Timeline for Probabilities in a riddle involving axiom of choice
Current License: CC BY-SA 4.0
7 events
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| Dec 14, 2022 at 13:30 | comment | added | Mateusz Kwaśnicki | More informally speaking: this is virtually the same as in the two envelopes paradox, where with probability $\tfrac12(\tfrac23)^n$ one envelope contains $2^n$ and the other one $2^{n - 1}$. You are given a randomly chosen envelope and you find it contains $2^N$. Are you going to switch? For each possible value of $N$ on average you should switch, but of course that's absurd. Here you get the same conclusion: for each Ann's choice of $f$, Bob almost surely wins. But does it mean he really wins almost surely? It is just for a different reason: lack of measurability rather than nonintegrability. | |
| Dec 14, 2022 at 13:28 | comment | added | Mateusz Kwaśnicki | On a formal level: I am no expert, but a "game with incomplete information" is a rigorously defined term in the theory of stochastic games, and, as far as I can tell, measurability assumption is unavoidable there. | |
| Dec 14, 2022 at 10:36 | comment | added | Denis | Yes I forgot to mention that we want to make this a game of incomplete information in order to make sense of it. So Ann chooses $f$, but Bob does not know what it is, nevertheless he has a probabilistic strategy working with probability 1. | |
| Dec 13, 2022 at 23:57 | comment | added | Mateusz Kwaśnicki | I tried to address this in the last paragraph. If the experiment is repeated, does Ann keep choosing the same function $f$? If she does, then of course you can rigorously say that Bob guesses correctly with probability one, but in this case picking $a = 0$ and "guessing" the real value $f(0)$ is an equally good strategy for Bob. If she does not, then we hit the question of measurability. | |
| Dec 13, 2022 at 23:48 | comment | added | Denis | My initial claim was about this "overall" formulation, and I understand now why it is flawed, but I was under the impression that specifying the problem as a game allows to avoid this issue, by explicitely quantifying on all functions in the formulation of the problem and specifying that the probabilities are computed only after this quantification. | |
| Dec 13, 2022 at 23:40 | comment | added | Denis | Thanks for your answer and for this interesting variant. Unfortunately I must have some kind of block in my thinking, I still don't see why it is not correct to talk about probabilities in a game formulation. In the game you describe, when Bob plays his last move (applying his strategy), he will win with probability 1, and this probability is only computed in the context of the moves already played, ie with a fixed function $f$. The only source of randomness here is Bob's strategy. Maybe the misunderstanding is on whether a stronger claim can be made about an "overall" probability for all f. | |
| Dec 13, 2022 at 22:13 | history | answered | Mateusz Kwaśnicki | CC BY-SA 4.0 |