Comment author:[deleted]
31 August 2011 09:51:03PM
*
3 points
[-]

I can't speak for the rest of your post, but

We can answer the question without knowing anything more about b, than that it is not 1/2. For any 0<=b1<1/2, since we have no other information, b=b1 and b=1-b1 must be treated as equally likely. Regardless of what the distribution of b1 is, this makes the probability the coin landed on heads 1/2.

is pretty clearly wrong. (In fact, it looks a lot like you're establishing a prior distribution, and that's uniquely a Bayesian feature.) The probability of an event (the result of the flip is surely an event, though I can't tell if you're claiming to the contrary or not) to a frequentist is the limit of the proportion of times the event occurred in independent trials as the number of trials tends to infinity. The probability the coin landed on heads is the one thing in the problem statement that can't be 1/2, because we know that the coin is biased. Your calculation above seems mostly ad hoc, as is your introduction of additional random variables elsewhere.

Comment author:nshepperd
01 September 2011 03:51:06AM
0 points
[-]

I think they are arguing that the "independent trials" that are happening here are instances of "being given a 'randomly' biased coin and seeing if a single flip turns up heads". But of course the techniques they are using are bayesian, because I'd expect a frequentist to say at this point "well, I don't know who's giving me the coins, how am I supposed to know the probability distribution for the coins?".

Comment author:JeffJo
01 September 2011 10:47:06AM
0 points
[-]

The random process a frequentist should repeat is flipping arandom biased coin, and getting a random bias b and either heads or tails. You are assuming it is flipping the *same biased coin with fixed bias B, and getting heads or tails.

The probability arandom biased coins lands heads is 1/2, from either point of view. And for nshepperd, the point is that a Frequentist doesn't need to know what the bias is. As long as we can't assume it is different for b1 and 1-b1, when you integrate over the unknown distribution (yes, you can do that in this case) the answer is 1/2.

Comment author:JeffJo
23 September 2011 08:27:40PM
0 points
[-]

Say a bag contains 100 unique coins that have been carefully tuned to be unfair when flipped. Each is stamped with an integer in the range 0 to 100 (50 is missing) representing its probability, in percent, of landing on heads. A single coin is withdrawn without revealing its number, and flipped. What is the probability that the result will be heads?

You are claiming that anybody who calls himself a Frequentist needs to know the number on the coin to answer this question. And that any attempt to represent the probability of drawing coin N is specifying a prior distribution, an act that is strictly prohibited for a Frequentist. Both claims are absurd. Prior distributions are a fact of the mathematics of probability, and belong to Frequentist and Bayesian alike. The only differences are (1) the Bayesian may use information differently to determine a prior, sometimes in situations where a Frequentist wouldn't see one at all; (2) The Bayesian will prefer solutions based explicitly on that prior, while the Frequentist will prefer solutions based on the how the prior affects repeated experiments; and (3) Some Frequentists might not realize when they have enough information to determine a prior, and/or its effects, that should satisfy them.

If both get answers, and they don't agree, somebody did something wrong.

The answer is 50%. The Bayesian says that, based on available information, neither result can be favored over the other so they must both have probability 50%. The Frequentist says that if you repeat the experiment 100^2 times, including the part where you draw a coin from the bag of 100 coins, you should count on getting each coin 100 times. And you should also count, for each coin, on getting heads in proportion to its probability. That way, you will count 5,000 heads in 10,000 trials, making the answer 50%. Both solutions are based on the same facts and assumptions, just organized differently.

The answer Eliezer_Yudkowsky attributes to Frequentists, for the simpler problem without the bag and stamped coins, is an incorrect Frequentist solution. Or at least, a correct solution to a different problem. One that corresponds to the different question "What proportion of the time will this coin come up heads?" I agree that some who claim to be Frequentists will answer that question. But the true Frequentist will answer the question that was asked: "What proportion of the time will the process of flipping a coin with unknown bias come up heads?" His repetitions must represent the bias for each flip as independent of any other flips, not the same bias each time. The bias B will come up just as often as the bias (1-B), so the number of heads will always be half the number of trials.

## Comments (190)

Old*3 points [-]I can't speak for the rest of your post, but

is pretty clearly wrong. (In fact, it looks a lot like you're establishing a prior distribution, and that's uniquely a Bayesian feature.) The probability of an event (the result of the flip is surely an event, though I can't tell if you're claiming to the contrary or not) to a frequentist is the limit of the proportion of times the event occurred in independent trials as the number of trials tends to infinity. The probability the coin landed on heads is the one thing in the problem statement that can't be 1/2, because we know that the coin is biased. Your calculation above seems mostly ad hoc, as is your introduction of additional random variables elsewhere.

However, I'm not a statistician.

I think they are arguing that the "independent trials" that are happening here are instances of "being given a 'randomly' biased coin and seeing if a single flip turns up heads". But of course the techniques they are using

arebayesian, because I'd expect a frequentist to say at this point "well, I don't know who's giving me the coins, how am I supposed to know the probability distribution for the coins?".The random process a frequentist should repeat is flipping

arandombiased coin, and getting a random bias b and either heads or tails. You are assuming it is flippingthe *samebiased coin with fixed bias B, and getting heads or tails.The probability

arandombiased coins lands heads is 1/2, from either point of view. And for nshepperd, the point is that a Frequentist doesn't need to know what the bias is. As long as we can't assume it is different for b1 and 1-b1, when you integrate over the unknown distribution (yes, you can do that in this case) the answer is 1/2.Say a bag contains 100 unique coins that have been carefully tuned to be unfair when flipped. Each is stamped with an integer in the range 0 to 100 (50 is missing) representing its probability, in percent, of landing on heads. A single coin is withdrawn without revealing its number, and flipped. What is the probability that the result will be heads?

You are claiming that anybody who calls himself a Frequentist needs to know the number on the coin to answer this question. And that any attempt to represent the probability of drawing coin N is specifying a prior distribution, an act that is strictly prohibited for a Frequentist. Both claims are absurd. Prior distributions are a fact of the mathematics of probability, and belong to Frequentist and Bayesian alike. The only differences are (1) the Bayesian may use information differently to determine a prior, sometimes in situations where a Frequentist wouldn't see one at all; (2) The Bayesian will prefer solutions based explicitly on that prior, while the Frequentist will prefer solutions based on the how the prior affects repeated experiments; and (3) Some Frequentists might not realize when they have enough information to determine a prior, and/or its effects, that should satisfy them.

If both get answers, and they don't agree, somebody did something wrong.

The answer is 50%. The Bayesian says that, based on available information, neither result can be favored over the other so they must both have probability 50%. The Frequentist says that if you repeat the experiment 100^2 times,

includingthe part where you draw a coin from the bag of 100 coins, you should count on getting each coin 100 times. And you should also count, for each coin, on getting heads in proportion to its probability. That way, you will count 5,000 heads in 10,000 trials, making the answer 50%. Both solutions are based on the same facts and assumptions, just organized differently.The answer Eliezer_Yudkowsky attributes to Frequentists, for the simpler problem without the bag and stamped coins, is an incorrect Frequentist solution. Or at least, a correct solution to a different problem. One that corresponds to the different question "What proportion of the time will

thiscoin come up heads?" I agree that some who claim to be Frequentists will answer that question. But the true Frequentist will answer the question that was asked: "What proportion of the time will the process of flipping a coin with unknown bias come up heads?" His repetitions must represent the bias for each flip as independent of any other flips, not the same bias each time. The bias B will come up just as often as the bias (1-B), so the number of heads will always be half the number of trials.