If croupier choose the players, then players learn that they were chosen, then croupier roll the dice, then players either get bajillion dollars each or die, then (if not snake eyes) croupier choose next players and so on—answer is 1⁄36.
If croupier choose the players, then roll the dice, then (if not snake eyes) croupier choose next players and so on, and only when dice come up snake eyes players learn that they were chosen, and then last players die and all other players get bajillion dollars each—answer is about 1⁄2.
(I’ll probably post an expanded/edited version as a full-post soon, just wanted to see what I could produce in a single sitting. Apologies, as I engage in some handwaving).
I think it’s worth breaking this down more.
In scenario 1, you learn that you are a participant who was selected in round r, where r is unknown.
In scenario 2, you learn that you are a participant who was selected full-stop.
If this was all you learned, then these two scenarios would be equivalent, as all selected participants were selected in some round.
However, I’ll argue in the next section that each scenario contains additional information about which round you should expect to be in.
Analysing Indexical Information
In scenario 1, the participants are informed at different times depending on their group.
When a participant is informed they were selected, they know:
a) that the previous rounds were already rejected and
b) that if they aren’t eliminated, then additional rounds will be run, but you aren’t in those future run
c) that the dice are about to be thrown for them
In scenario 2, all selected participants are informed at once. Participants still know:
a) that all rounds before them were rejected AND
b) if they weren’t marked to be eliminated, then additional rounds were run
c) that the dice were thrown for them
Again, aren’t these the same?
In the next section, I’ll argue that the exact mechanism of how and when people are informed affects the way that we handle reference classes.
Analysing Reference Classes
Reference classes are strange. It’s not clear what we mean by reference classes.
Should it be the people in the same round as us, or everyone who ends up in the snake eyes problem?
One answer would be that in both scenarios it should be everyone who ends up in the same problem, such that the reference classes are the same (following the argument that you have no non-anthropic information about which round you are it)
Another answer would be that in scenario 1, your reference class is everyone in the same round due to indexical information, but in scenario 2, it is everyone in the problem due to the lack of such information.
So how should we interpret the indexical information when figuring out your reference class?
First, let’s imagine that we’ve fixed round r. In that case, it’s obvious that your reference class is everyone in this round and that everyone in the round has a 1⁄36 chance of being eliminated regardless of who they are.
We then want to consider the case where you don’t know which round you are in. This requires us to assign a probability distribution to rounds.
We can choose to start simple again and assume that there are n rounds in total. We now have to choose whether to say that your chance of being in any particular round is 1/n or whether it is proportional to the number of people in that round?
I’m going to suggest that the indexical information (that previous rounds have passed and future rounds are yet to come) tells you that it should be the former. There’s a sense in which people with the same indexical information naturally belong in the same group and it would be weird to mix them all together as per the proportional case. This is somewhat handwavey, but hopefully, I can make a more rigorous argument in future work.
This leads to your chance of being in any particular round is 1/n. Again, since all rounds have the same probability of being eliminated, this doesn’t affect anything.
We can then extend to the case where we don’t know the number of rounds. With snake eyes, there’s a 1/36*(35/36)^(i-1) chance that it goes for i rounds. Again, there’s the question of weighting by pure probability of the game lasting that long or to also take into account the number of rounds. Luckily, we don’t have to decide as the probability of each of the cases we are weighting is 1⁄36, giving 1⁄36 overall.
In scenario 2, all selected participants are informed at once, so they don’t have access to the same indexical information, although they do have indexical information about all rounds having been run (essentially fixes n, even though this is unknown).
If we also fix the number of players as x, then your chance of being in round r with 2^(r-1) participants is 2^(r-1)/x, as you lack any information about which round you were in beyond the anthropic bias of being more likely to be in a larger round.
We can then aggregate over all possible n.
Again, there’s the question of whether we should weight purely by probabilities or take into account the number of people in each scenario. And again, I’m tempted to say that the indexical information indicates that we should treat a fixed n as a single group, and so weight purely by probabilities.
However, regardless, we can observe that we always have > 50% chance of being in the last round, so there’s > 50% chance of being in the last round overall.
Thoughts on aggregation:
I’m still confused here, but there seem to be two kinds of aggregation:
One where we group everything into one scenario
One where imagine ourselves to have a group of different scenarios
It depends.
