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.
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.