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drnickbone comments on Problematic Problems for TDT - Less Wrong
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Omega's choice of box depends on its assessment of the simulated agent's choosing probabilities. The tie-breaking rule (if there are several boxes with equal lowest choosing probability, then select the one with the lowest label) is to an extent arbitrary, but it is important that there is some deterministic tie-breaking rule.
I also agree this is entirely a maths problem for Omega or for anyone whose decisions aren't entangled with the problem (with a proof that Box 1 will contain the $1 million). The difficulty is that a TDT agent can't treat it as a straight maths problem which is unlinked to its own decisions.
Why is it important that there is a deterministic breaking rule ? When you would like random numbers, isn't it always better to have a distribution as close as random as possible, even if it is pseudo-random ?
That question is perhaps stupid, I have the impression that I am missing something important...
Remember it is Omega implementing the tie-breaker rule, since it defines the problem.
The consequence of the tie-breaker is that the choosing agent knows that Omega's box-choice was a simple deterministic function of a mathematical calculation (or a proof). So the agent's uncertainty about which box contains the money is pure logical uncertainty.
Whoops... I can't believe I missed that. You are obviously right.