to associate correlation with
  • causation. If a car behind us takes all the same turns that we do, we first think
  • it is following us (causation!). We next think that we are going to the same
  • place (i.e., there is a common cause behind each of our turns). But causeless
  • correlation violates our common sense. Thus, the Monty Hall paradox is just
  • like an optical illusion or a magic trick: it uses our own cognitive machinery
  • to deceive us.
  • >&nbsp; &nbsp; &nbsp; Why do I say that Monty Hall’s opening of Door 3 was a “transfer of
  • information”? It didn’t, after all, provide any evidence about whether your
  • initial choice of Door 1 was correct. You knew in advance that he was going
  • to open a door that hid a goat, and so he did. No one should ask you to change
  • your beliefs if you witness the inevitable. So how come your belief in Door 2
  • has gone up from one-third to two-thirds?
  • 2. [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • 3. [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
#5: Post edited by user avatar DNB‭ · 2021-08-24T06:06:13Z (about 3 years ago)
  • The second quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? The first quotation below uses "transfer". But how can probabilities "shift" or "transfer"?
  • This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
  • 1. Judea Pearl, The Book of Why (2018). Page 181 of 402.
  • >![](https://i.imgur.com/zBGwqvk.jpg)
  • >&nbsp; &nbsp; &nbsp; As you can see from Figure 6.1, Door Opened is a collider. Once we
  • obtain information on this variable, all our probabilities become conditional
  • on this information. But when we condition on a collider, we create a spurious
  • dependence between its parents. The dependence is borne out in the
  • probabilities: if you chose Door 1, the car location is twice as likely to be
  • behind Door 2 as Door 1; if you chose Door 2, the car location is twice as
  • likely to be behind Door 1.
  • >&nbsp; &nbsp; &nbsp; It is a bizarre dependence for sure, one of a type that most of us are
  • unaccustomed to. It is a dependence that has no cause. It does not involve
  • physical communication between the producers and us. It does not involve
  • mental telepathy. It is purely an artifact of Bayesian conditioning: a magical
  • transfer of information without causality. Our minds rebel at this possibility
  • because from earliest infancy, we have learned to associate correlation with
  • causation. If a car behind us takes all the same turns that we do, we first think
  • it is following us (causation!). We next think that we are going to the same
  • place (i.e., there is a common cause behind each of our turns). But causeless
  • correlation violates our common sense. Thus, the Monty Hall paradox is just
  • like an optical illusion or a magic trick: it uses our own cognitive machinery
  • to deceive us.
  • >&nbsp; &nbsp; &nbsp; Why do I say that Monty Hall’s opening of Door 3 was a “transfer of
  • information”? It didn’t, after all, provide any evidence about whether your
  • initial choice of Door 1 was correct. You knew in advance that he was going
  • to open a door that hid a goat, and so he did. No one should ask you to change
  • your beliefs if you witness the inevitable. So how come your belief in Door 2
  • has gone up from one-third to two-thirds?
  • 2. [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • 3. [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
  • The second quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? The first quotation below uses "transfer". But how can probabilities "shift" or "transfer"?
  • This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
  • 1. Judea Pearl, _The Book of Why _(2018). Page 181 of 402.
  • >![](https://i.imgur.com/zBGwqvk.jpg)
  • >&nbsp; &nbsp; &nbsp; As you can see from Figure 6.1, Door Opened is a collider. Once we
  • obtain information on this variable, all our probabilities become conditional
  • on this information. But when we condition on a collider, we create a spurious
  • dependence between its parents. The dependence is borne out in the
  • probabilities: if you chose Door 1, the car location is twice as likely to be
  • behind Door 2 as Door 1; if you chose Door 2, the car location is twice as
  • likely to be behind Door 1.
  • >&nbsp; &nbsp; &nbsp; It is a bizarre dependence for sure, one of a type that most of us are
  • unaccustomed to. It is a dependence that has no cause. It does not involve
  • physical communication between the producers and us. It does not involve
  • mental telepathy. It is purely an artifact of Bayesian conditioning: a magical
  • transfer of information without causality. Our minds rebel at this possibility
  • because from earliest infancy, we have learned to associate correlation with
  • causation. If a car behind us takes all the same turns that we do, we first think
  • it is following us (causation!). We next think that we are going to the same
  • place (i.e., there is a common cause behind each of our turns). But causeless
  • correlation violates our common sense. Thus, the Monty Hall paradox is just
  • like an optical illusion or a magic trick: it uses our own cognitive machinery
  • to deceive us.
  • >&nbsp; &nbsp; &nbsp; Why do I say that Monty Hall’s opening of Door 3 was a “transfer of
  • information”? It didn’t, after all, provide any evidence about whether your
  • initial choice of Door 1 was correct. You knew in advance that he was going
  • to open a door that hid a goat, and so he did. No one should ask you to change
  • your beliefs if you witness the inevitable. So how come your belief in Door 2
  • has gone up from one-third to two-thirds?
  • 2. [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • 3. [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
#4: Post edited by user avatar DNB‭ · 2021-08-24T06:05:53Z (about 3 years ago)
