The 8 rooms are definitely the unbiased sample (of your rooms with one red room subtracted).
I think you are making two mistakes:
First, I think you’re too focused on the nice properties of an unbiased sample. You can take an unbiased sample all you want, but if we know information in addition to the sample, our best estimate might not be the average of the sample! Suppose we have two urns, urn A has 10 red balls and 10 blue balls, while urn B has 5 red balls and 15 blue balls. We choose an urn by rolling a die, such that we have a 5⁄6 chance of choosing urn A and a 1⁄6 chance of choosing urn B. Then we take a fair, unbiased sample of 4 balls from whatever urn we chose. Suppose we draw out 1 red ball and 3 blue balls. Since this is an unbiased sample, does the process that you are calling “statistical analysis” have to estimate that we were drawing from urn B?
Second, you are trying too hard to make everything about the rooms. It’s like someone was doing the problem with two urns from the previous paragraph, but tried to mathematically arrive at the answer only as a function of the number of red balls drawn, without making any reference to the process that causes them to draw from urn A vs. urn B. And they come up with several different ideas about what the function could be, and they call those functions “the Two-Thirds-B-er method” and “the Four-Tenths-B-er method.” When really, both methods are incomplete because they fail to take into account what we know about how we picked the urn to draw from.
To answer the last part of your statement. If beauty randomly opens 8 doors and found them all red then she has a sample of pure red. By simple statistics she should give R=81 as the estimation. Halfer and thirders would both agree on that. If they do a bayesian analysis R=81 would also be the case with the highest probability. I’m not sure where 75 comes from I’m assuming by summing the multiples of probability and Rs in the bayesian analysis? But that value does not correspond to the estimation in statistics. Imagine you randomly draw 20 beans from a bag and they are all red, using statistics obviously you are not going to estimate the bag contains 90% red bean.
Think of it like this: if Beauty opens 8 doors and they’re all red, and then she goes to open a ninth door, how likely should she think it is to be red? 100%, or something smaller than 100%? For predictions, we use the average of a probability distribution, not just its highest point.
No problem, always good to have a discussion with someone serious about the subject matter.
First of all, you are right: statistic estimation and expected value in bayesian analysis are different. But that is not what I’m saying. What I’m saying is in a bayesian analysis with an uninformed prior (uniform) the case with highest probability should be the unbiased statistic estimation (it is not always so because round offs etc).
In the two urns example, I think what you meant is that using the sample of 4 balls a fair estimation would be 5 reds and 15 blues as in the case of B but bayesian analysis would give A as more likely? However this disagreement is due to the use of an informed prior, that you already know we are more likely to draw from A right from the beginning. Without knowing this bayesian would give B as the most likely case, same as statistic estimate.
Think of it like this: if Beauty opens 8 doors and they’re all red, and then she goes to open a ninth door, how likely should she think it is to be red? 100%, or something smaller than 100%? For predictions, we use the average of a probability distribution, not just its highest point.
Definitely something smaller than 100%. Just because beauty thinks r=81 is the most likely case doesn’t mean she think it is the only case. But that is not what the estimation is about. Maybe this question would be more relevant: If after opening 8 doors and they are all red and beauty have to guess R. what number should she guess (to be most likely correct)?
Sorry for the slow reply.
The 8 rooms are definitely the unbiased sample (of your rooms with one red room subtracted).
I think you are making two mistakes:
First, I think you’re too focused on the nice properties of an unbiased sample. You can take an unbiased sample all you want, but if we know information in addition to the sample, our best estimate might not be the average of the sample! Suppose we have two urns, urn A has 10 red balls and 10 blue balls, while urn B has 5 red balls and 15 blue balls. We choose an urn by rolling a die, such that we have a 5⁄6 chance of choosing urn A and a 1⁄6 chance of choosing urn B. Then we take a fair, unbiased sample of 4 balls from whatever urn we chose. Suppose we draw out 1 red ball and 3 blue balls. Since this is an unbiased sample, does the process that you are calling “statistical analysis” have to estimate that we were drawing from urn B?
Second, you are trying too hard to make everything about the rooms. It’s like someone was doing the problem with two urns from the previous paragraph, but tried to mathematically arrive at the answer only as a function of the number of red balls drawn, without making any reference to the process that causes them to draw from urn A vs. urn B. And they come up with several different ideas about what the function could be, and they call those functions “the Two-Thirds-B-er method” and “the Four-Tenths-B-er method.” When really, both methods are incomplete because they fail to take into account what we know about how we picked the urn to draw from.
Think of it like this: if Beauty opens 8 doors and they’re all red, and then she goes to open a ninth door, how likely should she think it is to be red? 100%, or something smaller than 100%? For predictions, we use the average of a probability distribution, not just its highest point.
No problem, always good to have a discussion with someone serious about the subject matter.
First of all, you are right: statistic estimation and expected value in bayesian analysis are different. But that is not what I’m saying. What I’m saying is in a bayesian analysis with an uninformed prior (uniform) the case with highest probability should be the unbiased statistic estimation (it is not always so because round offs etc).
In the two urns example, I think what you meant is that using the sample of 4 balls a fair estimation would be 5 reds and 15 blues as in the case of B but bayesian analysis would give A as more likely? However this disagreement is due to the use of an informed prior, that you already know we are more likely to draw from A right from the beginning. Without knowing this bayesian would give B as the most likely case, same as statistic estimate.
Definitely something smaller than 100%. Just because beauty thinks r=81 is the most likely case doesn’t mean she think it is the only case. But that is not what the estimation is about. Maybe this question would be more relevant: If after opening 8 doors and they are all red and beauty have to guess R. what number should she guess (to be most likely correct)?