Anthropics: A Short Note on the Fission Riddle

This is also interesting if you imagine that you and your other copies don't all have the same memories - instead, each person remembers how many times they've been copied. So the first copy never remembers getting copied, and then the second copy remembers getting copied once, etc. Now the probabilities are no longer so obvious, because the only straightforward probability is the 50% probability the two most recent yous have that they're the other one - everyone else is distinguishable. The "probability" is instead some measure of caring, or personal identity transfer (to the extent that one believes that there is some kind of identity-fluid that obeys locality and conservation when getting transferred around).

There's a standard way to elicit practical 'probabilities' via bets in these situations: play for charity (with independent bets). In order to have something to bet about, let's turn the memories back off, and ask people to bet on whether they're the original (perhaps you have a tattoo on the back of your neck that doesn't get copied), betting with their favorite charity's money. At what odds should they make these bets - when does the charity start making money if your decision procedure says to bet? It's at 1:N odds, not 1:2^N odds.

Again, we should contrast this with the case where you have some family heirloom in a box, and whenever you're duplicated you put the box at random in the room of one of your two causal descendants. If only the original gets duplicated 100 times, the box really is in the first copy's room with 50% probability and in the original's room with 1 in 2^100 probability. So if personal identity behaves like physical stuff that gets passed on to your duple descendants, changing how you make copies changes the distribution of the stuff.

Another variation might be to Sleeping Beautify this problem - change the duplication into memory erasure. This makes more sense if we use the variation where you remember how many times you've been copied. Suppose the scientists who perfected the memory erasure pill used on Sleeping Beauty are now working on a selective memory restoration pill. Every day, your memories of the last 24 hours will be erased. But every two days, the scientists will flip a coin - if heads, restore your memory from one day ago, if tails, restore your memory from two days ago. In this way, every two days your memories advance by one day.

This procedure doesn't seem, to me, to change any relevant probability about the situation. On each day, you know that you have a 50% chance of being the one with the causal descendant. So becoming the "oldest" means winning a series of coin flips. But as I noted above, in addition to probability, this problem is asking us about the measure of caring, or hypothetical identity-fluid, and the memory erasure version totally changes our intuitions about how that gets sloshed around.

What if both clone and original are told which one they are right after cloning? Then probability of being told that you are original twice is still 1/4.

Assume this real life scenario:

100 refugees are met by the king of host country, who says, only 1 of you will become our citizen and other 99 will be slaves. The procedure of selecting the citizen is as follows: we choose 2 of you randomly, then coin is tossed - the looser becomes slave and the winner goes for second round against another fellow randomly selected from remaining 98 and so on. The one who wins the last coin toss becomes citizen.

In this setting if you are selected 1st then you have close to 0 chance to become citizen, as if you are selected last you have 50%. The game is unfair for 1st guys being selected same way as it is unfair for the original in cloning scenario.

Probability is a model of anticipated experiences and can be tested by measuring frequency of an event they predict [LW · GW], whether in a real or in a thought experiment. What experiment would you set up to test your model that the odd of being the original is 1%?

Another way to answer to this problem is applying indexical information changes tracking (known to the experimenter).

1) If he presses the button every millisecond, all 101 copy will have the same memories and the same indexical information, so the answer is 1/101.

2) If he presses the button every hour, his timestap changes as well as his indexical information about each pair of copies, so his probability to be the last copy is 1/2^100.