As an instrumental rationalist, I would say that 0 is a better approximation than 1⁄2 ….
How is this train of thought “instrumental”? You aren’t making any choices or decisions outside of your own brain.
To make it a real instrumental example, consider whether or not you should go buy a meteorite shield. Lets say the shield costs S and if the meteorite hits you it costs M, and the true probability of the strike is p. So buying the shield is best if pM > S.
Now if you go with 0, you’ll never buy the shield, so if pM > S you have an expected loss of (pM—S) due to your approxamation.
If you go with 1⁄2 then you’ll buy the shield if M/2 > S. If M/2 > S and pM ⇐ S then you bought the shield when you shouldn’t have, and you lose and expected (S—pM).
So you see, it all depends on how big M is compared to S
M < S/p : 0 is the better instrumental approximation
M > S/p : 1⁄2 is better
In other words, if the risks (or payoffs) are small compared to the probabilities involved and the costs of shields, round to 0. Otherwise round to 1⁄2.
How is this train of thought “instrumental”? You aren’t making any choices or decisions outside of your own brain.
To make it a real instrumental example, consider whether or not you should go buy a meteorite shield. Lets say the shield costs S and if the meteorite hits you it costs M, and the true probability of the strike is p. So buying the shield is best if pM > S.
Now if you go with 0, you’ll never buy the shield, so if pM > S you have an expected loss of (pM—S) due to your approxamation.
If you go with 1⁄2 then you’ll buy the shield if M/2 > S. If M/2 > S and pM ⇐ S then you bought the shield when you shouldn’t have, and you lose and expected (S—pM).
So you see, it all depends on how big M is compared to S
M < S/p : 0 is the better instrumental approximation
M > S/p : 1⁄2 is better
In other words, if the risks (or payoffs) are small compared to the probabilities involved and the costs of shields, round to 0. Otherwise round to 1⁄2.