I'm struggling to reason with this problem:
A robot performs coin tossing. It is poorly designed, it produces a lot of sounds, lights, and vapors, and it takes one hour to toss a coin. Yet in the end, when the coin finally lands, it somehow has an equal probability of showing heads and tails. Two scientists, A and B, enjoy observing this robot, and by analyzing its unusual and faulty behavior, they became fairly decent at guessing whether the coin will land heads or tails half an hour before the coin is released from the robot's hand. The scientist A has 80% chance of successfully predicting the outcome, while the scientist B is successful at 60% of the time. The robot started its routine, and scientist A predicts the coin will land tails. The scientist B predicts the coin will land heads. Can you calculate the probability that the coin will land heads?
In the book I'm studying, the provided answer is: "The correct answer to this problem is 'No, I cannot.'"
I'm struggling to understand why that's the case. Can anyone help?
I regard this question as very subtle. The short answer is that no information is given as to whether the analysis of scientist-A is independent of the analysis of scientist-B. In order to explore the relevance of this issue, I feel that I need to start with a more classic conditional probability problem.
Person-A and Person-B are each weather forecasters, who forecast whether it will rain tomorrow.
Person-A is right 90% of the time.
Person-B is right 80% of the time.
On a given day, Person-A predicts that it will rain tomorrow and Person-B predicts that it will not rain tomorrow. What is the probability that it will rain?
In order to attack this problem, the first thing to do is notice that in general, there are $4$ possibilities, for such a scenario:
Possibility-1 : A and B are both right.
Supposedly, the probability of this happening is $(0.9) \times (0.8) = (0.72).$
Possibility-2 : A and B are both wrong.
Supposedly, the probability of this happening is $(0.1) \times (0.2) = (0.02).$
Possibility-3: A is right and B is wrong.
Supposedly, the probability of this happening is $(0.9) \times (0.2) = (0.18).$
Possibility-4: A is wrong and B is right.
Supposedly, the probability of this happening is $(0.1) \times (0.8) = (0.08).$
Then, supposedly, you can reason that Possibility-1 and Possibility-2 have been eliminated, since A and B disagree. Then, supposedly, the desired computation is
$$\frac{0.18}{0.18 + 0.08} = \frac{9}{13}.$$
Note two things about the treatment of this classic problem:
It is certainly true that you can not attack the problem without (somehow) computing the relative probability of Possiblity-3 vs Possibility-4 occurring.
The method used to compute the relative probabilities of Possibility-3 and Possibility-4 is only valid if you are given the premise that Person-A and Person-B are using forecasting methods that are somehow based on different aspects of the weather and are therefore (somehow) independent of each other.
Realistically, such independence isn't really possible with weather forecasting, as different aspects of the weather (i.e. temperature, barometric pressure, cloud formations) are all interrelated.
To take an extreme case, suppose that Person-A is always studying the exact same data that Person-B is studying, but Person-A is more thorough and studies additional data.
It is entirely plausible that Person-A will almost never be wrong, when Person-A and Person-B disagree.
As another example, suppose that the issue is diagnosing a heart ailment, where Person-A is a 2nd year medical student with poor study habits, and Person-B is a first year medical student with good study habits. When Person-A and Person-B disagree, how would you determine the relative probability that Person-A is right? How important is 2nd year / 1st year, vs bad study habits / good study habits?
With the original problem, you have the same computational difficulty. When the two scientists disagree about the robot's coin flip, you need data on the relative probabilities that each scientist is right, when there is disagreement.
The problem can not be attacked without such data, and the data is missing from the problem.