Post History
#2: Post edited
by
whybecause
·
2021-12-31T15:05:54Z (over 2 years ago)
- $P(C_3|D_2)$ is the probability of the car being behind door 3 given that Monty opened door 2. $p$ is the probability that Monty opens door 2, under the assumption that he has a choice between it and door 3. That is to say $p$ is the probability that Monty opens door 2 under the conditions that: (i) the car is behind door 1 and (ii) the player first picks door 1.
- If $p=1/2$ then this is just like the regular Monty Hall problem, and there $P(C_3|D_2)=2/3$.
- If $p=1$ then you lose information about door 3. Because whether the car is behind door 3 or not, when given the chance, Monty just always shows you door 2. The game becomes completely "insensitive" to where the car is, and therefore you should expect that the probability becomes close to 1/2 (or, in this extreme case where $p=1$ then the probability becomes _exactly_ 1/2). In effect, under this strategy and in this scenario, Monty is no longer giving you information about door 3.
- This explains the two extreme values that $p$ could take, and this agrees with what you see in
- $$ P(C_3|D_2) = \frac 1 {1+p} $$
- ---
On a more pedagogical note about intuitive explanations: Human intuition is imperfect and that's why we have algebra. So it's good to look for intuition when it's available, and so this question is fine. But be ready for the fact that sometimes you just have to rely on the algebra to get all the precise details correct. Algebra is effectively a tool which allows us to export our understanding into symbols, manipulate the symbols according to valid truth-preserving rules, and then translate the answer back into real-world meaning at the end. So in that way, algebra is a tool that we can use to get a correct answer even when it's impossible for us to intuitively understand the steps needed to generate the answer (because there are too many steps or too many variables for human intuition to be able to manage them all).
- $P(C_3|D_2)$ is the probability of the car being behind door 3 given that Monty opened door 2. $p$ is the probability that Monty opens door 2, under the assumption that he has a choice between it and door 3. That is to say $p$ is the probability that Monty opens door 2 under the conditions that: (i) the car is behind door 1 and (ii) the player first picks door 1.
- If $p=1/2$ then this is just like the regular Monty Hall problem, and there $P(C_3|D_2)=2/3$.
- If $p=1$ then you lose information about door 3. Because whether the car is behind door 3 or not, when given the chance, Monty just always shows you door 2. The game becomes completely "insensitive" to where the car is, and therefore you should expect that the probability becomes close to 1/2 (or, in this extreme case where $p=1$ then the probability becomes _exactly_ 1/2). In effect, under this strategy and in this scenario, Monty is no longer giving you information about door 3.
- This explains the two extreme values that $p$ could take, and this agrees with what you see in
- $$ P(C_3|D_2) = \frac 1 {1+p} $$
- ---
- On a more pedagogical note about intuitive explanations: Human intuition is imperfect and that's why we have algebra. So it's good to look for intuition when it's available, and so this question is fine. In this particular case, it is actually possible to get some amount of intuition.
- But be ready for the fact that sometimes you just have to rely on the algebra to get all the precise details correct. Algebra is effectively a tool which allows us to export our understanding into symbols, manipulate the symbols according to valid truth-preserving rules, and then translate the answer back into real-world meaning at the end. So in that way, algebra is a tool that we can use to get a correct answer even when it's impossible for us to intuitively understand the steps needed to generate the answer (because there are too many steps or too many variables for human intuition to be able to manage them all).
#1: Initial revision
by
whybecause
·
2021-12-31T15:05:11Z (over 2 years ago)
$P(C_3|D_2)$ is the probability of the car being behind door 3 given that Monty opened door 2. $p$ is the probability that Monty opens door 2, under the assumption that he has a choice between it and door 3. That is to say $p$ is the probability that Monty opens door 2 under the conditions that: (i) the car is behind door 1 and (ii) the player first picks door 1.
If $p=1/2$ then this is just like the regular Monty Hall problem, and there $P(C_3|D_2)=2/3$.
If $p=1$ then you lose information about door 3. Because whether the car is behind door 3 or not, when given the chance, Monty just always shows you door 2. The game becomes completely "insensitive" to where the car is, and therefore you should expect that the probability becomes close to 1/2 (or, in this extreme case where $p=1$ then the probability becomes _exactly_ 1/2). In effect, under this strategy and in this scenario, Monty is no longer giving you information about door 3.
This explains the two extreme values that $p$ could take, and this agrees with what you see in
$$ P(C_3|D_2) = \frac 1 {1+p} $$
---
On a more pedagogical note about intuitive explanations: Human intuition is imperfect and that's why we have algebra. So it's good to look for intuition when it's available, and so this question is fine. But be ready for the fact that sometimes you just have to rely on the algebra to get all the precise details correct. Algebra is effectively a tool which allows us to export our understanding into symbols, manipulate the symbols according to valid truth-preserving rules, and then translate the answer back into real-world meaning at the end. So in that way, algebra is a tool that we can use to get a correct answer even when it's impossible for us to intuitively understand the steps needed to generate the answer (because there are too many steps or too many variables for human intuition to be able to manage them all).