No. The probability of K balls all being the good balls (assuming you’re drawing with replacement) is ((N-1)/N)^K. So the expected value is ((N-1)/N)^K x (KxA) - (1 - ((N-1)/N)^K) x B
A low probability of large bad effects can swamp a high probability of small good effects, but it doesn’t have to, so you can have the high probability of small good effects dominate.
Let me be concrete: imagine you have a one in a hundred chance of a bad outcome of utility −100 (where if it happens all good effects get wiped out), and with the rest of the probability you get a good outcome of utility 2 (and the utility of these good outcomes stacks with how many times they happen). Then the expected utility of doing this once is 2 x 0.99 − 100 x 0.01 = 0.98 > 0, but the expected utility of doing it one thousand times is 2 x 1000 x (0.99 ^ 1000) − 100 x (1 − 0.99^1000) = 2000 x 0.000043 − 100 x 0.999957 = 0.086 − 99.9957 < 0.
True, but this doesn’t apply to the original reasoning in the post—he assumes constant probability while you need increasing probability (as with the balls) to make the math work.
Or decreasing benefits, which probably is the case in the real world.
My comment involves a constant probability of the bad outcome with each draw, and no decreasing benefits. I think this is a good exposition of this portion of the post (which I wrote), if you assume that each unit of bio progress is equally good, but that the goods don’t materialize if we all die of a global pandemic:
Suppose instead that the benefits of thing X grow proportionally to how much it happens: for example, maybe every person who learns about biology makes roughly the same amount of incremental progress in learning how to cure disease and make humans healthier. Also suppose that every person who does thing X has a small probability of causing bad effect Y for everyone that negates all the benefits of X: for example, perhaps 0.01% of people would cause a global pandemic killing everyone if they learned enough about biology.
Yeah, but the expected value would still be K(N−1NA−BN).
No. The probability of K balls all being the good balls (assuming you’re drawing with replacement) is ((N-1)/N)^K. So the expected value is ((N-1)/N)^K x (KxA) - (1 - ((N-1)/N)^K) x B
OK, that’s fair, I should have written down the precise formula rather than an approximation. My point though is that your statement
is wrong because a low probability of large bad effects can swamp a high probability of small good effects in expected value calculations.
A low probability of large bad effects can swamp a high probability of small good effects, but it doesn’t have to, so you can have the high probability of small good effects dominate.
Let me be concrete: imagine you have a one in a hundred chance of a bad outcome of utility −100 (where if it happens all good effects get wiped out), and with the rest of the probability you get a good outcome of utility 2 (and the utility of these good outcomes stacks with how many times they happen). Then the expected utility of doing this once is 2 x 0.99 − 100 x 0.01 = 0.98 > 0, but the expected utility of doing it one thousand times is 2 x 1000 x (0.99 ^ 1000) − 100 x (1 − 0.99^1000) = 2000 x 0.000043 − 100 x 0.999957 = 0.086 − 99.9957 < 0.
OK, that makes sense.
True, but this doesn’t apply to the original reasoning in the post—he assumes constant probability while you need increasing probability (as with the balls) to make the math work.
Or decreasing benefits, which probably is the case in the real world.
Edit: misred the previous comment, see below
My comment involves a constant probability of the bad outcome with each draw, and no decreasing benefits. I think this is a good exposition of this portion of the post (which I wrote), if you assume that each unit of bio progress is equally good, but that the goods don’t materialize if we all die of a global pandemic: