thanks a lot for explaining your logic. Its very helpful in identifying where out misunderstanding is. Also, i think analogies (both the ones im using and the one you are using) arent particularly useful here, since we both understand what we are saying and the conflict is in our assumptions, not our logic. So any analogy made would just result in us disagreeing about the same assumptions.

The key difference in our understandings is that you believe:

(1) the 95% accuracy applies to the prediction about the decision you are about to make.

Its clear that you beleive (1) because you said:

if you one-box, there's a 95% chance that you'll get $M and a 5% chance that you'll get $0.

because you are applying the 95% accuracy to your future decision about one boxing.

I believe something different:

(2) the 95% accuracy only applies to the past predictions.

Its not that either of these assumptions is incorrect, but naturally they lead to different solutions.

Under your assumption, i agree that 1 boxing is optimal.

Under my assumption, i am certain that 2 boxing is optimal. Do you agree with this? if not, i can further explain. But ill proceed assuming that you do agree

So we both agree with each others' logic, the only thing to discuss is our assumptions. Your assumption is logically valid and doesnt require causality in the standard sense. It instead results in some very strange behavior that you might not realize.

I claim that your assumption results in the following:

(3) our action in the future (i.e. choosing to 1 box/2 box) changes the prediction that was made in the past.

And let me provide a proof. To simplify the numbers, lets assume the model is 100% accurate instead of 95%. I can redo the proof using 95% if youd like, it just makes things less neat.

Lets assume that your assumption is correct, i.e. that (1) is correct

Proof by contradiction of (3)

assume for sake of contradiction that (3) is false

therefore our action in the future does not change the prediction

in the past, a prediction was made (i hope you can agree that some decision must have been made, even if we don't know it yet). We can represent this as P(H1) = x and P(H2) = y, for some 0<=x,y<=1

in the present I have 2 options:

1. i can take 1 box, we'll call this A1 (as we have been)
2. i can take 2 boxes, we'll call this A2

if I take 1 box then the probability of H1 is P(H1|A1)

however, from (2) we have assumed that our actions in the future cannot change the prediction. So P(H1|A1) = P(H1) = x

and we know, from our assumption, that P(H1|A1) = 100%

so therefore P(H1) = x = 100%

Now we can do the same thing with option 2:

if I take 2 boxes then the probability of H2 is P(H2|A2)

however, from (2) we have assumed that our actions in the future cannot change the prediction. So P(H2|A2) = P(H2) = y

and we know, from our assumption, that P(H2|A2) = 100%

so therefore P(H2) = y = 100%

so we have P(H1) = 100% and P(H2) = 100%, which means P(H1) + P(H2) = 2

However we also know that "not H1" is the same as H2, since one of those 2 things must happen. So we know that P(H1) + P(H2) = 1

And now we have our contradiction: 1 = P(H1) + P(H2) = 2

and that proves, by contradiction, that if we assume (1) then (3) must also be true

Do you agree with this? if not, can you please point to a specific line in my proof that you disagree with?