Here the optimal strategy is to choose “yea” with a certain probability p, which I don’t have time to calculate right now
The expected value is $1000 (10 * p − 10 p^ 10). Maximums and minimums of functions may occur when the derivative is zero, or at boundaries.
The derivative is $1000(10 − 100 p^ 9). This is zero when p = 0.1^(1/9) ~= 0.774.
The boundaries of 0 and 1 are minima, and this is a maximum.
EDIT: Huh. This simple calculation that mildly adds to the parent is worth more karma than the parent? I thought the parent really got to the heart of things with:
“(because there’s no reliable way to account for the decisions of others if they depend on yours)” Of course, TDT and UDT are attempts to do just that in some circumstances.
Seems to me that you’d want to add up the probabilities of each of the 10 outcomes, 0*p^10*(10!/(10!*0!)) + 9000*p^9*(1-p)*(10!/(9!*1!)) + 8000*p^8*(1-p)^2*(10!/(8!*2!)) + 7000*p^7*(1-p)^3*(10!/(7!*3!))… This also has a maximum at p~= 0.774, with expected value of $6968. This verifies that your shortcut was correct.
James’ equation gives a bigger value, because he doesn’t account for the fact that the lost payoff is always the maximum $10,000. His equation would be the correct one to use, if the problem were with 20 people, 10 of which determine the payoff and the other 10 whether the payoff is payed and they all have to use the same probability.
The expected value is $1000 (10 * p − 10 p^ 10). Maximums and minimums of functions may occur when the derivative is zero, or at boundaries.
The derivative is $1000(10 − 100 p^ 9). This is zero when p = 0.1^(1/9) ~= 0.774. The boundaries of 0 and 1 are minima, and this is a maximum.
EDIT: Huh. This simple calculation that mildly adds to the parent is worth more karma than the parent? I thought the parent really got to the heart of things with: “(because there’s no reliable way to account for the decisions of others if they depend on yours)” Of course, TDT and UDT are attempts to do just that in some circumstances.
Shouldn’t the expected value be $1000 (10p)*(1-p^10) or $1000 (10p − 10p^11) ? (p now maximised at 0.7868… giving EV $7.15K)
That does look right.
Seems to me that you’d want to add up the probabilities of each of the 10 outcomes, 0*p^10*(10!/(10!*0!)) + 9000*p^9*(1-p)*(10!/(9!*1!)) + 8000*p^8*(1-p)^2*(10!/(8!*2!)) + 7000*p^7*(1-p)^3*(10!/(7!*3!))… This also has a maximum at p~= 0.774, with expected value of $6968. This verifies that your shortcut was correct.
James’ equation gives a bigger value, because he doesn’t account for the fact that the lost payoff is always the maximum $10,000. His equation would be the correct one to use, if the problem were with 20 people, 10 of which determine the payoff and the other 10 whether the payoff is payed and they all have to use the same probability.