Now on to Monthy Hall the linked explanation is not that intuitive to me.
To me the intuitive explanation is that if I chose a goat, switching gives me 100% possibility to get a car and not switching 0%, if I chose a car, switching gives me 0% possibility and switching 100%, thus my original 2⁄3 chance to win a goat wins me a car with the same 2⁄3 chance if I switch.
I don’t know if it is Bayesian what I am doing… let’s play with 4 doors, 3 goats 1 cup, er, car. If I chose a goat, 75% chance, switching gives me an 50% chance so it is 37.5% if I chose a car, 25% chance, switching gives me 0%. So the switch gives me 37.5% in the four-door game. If I chose a goat, 75%, staying gives me 0%, if I chose a car, 25%, staying gives me 100%, that is just 25%. Still switch.
Am I doing it right? The line of reasoning being “If my prior is right… the evidence does this. If my prior is wrong, the evidence does that. Add it up.” (probability of correct prior judgement probability of new judgement) + (probability of incorrect prior judgement probability of new judgement) something like this… and some of these four is apparently always 0 and 100% ?
Your argument does all the things that are necessary to solve Monty Hall, but it doesn’t consider some things that could be necessary. (Now, maybe you would have realized that those things need to be considered, if they were necessary. I am just explaining how things can get tricky.)
Suppose instead of Monty Hall, we have Forgetful Monty Hall. Forgetful Monty Hall does not remember where the car is, so he opens a door (that you have not picked) at random, and luckily there is a goat behind it!
Here your line of reasoning still seems to apply: if you chose a goat, switching is 100% and staying 0%, while if you chose a car, switching is 0% and staying 100%. So shouldn’t switching still win with probability 2/3?
An extra thing happened, though. In the “if your prior was right” a.k.a. “if you chose a car” case, it’s not surprising that Forgetful Monty Hall opened a door with a goat. In the other case, if you chose a goat, then Forgetful Monty Hall had a 1 in 2 chance of opening the door with a car by mistake. He didn’t, so the probability you chose a goat should be penalized by that factor of 2. The 1:2 prior becomes 1:1, and then your argument (correctly) tells us that switching and staying are both 50%.
One final comment.
I don’t know if it is Bayesian what I am doing...
Here we are dealing with a problem that can be solved exactly. Any mathematician, Bayesian or otherwise, ought to agree with your answer. When solving harder problems, we might get something that cannot be solved exactly. For instance, I chose the 5 integers 4,3,2,4,3 and then chose the 5 integers 2,2,1,5,1. How likely is it that they came from the same distribution?
This is not a question we can answer and so we instead answer a different question or answer this question with simplifying assumptions. When we do this, if we end up talking about “conjugate priors” it is Bayesian; if we end up talking about “null hypothesis testing” it is not Bayesian.
(A clever trick of demagoguery is to take a question that can be solved exactly and point out that null hypothesis testing solves it incorrectly, and thus Bayesian methods are superior. This is silly! Obviously if you can solve a question exactly, you do so.)
Of course, I don’t know how I missed that…
Now on to Monthy Hall the linked explanation is not that intuitive to me.
To me the intuitive explanation is that if I chose a goat, switching gives me 100% possibility to get a car and not switching 0%, if I chose a car, switching gives me 0% possibility and switching 100%, thus my original 2⁄3 chance to win a goat wins me a car with the same 2⁄3 chance if I switch.
I don’t know if it is Bayesian what I am doing… let’s play with 4 doors, 3 goats 1 cup, er, car. If I chose a goat, 75% chance, switching gives me an 50% chance so it is 37.5% if I chose a car, 25% chance, switching gives me 0%. So the switch gives me 37.5% in the four-door game. If I chose a goat, 75%, staying gives me 0%, if I chose a car, 25%, staying gives me 100%, that is just 25%. Still switch.
Am I doing it right? The line of reasoning being “If my prior is right… the evidence does this. If my prior is wrong, the evidence does that. Add it up.” (probability of correct prior judgement probability of new judgement) + (probability of incorrect prior judgement probability of new judgement) something like this… and some of these four is apparently always 0 and 100% ?
Your argument does all the things that are necessary to solve Monty Hall, but it doesn’t consider some things that could be necessary. (Now, maybe you would have realized that those things need to be considered, if they were necessary. I am just explaining how things can get tricky.)
Suppose instead of Monty Hall, we have Forgetful Monty Hall. Forgetful Monty Hall does not remember where the car is, so he opens a door (that you have not picked) at random, and luckily there is a goat behind it!
Here your line of reasoning still seems to apply: if you chose a goat, switching is 100% and staying 0%, while if you chose a car, switching is 0% and staying 100%. So shouldn’t switching still win with probability 2/3?
An extra thing happened, though. In the “if your prior was right” a.k.a. “if you chose a car” case, it’s not surprising that Forgetful Monty Hall opened a door with a goat. In the other case, if you chose a goat, then Forgetful Monty Hall had a 1 in 2 chance of opening the door with a car by mistake. He didn’t, so the probability you chose a goat should be penalized by that factor of 2. The 1:2 prior becomes 1:1, and then your argument (correctly) tells us that switching and staying are both 50%.
One final comment.
Here we are dealing with a problem that can be solved exactly. Any mathematician, Bayesian or otherwise, ought to agree with your answer. When solving harder problems, we might get something that cannot be solved exactly. For instance, I chose the 5 integers 4,3,2,4,3 and then chose the 5 integers 2,2,1,5,1. How likely is it that they came from the same distribution?
This is not a question we can answer and so we instead answer a different question or answer this question with simplifying assumptions. When we do this, if we end up talking about “conjugate priors” it is Bayesian; if we end up talking about “null hypothesis testing” it is not Bayesian.
(A clever trick of demagoguery is to take a question that can be solved exactly and point out that null hypothesis testing solves it incorrectly, and thus Bayesian methods are superior. This is silly! Obviously if you can solve a question exactly, you do so.)