P(original) = 2⁄3
P(heads) = 1⁄3
P(heads|original) = 1⁄2
Value of bet is (1/3)*5 = 1.667 < 2, so don’t take the bet.
The policy of not betting maximizes the total payoff to all clones (6 vs. 5). The policy of betting maximizes the average payoff-per-clone over worlds (2.5 vs 2). If I don’t care at all about my (possible) future clones, I should pre-commit to taking the bet, since this maximizes the payoff of the original. However, after waking up, I have to consider the possibility (with its associated probability) that I am the clone, so that changes my answer.
> However, after waking up, I have to consider the possibility (with its associated probability) that I am the clone, so that changes my answer.
The original and the clone are treated exactly the same regarding gold bars and bets. And both of them are offer after waking up to you, regardless if you are the orignal or the clone. You are just trying to maximize your own gain. Do you still consider not taking the bet a better decision?
Say you can have several gold bars in your pocked when going into the experiment. And the mad scientist knows this. To make sure you have no new information when looking into your pocked, he will place the same amount of gold bars into the clone’s pocket before waking him up. And you know this. So if you had 3 bars before, after waking up you will for sure find 3 gold bars in your pocket.
The rest of the problem is the same. You will be given 2 bars and offered the bet. Does this changes anything? Would you still reject the bet?
I don’t see why that would change anything. (This is all assuming my utility in gold bars is linear. Otherwise I’d have to worry about being flat broke and unable to feed myself if I turn out to be the clone.)
OK. Then consider doing this: After the first experiment, take part in the same experiment again, then again and again. You can keep the bars you earned in your pocket.
Say you participated for 100 iterations, by not entering the bet you would have 200 bars. By entering the bet do you expect to have approximately 100*1/3*5=167 bars?
P(original) = 2⁄3
P(heads) = 1⁄3
P(heads|original) = 1⁄2
Value of bet is (1/3)*5 = 1.667 < 2, so don’t take the bet.
The policy of not betting maximizes the total payoff to all clones (6 vs. 5). The policy of betting maximizes the average payoff-per-clone over worlds (2.5 vs 2). If I don’t care at all about my (possible) future clones, I should pre-commit to taking the bet, since this maximizes the payoff of the original. However, after waking up, I have to consider the possibility (with its associated probability) that I am the clone, so that changes my answer.
> However, after waking up, I have to consider the possibility (with its associated probability) that I am the clone, so that changes my answer.
The original and the clone are treated exactly the same regarding gold bars and bets. And both of them are offer after waking up to you, regardless if you are the orignal or the clone. You are just trying to maximize your own gain. Do you still consider not taking the bet a better decision?
After waking up, I hold P(heads) to be 1⁄3, so the bet is negative EV for me personally.
Say you can have several gold bars in your pocked when going into the experiment. And the mad scientist knows this. To make sure you have no new information when looking into your pocked, he will place the same amount of gold bars into the clone’s pocket before waking him up. And you know this. So if you had 3 bars before, after waking up you will for sure find 3 gold bars in your pocket.
The rest of the problem is the same. You will be given 2 bars and offered the bet. Does this changes anything? Would you still reject the bet?
I don’t see why that would change anything. (This is all assuming my utility in gold bars is linear. Otherwise I’d have to worry about being flat broke and unable to feed myself if I turn out to be the clone.)
OK. Then consider doing this: After the first experiment, take part in the same experiment again, then again and again. You can keep the bars you earned in your pocket.
Say you participated for 100 iterations, by not entering the bet you would have 200 bars. By entering the bet do you expect to have approximately 100*1/3*5=167 bars?
That’s the average of what all the clones will have at the end. The original should have about 250.