What do I claim explains this? The simple fact that you only offer the double computer one bet, not two
What makes you so confident that the double computer’s offered one bet and the two separate computers are offered two? Because the two separate computers might make two different choices? Only if there’s some other difference (e.g. you’ve fitted separate random number generators to the two separate computers, and the double computer only has one). Because you’re paying out twice as much if you lose? Presumably you put twice as much current into the double computer as to each of the single computers. Because you, as the experimenter, are offering different experiences to the two separate computers? That just repeats the same problem one level higher (considering that there are probably Everett branches in which you took both different actions with the double computer).
The computer doesn’t hold cash (clearly), it has account # and password of a bank account (or private key to bitcoin addresses if it’s a particularly risky computer). The two thin computers therefore only have half as much money to bet. (Or they’ll overdraw their account.)
How does your dollar get into the computers? Is there some dollar-input peripheral that the double computer only has one of and the two separate computers have one each? If so, isn’t that the difference there?
Sure, one could specify it like that. But a dollar sliced down the middle is not legal tender.
My original intent was that I’d simply talk to the computer on an input/output peripheral and make a bet that way. Two peripherals would mean two bets.
But the difference isn’t in the computers, it’s in the peripherals. If you connect two peripherals to the two computers and one to the double computer then you can make two bets with the two computers and one with the double computer. If you connect one peripheral to the two computers and two peripherals to the double computer then you can make one bet with the two computers and two bets with the double computer. You haven’t demonstrated any difference between the actual computers.
It’s the same either way. The probability assignment might depend on how many dollar-inputs you connect, but it’s the same whether you connect them to two computers or one double computer.
The bet goes like this: you pay me X dollars, and if the coin is tails, I’ll pay you $1. I offer the bet to each terminal.
At what X will the robot accept? Well, it will accept when ($1 - $X) * P(tails) - $X * P(heads) > 0. This simplifies to P(tails) > X.
So for example, if there’s no duplication—I just flip a coin and offer the bet to a computer—the computer will accept for X < 0.5, reflecting a probability of 0.5.
If I duplicate the robot in the tails case, then it accepts the same bet (assuming complete selfishness), but now it does best if it accepts when X < 2⁄3, indicating P(tails) = 2⁄3.
If I attach an extra peripheral in the tails-case (and the computer doesn’t notice), the bet is now: pay me X dollars if heads and 2X dollars if tails, for a payout of $2 if tails.
The robot will then accept if ($2 - $2X) * P(tails) - $X * P(heads) > 0. We can us that P(T) = 1-P(H) to expand this to 2/X > 1/P(tails)+1. Now if P(tails) = 1⁄2, the correct strategy is to accept when X<2/3.
In short, the probability assignment does not depend on how many peripherals the computer has, the different betting behavior follows straightforwardly from the payoffs of different betting strategies changing.
If you simulate someone on one thin computer connected to one dollar-input peripheral if heads, and one thin computer connected to one dollar-input peripheral if tails, and you offer them a bet on your coinflip, the correct probability is 1⁄2.
If you ran one thin computer connected to one dollar-input peripheral if heads, and two thin computers connected to two dollar-input peripherals if tails, the correct probability is 2⁄3.
If you ran one thin computer connected to one dollar-input peripheral if heads, and one thick computer connected to two dollar-input peripherals if tails, the correct probability is 2⁄3.
There’s no difference between two thin computers and one thick computer—only a difference between two dollar-input peripherals or one dollar-input peripheral.
What makes you so confident that the double computer’s offered one bet and the two separate computers are offered two? Because the two separate computers might make two different choices? Only if there’s some other difference (e.g. you’ve fitted separate random number generators to the two separate computers, and the double computer only has one). Because you’re paying out twice as much if you lose? Presumably you put twice as much current into the double computer as to each of the single computers. Because you, as the experimenter, are offering different experiences to the two separate computers? That just repeats the same problem one level higher (considering that there are probably Everett branches in which you took both different actions with the double computer).
Because if I put one dollar on each bet, I can only lose one dollar to the double computer, but can lose two dollars to the two computers.
The computer doesn’t hold cash (clearly), it has account # and password of a bank account (or private key to bitcoin addresses if it’s a particularly risky computer). The two thin computers therefore only have half as much money to bet. (Or they’ll overdraw their account.)
Sure, let’s go with that.
How does your dollar get into the computers? Is there some dollar-input peripheral that the double computer only has one of and the two separate computers have one each? If so, isn’t that the difference there?
Sure, one could specify it like that. But a dollar sliced down the middle is not legal tender.
My original intent was that I’d simply talk to the computer on an input/output peripheral and make a bet that way. Two peripherals would mean two bets.
But the difference isn’t in the computers, it’s in the peripherals. If you connect two peripherals to the two computers and one to the double computer then you can make two bets with the two computers and one with the double computer. If you connect one peripheral to the two computers and two peripherals to the double computer then you can make one bet with the two computers and two bets with the double computer. You haven’t demonstrated any difference between the actual computers.
Think about the subjective probability assignments of the computers that’s needed to generate correct betting behavior.
It’s the same either way. The probability assignment might depend on how many dollar-inputs you connect, but it’s the same whether you connect them to two computers or one double computer.
Ok, I’ll do the math, then I’m tapping out.
The bet goes like this: you pay me X dollars, and if the coin is tails, I’ll pay you $1. I offer the bet to each terminal.
At what X will the robot accept? Well, it will accept when ($1 - $X) * P(tails) - $X * P(heads) > 0. This simplifies to P(tails) > X.
So for example, if there’s no duplication—I just flip a coin and offer the bet to a computer—the computer will accept for X < 0.5, reflecting a probability of 0.5.
If I duplicate the robot in the tails case, then it accepts the same bet (assuming complete selfishness), but now it does best if it accepts when X < 2⁄3, indicating P(tails) = 2⁄3.
If I attach an extra peripheral in the tails-case (and the computer doesn’t notice), the bet is now: pay me X dollars if heads and 2X dollars if tails, for a payout of $2 if tails.
The robot will then accept if ($2 - $2X) * P(tails) - $X * P(heads) > 0. We can us that P(T) = 1-P(H) to expand this to 2/X > 1/P(tails)+1. Now if P(tails) = 1⁄2, the correct strategy is to accept when X<2/3.
In short, the probability assignment does not depend on how many peripherals the computer has, the different betting behavior follows straightforwardly from the payoffs of different betting strategies changing.
If you simulate someone on one thin computer connected to one dollar-input peripheral if heads, and one thin computer connected to one dollar-input peripheral if tails, and you offer them a bet on your coinflip, the correct probability is 1⁄2. If you ran one thin computer connected to one dollar-input peripheral if heads, and two thin computers connected to two dollar-input peripherals if tails, the correct probability is 2⁄3. If you ran one thin computer connected to one dollar-input peripheral if heads, and one thick computer connected to two dollar-input peripherals if tails, the correct probability is 2⁄3.
There’s no difference between two thin computers and one thick computer—only a difference between two dollar-input peripherals or one dollar-input peripheral.