I’m glad that someone engages with non-Lewisian halfism but this is clearly wrong on a very basic level. To understand why, let’s consider a simpler problem:
Two coins are tossed. Then you are told state of one of the coins, but you don’t know whether it’s the first coin or the second. Then you are told whether it was the first coin or the second. What should be your credence that the state of both coins are the same 1. before you were told the state of one of the coins? 2. after you were told the state of one of the coins? 3. after you were told which coin was it?
You are creating a related but different and also complicated problem: the Two Child Problem, which is notoriously ambiguous. “Then you are told the state of one of the coins” can have many meanings.
If I ask the experimenter “choose one of the coins randomly and tell me what it is” then I am not able to update my probability. It will still be 1⁄2 that the coins are the same.
If I ask the experimenter “is there at least one heads?” then I will be able to update. If they say yes I can update to 1⁄3, if they say no I can update to 1.
Frankly, it seems as if Conitzer forgot that credence can change for reasons not related to memory loss. That one can simply receive new evidence and change their credence based on that.
Conitzer’s problem can be simplified further by letting Beauty flip a coin herself on Monday and Tuesday.
She wakes up Monday and flips a coin. She wakes up Tuesday and flips a coin. That’s it.
After flipping a coin, what should her credence be that the coin flips are the same?
Do you disagree now that the answer is 1/2?
I think it is clearly 1⁄2 precisely because there is no new evidence. The violation of the Reflection Principle is secondary. More importantly, something has gone wrong if we think she can flip a coin and update the probability of the coins being the same.
She didn’t know the sequence in advance, like, for example, she knew that she is to be awakened on Monday. She made a guess and managed to guess right. The difference is that on a repetition of the probability experiment, she is awakened on Monday in every iteration of it, but the sequence of the tosses is the same as she precommited to only in a smal fraction of all iterations of the experiment.
I agree, but she doesn’t get to observe the sequence of tosses in the experiment. She isn’t even able to observe that a sequence of tosses happens “at least once” in the experiment. That’s what Conitzer shows in his problem.
She can’t update her probability based on observing a rare event C (as you have defined it), because she can’t observe C in the first place.
A version without amnesia is not exactly the same situation, but something similar can happen. Suppose the experimenter will flip a coin, on heads they will flip a new sequence of 1000, on tails they will flip 2 new sequences of 1000. I ask the experimenter “randomly choose one of the sequences and tell me the result”, and they tell me the result was 1000 heads in a row. A sequence of 1000 heads in a row is more likely to have occurred at least once if they flipped 2 sequences. But this does not allow me to update my probability of the number of sequences, because I have not learned “there is at least one sequence of 1000 heads.”
You are creating a related but different and also complicated problem: the Two Child Problem, which is notoriously ambiguous. “Then you are told the state of one of the coins” can have many meanings.
If I ask the experimenter “choose one of the coins randomly and tell me what it is” then I am not able to update my probability. It will still be 1⁄2 that the coins are the same.
If I ask the experimenter “is there at least one heads?” then I will be able to update. If they say yes I can update to 1⁄3, if they say no I can update to 1.
What if no one is answering your questions? You are just shown one coin with no insight about the algorithm according to which the experimenter showed you it, other that this is the outcome of the coin toss. There is actually the least presumptious way to reason about such things. And this is the one I described.
But nevermind that. Let’s not go on an unnecessary tangent about this problem. For the sake of the argument I’m happy to concede that both 1⁄2 and 1⁄3 are reasonable answers to it. However Conitzer’s reasoning would imply that only 1⁄2 is the correct answer. As a matter of fact he simply assumes that 1⁄2 has to be correct, refusing to entertain the idea that it’s not the case. Not unlike Briggs in Technicolor.
Conitzer’s problem can be simplified further by letting Beauty flip a coin herself on Monday and Tuesday.
She wakes up Monday and flips a coin. She wakes up Tuesday and flips a coin. That’s it.
After flipping a coin, what should her credence be that the coin flips are the same?
Do you disagree now that the answer is 1/2?
If she just flipped the coin then the answer is 1⁄2. If she observed the event “the coin is Tails” then the answer is 1⁄3. If she observed the event “the coin is Heads” the answer is 1⁄3. But doesn’t she always observes one of these events in every iteration of the experiment? No, she doesn’t.
This is the same situation as with Technicolor Sleeping Beauty. She observes the event “Blue” instead of “Blue or Red” only when she’s configured her event space in a specific way, by precommited to this outcome in particular. Likewise here. When the Beauty precommited to Tails, flips the coin and sees that the coin is indeed Tails she has observed the event “the coin is Tails”, when she did no precommitments or the coin turned out to be the other side—she observed the event “the coin is Heads or Tails”.
I think it is clearly 1⁄2 precisely because there is no new evidence. The violation of the Reflection Principle is secondary. More importantly, something has gone wrong if we think she can flip a coin and update the probability of the coins being the same.
