The Farrington Cases
Another nice problem from Brian Weatherson's weblog: Farrington is 50% confident that it's after 4:30, and 50% confident that a certain coin landed tails. Now he comes to know that iff the coin landed tails, some researchers create a brain-in-a-vat duplicate of himself at exactly 4:30 today. What are the probabilities he should assign to the 5 open possibilities:
- P1: It is before 4:30 and the coin landed heads.
- P2: It is before 4:30 and the coin landed tails.
- P3: It is after 4:30 and the coin landed heads.
- P4: It is after 4:30 and the coin landed tails and he is the original Farrington.
- P5: It is after 4:30 and the coin landed tails and he is the duplicate Farrington.
I'm with Marc Zazzaro that it should be 1/4, 1/4, 1/4, 1/8, 1/8. Unfortunately, I don't have a good argument. Here is a not-so-good one:
Suppose Farrington is almost certain that it's before 4:30, because his watch and all the clocks around him show that it is only shortly after noon. Let's say his credence for it being after 4:30 is something like 0.001. Now he is informed that those researchers are not going to create a single Farrington duplicate, but rather a million of them. Should he then conclude that his watch is wrong because it is very probably after 4:30 now? This would appear very odd to me. But if in the original case we assigned equal credence to all 5 cases, this would be the rational reaction.
Brian argues for assigning 1/5, 1/5, 1/5, 1/5, 1/5 from the assumption that if Farrington were authoritatively told that he is the original Farrington, he should conditionalize to a distribution of 1/4, 1/4, 1/4, 1/4, 0. I find this quite implausible though: Suppose you don't know whether it's before or after 4:30. Now you're told that with a probability of 0.99 you will turn into a frog at 4:30. Moreover, you convince yourself that you are not a frog. This should dramatically increase your belief that it is before 4:30.
Cian Dorr's argument for assigning 1/5, 1/5, 1/5, 1/5, 1/5 is possibly more interesting:
Consider a variant case, where what the Times reports is that the duplication was to occur immediately after the coin toss, and was to occur no matter what the outcome of the toss might be. In this case there are eight possibilities (heads/tails, pre-4.30/post-4.30, original/duplicate) and they should all get equal probability. If, in the variant case, someone authorative then told Farrington 'You are the original, unless it is after 4.30 and the coin landed tails', he should conditionalise, ending up assigning equal probability to the five remaining cases. But the original case seems relevantly similar to this---it's as if he gets the two pieces of information together, rather than sequentially---so there too he should assign equal probability.
I agree with everything except that the original case is relevantly similar. Consider the following two cases.
Case 1. There are two fair coins, both of which currently show heads. I toss the first coin. If it lands heads I will stop. If it lands tails I will also toss the second coin. I hope you agree that we should assign the following probabilities:
| Coin 1 | Coin 2 | Probability |
|---|---|---|
| heads | heads | 1/2 |
| heads | tails | 0 |
| tails | heads | 1/4 |
| tails | tails | 1/4 |
Case 2. This time assume that both coins are tossed. You'll assign equal probability of 1/4 to all the above possibilities. But now I also tell you authoritatively that coin 1 will land tails or coin 2 will land heads. That is, I rule out possibility 2. You should conditionalize on this information and assign equal probability to the remaining options, which results in
| Coin 1 | Coin 2 | Probability |
|---|---|---|
| heads | heads | 1/3 |
| heads | tails | 0 |
| tails | heads | 1/3 |
| tails | tails | 1/3 |
In a sense the available information is the same in both cases: I've ruled out possibility 2 and left all other possibilities open. But still the rational credence functions differ.