So in applying UDT in real life, it's necessary to stipulate the problem statement, the boundary event in which all relevant possibilities are contained, and over which we compute expected utility.
Yes, of course. But you perform normal Bayesian updates for everything else (everything you hold fixed). Holding something fixed and not updating leads to errors.
Simple example: An urn with either 90% red and 10% blue balls or 90% blue and 10% red balls (0.5 prior for either). You have drawn a red ball and put it back. What's the updateless expected utility of drawing another ball, assuming you get 1 util for drawing a ball in the same color and -2 utils for drawing a ball in a different color? Calculating as getting 1 util for red balls and -2 for blue, but not updating on the observation of having drawn a red ball suggests that it's -0.5, when in fact it's 0.46.
EDIT: miscalculated the utilities, but the general thrust is the same.
P(RedU)=P(BlueU)=P(red)=P(blue)=0.5
P(red|RedU)=P(RedU|red)=P(blue|BlueU)=P(BlueU|blue)=0.9
P(blue|RedU)=P(RedU|blue)=P(BlueU|red)=P(Red|BlueU)=0.1
U_updating=P(RedU|red)*P(red|RedU)*1 + P(BlueU|red)*Pred(|BlueU)*1 - P(RedU|red)*P(blue|RedU)*2 - P(BlueU|red)*P(blue|BlueU)*2 = 0.9*0.9+0.1*0.1-0.9*0.1*2*2= 0.46
U_semi_updateless=P(red)*1-P(blue)*2=-0.5
U_updateless= P(red)(P(RedU|red)*P(red|RedU)*1 + P(BlueU|red)*Pred(|BlueU)*1 - P(RedU|red)*P(blue|RedU)*2 - P(BlueU|red)*P(blue|BlueU)*2) +P(blue)(P(BlueU|blue)*P(blue|BlueU)*1 + P(RedU|blue)*P(blue|RedU)*1 - P(BlueU|blue)*P(red|BlueU)*2 - P(RedU|blue)*P(red|RedU)*2) =0.5*(0.9*0.9+0.1*0.1-0.9*0.1*2*2)+0.5* (0.9*0.9+0.1*0.1-0.9*0.1*2*2)=0.46
(though normally you'd probably come up with U_updateless in a differently factored form)
EDIT3: More sensible/readable factorization of U_updateless:
P(RedU)((P(red|RedU)(P(red|RedU)*1-P(blue|RedU)*2)+(P(blue|RedU)(P(blue|RedU)*1-P(red|RedU)*2)) + P(BlueU)((P(blue|BlueU)(P(blue|BlueU)*1-P(red|BlueU)*2)+(P(red|BlueU)(P(red|BlueU)*1-P(blue|BlueU)*2))
Holding something fixed and not updating leads to errors.
No, controlling something and updating it away leads to errors. Fixed terms in expected utility don't influence optimality, you just lose ability to consider the influence of various strategies on them. Here, the strategies under considerations don't have any relevant effects outside the problem statement.
(I'll look into your example another time.)
Consider the following thought experiment ("Counterfactual Calculation"):
Should you write "even" on the counterfactual test sheet, given that you're 99% sure that the answer is "even"?
This thought experiment contrasts "logical knowledge" (the usual kind) and "observational knowledge" (what you get when you look at a calculator display). The kind of knowledge you obtain by observing things is not like the kind of knowledge you obtain by thinking yourself. What is the difference (if there actually is a difference)? Why does observational knowledge work in your own possible worlds, but not in counterfactuals? How much of logical knowledge is like observational knowledge, and what are the conditions of its applicability? Can things that we consider "logical knowledge" fail to apply to some counterfactuals?
(Updateless analysis would say "observational knowledge is not knowledge" or that it's knowledge only in the sense that you should bet a certain way. This doesn't analyze the intuition of knowing the result after looking at a calculator display. There is a very salient sense in which the result becomes known, and the purpose of this thought experiment is to explore some of counterintuitive properties of such knowledge.)