Of course something has gone wrong. This is what you get when you add amnesia to probability theory problems—it messes the event space in a counterintuitive way. By default you are able only to observe the most general events which have probability one. Like “I’m awake at least once in the experiment” or “the room is either Blue or Red at least once in the experiment”. To observe more specific events you need precommitments.
To see that this actually works, check the betting arguments for rare event and technicolor sleeping beauty from the post. Likewise we can construct a betting argument for the Conitzer’s example, where you go through an iterated experiment and are asked to make one per experiment bet on the fact that both coins did not produce the same outcome, with betting odds a bit worse than 1:1. The optimal betting strategy is not always refusing the bet, as it would’ve been if the probability actually always was 1⁄2 and you did not get any new evidence.
She isn’t even able to observe that a sequence of tosses happens “at least once” in the experiment.
She is able to to observe that a particular sequence of tosses happens “at least once” in the experiment only if she has precommited to guessing this particular sequence. Otherwise she, indeed does not observe this event.
You are creating a related but different and also complicated problem: the Two Child Problem, which is notoriously ambiguous. “Then you are told the state of one of the coins” can have many meanings.
If I ask the experimenter “choose one of the coins randomly and tell me what it is” then I am not able to update my probability. It will still be 1⁄2 that the coins are the same.
If I ask the experimenter “is there at least one heads?” then I will be able to update. If they say yes I can update to 1⁄3, if they say no I can update to 1.
Conitzer’s problem can be simplified further by letting Beauty flip a coin herself on Monday and Tuesday.
She wakes up Monday and flips a coin. She wakes up Tuesday and flips a coin. That’s it.
After flipping a coin, what should her credence be that the coin flips are the same?
Do you disagree now that the answer is 1/2?
I think it is clearly 1⁄2 precisely because there is no new evidence. The violation of the Reflection Principle is secondary. More importantly, something has gone wrong if we think she can flip a coin and update the probability of the coins being the same.
I agree, but she doesn’t get to observe the sequence of tosses in the experiment. She isn’t even able to observe that a sequence of tosses happens “at least once” in the experiment. That’s what Conitzer shows in his problem.
She can’t update her probability based on observing a rare event C (as you have defined it), because she can’t observe C in the first place.
A version without amnesia is not exactly the same situation, but something similar can happen. Suppose the experimenter will flip a coin, on heads they will flip a new sequence of 1000, on tails they will flip 2 new sequences of 1000. I ask the experimenter “randomly choose one of the sequences and tell me the result”, and they tell me the result was 1000 heads in a row. A sequence of 1000 heads in a row is more likely to have occurred at least once if they flipped 2 sequences. But this does not allow me to update my probability of the number of sequences, because I have not learned “there is at least one sequence of 1000 heads.”
What if no one is answering your questions? You are just shown one coin with no insight about the algorithm according to which the experimenter showed you it, other that this is the outcome of the coin toss. There is actually the least presumptious way to reason about such things. And this is the one I described.
But nevermind that. Let’s not go on an unnecessary tangent about this problem. For the sake of the argument I’m happy to concede that both 1⁄2 and 1⁄3 are reasonable answers to it. However Conitzer’s reasoning would imply that only 1⁄2 is the correct answer. As a matter of fact he simply assumes that 1⁄2 has to be correct, refusing to entertain the idea that it’s not the case. Not unlike Briggs in Technicolor.
If she just flipped the coin then the answer is 1⁄2. If she observed the event “the coin is Tails” then the answer is 1⁄3. If she observed the event “the coin is Heads” the answer is 1⁄3. But doesn’t she always observes one of these events in every iteration of the experiment? No, she doesn’t.
This is the same situation as with Technicolor Sleeping Beauty. She observes the event “Blue” instead of “Blue or Red” only when she’s configured her event space in a specific way, by precommited to this outcome in particular. Likewise here. When the Beauty precommited to Tails, flips the coin and sees that the coin is indeed Tails she has observed the event “the coin is Tails”, when she did no precommitments or the coin turned out to be the other side—she observed the event “the coin is Heads or Tails”.
Of course something has gone wrong. This is what you get when you add amnesia to probability theory problems—it messes the event space in a counterintuitive way. By default you are able only to observe the most general events which have probability one. Like “I’m awake at least once in the experiment” or “the room is either Blue or Red at least once in the experiment”. To observe more specific events you need precommitments.
To see that this actually works, check the betting arguments for rare event and technicolor sleeping beauty from the post. Likewise we can construct a betting argument for the Conitzer’s example, where you go through an iterated experiment and are asked to make one per experiment bet on the fact that both coins did not produce the same outcome, with betting odds a bit worse than 1:1. The optimal betting strategy is not always refusing the bet, as it would’ve been if the probability actually always was 1⁄2 and you did not get any new evidence.
She is able to to observe that a particular sequence of tosses happens “at least once” in the experiment only if she has precommited to guessing this particular sequence. Otherwise she, indeed does not observe this event.