If croupier choose the players, then players learn that they were chosen, then croupier roll the dice, then players either get bajillion dollars each or die, then (if not snake eyes) croupier choose next players and so on—answer is 1⁄36.
If croupier choose the players, then roll the dice, then (if not snake eyes) croupier choose next players and so on, and only when dice come up snake eyes players learn that they were chosen, and then last players die and all other players get bajillion dollars each—answer is about 1⁄2.
(I’ll probably post an expanded/edited version as a full-post soon, just wanted to see what I could produce in a single sitting. Apologies, as I engage in some handwaving).
I think it’s worth breaking this down more.
In scenario 1, you learn that you are a participant who was selected in round r, where r is unknown.
In scenario 2, you learn that you are a participant who was selected full-stop.
If this was all you learned, then these two scenarios would be equivalent, as all selected participants were selected in some round.
However, I’ll argue in the next section that each scenario contains additional information about which round you should expect to be in.
Analysing Indexical Information
In scenario 1, the participants are informed at different times depending on their group.
When a participant is informed they were selected, they know:
a) that the previous rounds were already rejected and
b) that if they aren’t eliminated, then additional rounds will be run, but you aren’t in those future run
c) that the dice are about to be thrown for them
In scenario 2, all selected participants are informed at once. Participants still know:
a) that all rounds before them were rejected AND
b) if they weren’t marked to be eliminated, then additional rounds were run
c) that the dice were thrown for them
Again, aren’t these the same?
In the next section, I’ll argue that the exact mechanism of how and when people are informed affects the way that we handle reference classes.
Analysing Reference Classes
Reference classes are strange. It’s not clear what we mean by reference classes.
Should it be the people in the same round as us, or everyone who ends up in the snake eyes problem?
One answer would be that in both scenarios it should be everyone who ends up in the same problem, such that the reference classes are the same (following the argument that you have no non-anthropic information about which round you are it)
Another answer would be that in scenario 1, your reference class is everyone in the same round due to indexical information, but in scenario 2, it is everyone in the problem due to the lack of such information.
So how should we interpret the indexical information when figuring out your reference class?
First, let’s imagine that we’ve fixed round r. In that case, it’s obvious that your reference class is everyone in this round and that everyone in the round has a 1⁄36 chance of being eliminated regardless of who they are.
We then want to consider the case where you don’t know which round you are in. This requires us to assign a probability distribution to rounds.
We can choose to start simple again and assume that there are n rounds in total. We now have to choose whether to say that your chance of being in any particular round is 1/n or whether it is proportional to the number of people in that round?
I’m going to suggest that the indexical information (that previous rounds have passed and future rounds are yet to come) tells you that it should be the former. There’s a sense in which people with the same indexical information naturally belong in the same group and it would be weird to mix them all together as per the proportional case. This is somewhat handwavey, but hopefully, I can make a more rigorous argument in future work.
This leads to your chance of being in any particular round is 1/n. Again, since all rounds have the same probability of being eliminated, this doesn’t affect anything.
We can then extend to the case where we don’t know the number of rounds. With snake eyes, there’s a 1/36*(35/36)^(i-1) chance that it goes for i rounds. Again, there’s the question of weighting by pure probability of the game lasting that long or to also take into account the number of rounds. Luckily, we don’t have to decide as the probability of each of the cases we are weighting is 1⁄36, giving 1⁄36 overall.
In scenario 2, all selected participants are informed at once, so they don’t have access to the same indexical information, although they do have indexical information about all rounds having been run (essentially fixes n, even though this is unknown).
If we also fix the number of players as x, then your chance of being in round r with 2^(r-1) participants is 2^(r-1)/x, as you lack any information about which round you were in beyond the anthropic bias of being more likely to be in a larger round.
We can then aggregate over all possible n.
Again, there’s the question of whether we should weight purely by probabilities or take into account the number of people in each scenario. And again, I’m tempted to say that the indexical information indicates that we should treat a fixed n as a single group, and so weight purely by probabilities.
However, regardless, we can observe that we always have > 50% chance of being in the last round, so there’s > 50% chance of being in the last round overall.
Thoughts on aggregation:
I’m still confused here, but there seem to be two kinds of aggregation:
One where we group everything into one scenario
One where imagine ourselves to have a group of different scenarios
I don’t know what the difference is yet.
I’m already wanting to update my analysis.
Informing everyone together means that they should be treated as one group regardless of what round they were in.
Informing one round at a time means dealing with anthropics.