Comment author:JeffJo
23 September 2011 08:27:40PM
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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 A single coin is withdrawn without revealing its number, and flipped.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)

OldSay 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.