Wouldn't any moderately complex optimiser be likely to have some model of a probability spread, or distribution, rather than holding all its knowledge to a single number? Intuitively it seems to me to be useful to be able to model the difference between, say, on one hand a probability density that's (roughly) Gaussian-shaped with a mean of 0.5 and an sd of 0.02, and on the other hand a delta function at 0.5. With both, your single-value estimate of the probability is 0.5, but your certainty in that estimate is different.
For such a machine one wouldn't specify "achieve 'G' with probability 'p'", but rather "achieve 'G' such that the probability density below 'p' is < f", where f is some specified (small) fraction of the total. That seems rather more straightforward in many ways.
[Final Update: Back to 'Discussion'; stroked out the initial framing which was misleading.]
[Update: Moved to 'Main'. Also, judging by the comments, it appears that most have misunderstood the puzzle and read way too much into it; user 'Manfred' seems to have got the point.][Note: This little puzzle is my first article. Preliminary feedback suggests some of you might enjoy it while others might find it too obvious, hence the cautious submission to 'Discussion'; will move it to 'Main' if, and only if, it's well-received.]In his recent paper "The Superintelligent Will: Motivation and Instrumental Rationality in Advanced Artificial Agents", Nick Bostrom states:Let us take it on from here.It is tempting to say that a machine can never halt after achieving its goal because it cannot know with full certainty whether it has achieved its goal; it will continually verify, possibly to increasing degrees of certainty, whether it has achieved its goal, but never halt as such.
What if, from a naive goal G, the machine's goal were then redefined as "achieve 'G' with 'p' probability" for some p < 1? It appears this also would not work, given the machine would never be fully certain of being p certain of having achieved G. (and so on...)
Yet one can specify a set of conditions for which a program will terminate, so how is the argument above fallacious?
Solution in ROT13: Va beqre gb unyg fhpu na ntrag qbrfa'g arrq gb *xabj* vg'f c pregnva, vg bayl arrqf gb *or* c pregnva; nf gur pbaqvgvba vf rapbqrq, gur unygvat jvyy or gevttrerq bapr gur ntrag ragref gur fgngr bs c pregnvagl, ertneqyrff bs jurgure vg unf (shyy) xabjyrqtr bs vgf fgngr.