I’ve just read Initation Ceremony. Is this really where Bayesian probability begins? Because I don’t claim to understand it, but I worked it out easy enough, just not mentally but with calc.exe, using my usual method of assuming a sample of 100. So there are 100 people, 75 W and 25 M, 75x0.75=56.26 VW and 25x0.5 = 12.5 VM so our ratio is 12.5 to 56.26 so a 22.2% chance (Because only the Sith deal in incomprehensible verbal-math like ” two to nine, or a probability of two-elevenths”. Percentages are IMHO way more intuitive. I use a sample size of 100 precisely because then I can say of 100 people 56.26 are VW and thus 56.26% of a sample of any size.)
At what point “okay, let’s calculate on a sample of 100” breaks down and I really need to learn the Bayes Theorem and its applications? Note: the sample-100 method works well with the other example of diagnostic methods giving false positives for rare illnesses.
It is also possible percentages are not as intuitive to others as to me. To me 22% is visualized as drawing a 10 by 10 square on a grid paper and paint 22 of the constituent squares black, then throwing darts on the square. Assuming darts cannot land outside the cube.
Yes, if you are not prone to silly order-of-magnitude errors in mental arithmethics. For example if it is intuitive and fast for you that the square root of 40000 is 200 and never make the mistake of thinking for a second it is 2000 or 20. I do. Not sure why.
For numerical calculations, your method doesn’t ever really break down and, moreover, Brennan is essentially doing the same thing you are, but with a sample of 16 people instead of 100 to make the math simple enough to do mentally.
A more Bayes-theorem styled calculation tells us that we have 1:3 odds initially (as there are 1⁄4 men and 3⁄4 women) and the Virtuist evidence updates it by a factor of 2:3 (as Virtuists are 2⁄4 of men and 3⁄4 of women), so we end with 2:9 odds. I think this is easier than what either you or Brennan are doing, but it’s a matter of taste and of what’s more intuitive. (Which certainly varies from person to person; I find two-elevenths easier to grasp than 22.2%)
Doing Bayesian calculations formally is more important where you are doing symbolic calculations, especially with continuous probability distributions.
Edit: Also, 22.2% is wrong, which I didn’t realize at first; it’s 2⁄9, not 2⁄11. You want to compute 12.5/(12.5+56.25) instead.
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.)
I’ve just read Initation Ceremony. Is this really where Bayesian probability begins? Because I don’t claim to understand it, but I worked it out easy enough, just not mentally but with calc.exe, using my usual method of assuming a sample of 100. So there are 100 people, 75 W and 25 M, 75x0.75=56.26 VW and 25x0.5 = 12.5 VM so our ratio is 12.5 to 56.26 so a 22.2% chance (Because only the Sith deal in incomprehensible verbal-math like ” two to nine, or a probability of two-elevenths”. Percentages are IMHO way more intuitive. I use a sample size of 100 precisely because then I can say of 100 people 56.26 are VW and thus 56.26% of a sample of any size.)
At what point “okay, let’s calculate on a sample of 100” breaks down and I really need to learn the Bayes Theorem and its applications? Note: the sample-100 method works well with the other example of diagnostic methods giving false positives for rare illnesses.
It is also possible percentages are not as intuitive to others as to me. To me 22% is visualized as drawing a 10 by 10 square on a grid paper and paint 22 of the constituent squares black, then throwing darts on the square. Assuming darts cannot land outside the cube.
For calculations of conditional probabilities I’ve found an initial sample size of 10,000 is more manageable. But that’s just me.
Yes, if you are not prone to silly order-of-magnitude errors in mental arithmethics. For example if it is intuitive and fast for you that the square root of 40000 is 200 and never make the mistake of thinking for a second it is 2000 or 20. I do. Not sure why.
For numerical calculations, your method doesn’t ever really break down and, moreover, Brennan is essentially doing the same thing you are, but with a sample of 16 people instead of 100 to make the math simple enough to do mentally.
A more Bayes-theorem styled calculation tells us that we have 1:3 odds initially (as there are 1⁄4 men and 3⁄4 women) and the Virtuist evidence updates it by a factor of 2:3 (as Virtuists are 2⁄4 of men and 3⁄4 of women), so we end with 2:9 odds. I think this is easier than what either you or Brennan are doing, but it’s a matter of taste and of what’s more intuitive. (Which certainly varies from person to person; I find two-elevenths easier to grasp than 22.2%)
Doing Bayesian calculations formally is more important where you are doing symbolic calculations, especially with continuous probability distributions.
Edit: Also, 22.2% is wrong, which I didn’t realize at first; it’s 2⁄9, not 2⁄11. You want to compute 12.5/(12.5+56.25) instead.
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.)