  • After $n - 2$ unchosen doors are opened, how does the probability of the $n - 2$ unchosen doors "shift" to the lone unchosen door?
  • After $n - 2$ unchosen doors are opened, how does the probability of the $n - 2$ unchosen doors "shift" or "transfer" to the lone unchosen door?
#3: Post edited by user avatar DNB‭ · 2021-08-24T06:05:36Z (about 3 years ago)
  • The first quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? But how can probabilities "shift" like this"?
  • This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
  • [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
  • The second quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? The first quotation below uses "transfer". But how can probabilities "shift" or "transfer"?
  • This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
  • 1. Judea Pearl, The Book of Why (2018). Page 181 of 402.
  • >![](https://i.imgur.com/zBGwqvk.jpg)
  • >&nbsp; &nbsp; &nbsp; As you can see from Figure 6.1, Door Opened is a collider. Once we
  • obtain information on this variable, all our probabilities become conditional
  • on this information. But when we condition on a collider, we create a spurious
  • dependence between its parents. The dependence is borne out in the
  • probabilities: if you chose Door 1, the car location is twice as likely to be
  • behind Door 2 as Door 1; if you chose Door 2, the car location is twice as
  • likely to be behind Door 1.
  • >&nbsp; &nbsp; &nbsp; It is a bizarre dependence for sure, one of a type that most of us are
  • unaccustomed to. It is a dependence that has no cause. It does not involve
  • physical communication between the producers and us. It does not involve
  • mental telepathy. It is purely an artifact of Bayesian conditioning: a magical
  • transfer of information without causality. Our minds rebel at this possibility
  • because from earliest infancy, we have learned to associate correlation with
  • causation. If a car behind us takes all the same turns that we do, we first think
  • it is following us (causation!). We next think that we are going to the same
  • place (i.e., there is a common cause behind each of our turns). But causeless
  • correlation violates our common sense. Thus, the Monty Hall paradox is just
  • like an optical illusion or a magic trick: it uses our own cognitive machinery
  • to deceive us.
  • >&nbsp; &nbsp; &nbsp; Why do I say that Monty Hall’s opening of Door 3 was a “transfer of
  • information”? It didn’t, after all, provide any evidence about whether your
  • initial choice of Door 1 was correct. You knew in advance that he was going
  • to open a door that hid a goat, and so he did. No one should ask you to change
  • your beliefs if you witness the inevitable. So how come your belief in Door 2
  • has gone up from one-third to two-thirds?
  • 2. [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • 3. [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
#2: Post edited by user avatar DNB‭ · 2021-08-10T00:49:48Z (over 3 years ago)
  • After $n - 2$ doors are opened, how does the probability of the $n - 2$ doors "shift" to the lone unopened door?
  • After $n - 2$ unchosen doors are opened, how does the probability of the $n - 2$ unchosen doors "shift" to the lone unchosen door?
  • [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
  • The first quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? But how can probabilities "shift" like this"?
  • This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
  • [The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)
  • > Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
  • >
  • > Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
  • >
  • > ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
  • >
  • > When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
  • [by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)
  • ![](https://i.imgur.com/kkSCt8Q.jpg)
  • > At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
  • ![](https://i.imgur.com/Ng9McDg.jpg)
  • >Now let’s have Monty open door C again and see how that affects the probabilities.
  • ![](https://i.imgur.com/NBANjqx.jpg)
  • >### The Hundred Doors Solution to The Monty Hall Problem.
  • >
  • >In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
  • ![](https://i.imgur.com/7xVPawO.png)
  • >Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.
  • ![](https://i.imgur.com/eHfScEF.png)
#1: Initial revision by user avatar DNB‭ · 2021-08-10T00:43:39Z (over 3 years ago)
After $n - 2$ doors are opened, how does the probability of the $n - 2$ doors "shift" to the lone unopened door?
[The Time Everyone “Corrected” the World’s Smartest Woman](https://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/)

> Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
> 
> ![](https://pix-media.priceonomics-media.com/blog/907/doors1.png)
> 
> Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
> 
> ![](https://pix-media.priceonomics-media.com/blog/907/doorsx-edout.png)
> 
> When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely **"shifts" [boldening mine]** that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.

[by Robert Lea | Medium](https://medium.com/@roblea_63049/exploring-one-of-mathematics-most-enigmatic-puzzles-the-monty-hall-problem-f7b0beb8e1f1)

![](https://i.imgur.com/kkSCt8Q.jpg)

> At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B **must carry the full load [boldening mine]** of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.

![](https://i.imgur.com/Ng9McDg.jpg)

>Now let’s have Monty open door C again and see how that affects the probabilities.

![](https://i.imgur.com/NBANjqx.jpg)

>### The Hundred Doors Solution to The Monty Hall Problem.
>
>In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.

![](https://i.imgur.com/7xVPawO.png)

>Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.

![](https://i.imgur.com/eHfScEF.png)